Why can I use a list index as an indexing variable in a for loop?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0
1
2
2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python for-loop
New contributor
|
show 2 more comments
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0
1
2
2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python for-loop
New contributor
7
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
4
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
8
This would be a great question for an awful coding interview
– Nathan
1 hour ago
1
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
1
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago
|
show 2 more comments
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0
1
2
2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python for-loop
New contributor
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0
1
2
2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python for-loop
python for-loop
New contributor
New contributor
edited 40 mins ago
Amir A. Shabani
457616
457616
New contributor
asked 1 hour ago
Kundan VermaKundan Verma
712
712
New contributor
New contributor
7
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
4
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
8
This would be a great question for an awful coding interview
– Nathan
1 hour ago
1
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
1
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago
|
show 2 more comments
7
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
4
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
8
This would be a great question for an awful coding interview
– Nathan
1 hour ago
1
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
1
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago
7
7
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
4
4
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
8
8
This would be a great question for an awful coding interview
– Nathan
1 hour ago
This would be a great question for an awful coding interview
– Nathan
1 hour ago
1
1
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
1
1
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago
|
show 2 more comments
5 Answers
5
active
oldest
votes
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1])
in our particular case.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
53 mins ago
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
56 mins ago
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable.
It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.
So first a[-1]
gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
1 hour ago
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
1 hour ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
1 hour ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
44 mins ago
|
show 3 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55644201%2fwhy-can-i-use-a-list-index-as-an-indexing-variable-in-a-for-loop%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1])
in our particular case.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
53 mins ago
add a comment |
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1])
in our particular case.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
53 mins ago
add a comment |
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1])
in our particular case.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1])
in our particular case.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
edited 9 mins ago
answered 1 hour ago
TrebledJTrebledJ
3,76421329
3,76421329
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
53 mins ago
add a comment |
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
53 mins ago
2
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
1 hour ago
2
2
Essentially, the important thing to note here is that Python considers
a[-1]
to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1
is valid grammar). Thus a[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.– Christian Dean
53 mins ago
Essentially, the important thing to note here is that Python considers
a[-1]
to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1
is valid grammar). Thus a[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.– Christian Dean
53 mins ago
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
answered 1 hour ago
recnacrecnac
1,243123
1,243123
add a comment |
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
56 mins ago
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
56 mins ago
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
edited 53 mins ago
Amir A. Shabani
457616
457616
answered 1 hour ago
NathanNathan
2,02511326
2,02511326
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
56 mins ago
add a comment |
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
56 mins ago
Somehow I thought the variable used in
for loop
is immutable. +1 for pointing it out.– Amir A. Shabani
56 mins ago
Somehow I thought the variable used in
for loop
is immutable. +1 for pointing it out.– Amir A. Shabani
56 mins ago
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
answered 1 hour ago
blhsingblhsing
43.5k41744
43.5k41744
add a comment |
add a comment |
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable.
It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.
So first a[-1]
gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
1 hour ago
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
1 hour ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
1 hour ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
44 mins ago
|
show 3 more comments
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable.
It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.
So first a[-1]
gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
1 hour ago
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
1 hour ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
1 hour ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
44 mins ago
|
show 3 more comments
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable.
It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.
So first a[-1]
gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable.
It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.
So first a[-1]
gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
edited 26 mins ago
answered 1 hour ago
Tom KarzesTom Karzes
11.2k1926
11.2k1926
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
1 hour ago
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
1 hour ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
1 hour ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
44 mins ago
|
show 3 more comments
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
1 hour ago
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
1 hour ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
1 hour ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
44 mins ago
How does it actually
gets set to 0, then 1, then 2 and finally 2
? That's the confusing part!– Amir A. Shabani
1 hour ago
How does it actually
gets set to 0, then 1, then 2 and finally 2
? That's the confusing part!– Amir A. Shabani
1 hour ago
1
1
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
Agree, I don't really understand by reading this answer
– gameon67
1 hour ago
I expected it would be like saying
for 3 in a: print(a[-1])
(because a[-1]
was 3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.– Nathan
1 hour ago
I expected it would be like saying
for 3 in a: print(a[-1])
(because a[-1]
was 3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.– Nathan
1 hour ago
It's set by the
for
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.– Tom Karzes
1 hour ago
It's set by the
for
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.– Tom Karzes
1 hour ago
1
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element
a[-1]
evaluates to, rather, it's using the index a[-1]
itself. If you rephrase that statement, your answer becomes much more clear.– Christian Dean
44 mins ago
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element
a[-1]
evaluates to, rather, it's using the index a[-1]
itself. If you rephrase that statement, your answer becomes much more clear.– Christian Dean
44 mins ago
|
show 3 more comments
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55644201%2fwhy-can-i-use-a-list-index-as-an-indexing-variable-in-a-for-loop%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
7
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
1 hour ago
4
I don't know why you would ever want to do this, but now I know you can
– Nathan
1 hour ago
8
This would be a great question for an awful coding interview
– Nathan
1 hour ago
1
OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.
– Christian Dean
48 mins ago
1
This question first got two down votes :) I guess you shouldn't panic by down votes!
– Amir A. Shabani
27 mins ago