Grep remove line with 0 but not 0.2?
I have a file similar to this:
0
0
0.02
0
0
0
0
I want to remove all the lines with a single zero.
I was thinking to use something like:
grep -v "0"
But this also removes the instances of 0.2
.
I also saw there is a whole word option -w
but this doesn't seem to work either.
How can I remove all lines which are just a single 0 but keep lines which may contain a 0?
grep
add a comment |
I have a file similar to this:
0
0
0.02
0
0
0
0
I want to remove all the lines with a single zero.
I was thinking to use something like:
grep -v "0"
But this also removes the instances of 0.2
.
I also saw there is a whole word option -w
but this doesn't seem to work either.
How can I remove all lines which are just a single 0 but keep lines which may contain a 0?
grep
add a comment |
I have a file similar to this:
0
0
0.02
0
0
0
0
I want to remove all the lines with a single zero.
I was thinking to use something like:
grep -v "0"
But this also removes the instances of 0.2
.
I also saw there is a whole word option -w
but this doesn't seem to work either.
How can I remove all lines which are just a single 0 but keep lines which may contain a 0?
grep
I have a file similar to this:
0
0
0.02
0
0
0
0
I want to remove all the lines with a single zero.
I was thinking to use something like:
grep -v "0"
But this also removes the instances of 0.2
.
I also saw there is a whole word option -w
but this doesn't seem to work either.
How can I remove all lines which are just a single 0 but keep lines which may contain a 0?
grep
grep
asked 27 mins ago
Philip KirkbridePhilip Kirkbride
2,47313084
2,47313084
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
grep -vx 0
From man grep
:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w
fails because the first 0
in 0.02
is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v
, i.e. grep -w "0"
.
add a comment |
With grep:
grep -v "^0$" file
^
means beginning of the line, $
means end of the line.
add a comment |
- b - word border
grep -v "b0b"
- match beginning of line, your pattern and end of line
grep -v "^0$"
- or as @Sparhawk suggested -vx lineregexp
-w works, but in your case 0.2 are two words because dot character is a word separator.
grep -v "b0b"
doesn't really work here. What version of grep do you use?
– Arkadiusz Drabczyk
12 mins ago
add a comment |
grep
's -w
is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0
in 0.02
it had asserted the negation logic to remove the line.
Using sed
is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
add a comment |
When the lines you want to delete only contain a 0
followed by the next line you can select those lines by issuing the following command:
grep "0$"
This will only print the occurrences of 0
that are at the end of a line.
I hope this helps!
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
grep -vx 0
From man grep
:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w
fails because the first 0
in 0.02
is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v
, i.e. grep -w "0"
.
add a comment |
grep -vx 0
From man grep
:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w
fails because the first 0
in 0.02
is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v
, i.e. grep -w "0"
.
add a comment |
grep -vx 0
From man grep
:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w
fails because the first 0
in 0.02
is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v
, i.e. grep -w "0"
.
grep -vx 0
From man grep
:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w
fails because the first 0
in 0.02
is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v
, i.e. grep -w "0"
.
answered 19 mins ago
SparhawkSparhawk
9,74264094
9,74264094
add a comment |
add a comment |
With grep:
grep -v "^0$" file
^
means beginning of the line, $
means end of the line.
add a comment |
With grep:
grep -v "^0$" file
^
means beginning of the line, $
means end of the line.
add a comment |
With grep:
grep -v "^0$" file
^
means beginning of the line, $
means end of the line.
With grep:
grep -v "^0$" file
^
means beginning of the line, $
means end of the line.
answered 18 mins ago
Arkadiusz DrabczykArkadiusz Drabczyk
7,95521734
7,95521734
add a comment |
add a comment |
- b - word border
grep -v "b0b"
- match beginning of line, your pattern and end of line
grep -v "^0$"
- or as @Sparhawk suggested -vx lineregexp
-w works, but in your case 0.2 are two words because dot character is a word separator.
grep -v "b0b"
doesn't really work here. What version of grep do you use?
– Arkadiusz Drabczyk
12 mins ago
add a comment |
- b - word border
grep -v "b0b"
- match beginning of line, your pattern and end of line
grep -v "^0$"
- or as @Sparhawk suggested -vx lineregexp
-w works, but in your case 0.2 are two words because dot character is a word separator.
grep -v "b0b"
doesn't really work here. What version of grep do you use?
– Arkadiusz Drabczyk
12 mins ago
add a comment |
- b - word border
grep -v "b0b"
- match beginning of line, your pattern and end of line
grep -v "^0$"
- or as @Sparhawk suggested -vx lineregexp
-w works, but in your case 0.2 are two words because dot character is a word separator.
- b - word border
grep -v "b0b"
- match beginning of line, your pattern and end of line
grep -v "^0$"
- or as @Sparhawk suggested -vx lineregexp
-w works, but in your case 0.2 are two words because dot character is a word separator.
answered 15 mins ago
Jakub JindraJakub Jindra
817
817
grep -v "b0b"
doesn't really work here. What version of grep do you use?
– Arkadiusz Drabczyk
12 mins ago
add a comment |
grep -v "b0b"
doesn't really work here. What version of grep do you use?
– Arkadiusz Drabczyk
12 mins ago
grep -v "b0b"
doesn't really work here. What version of grep do you use?– Arkadiusz Drabczyk
12 mins ago
grep -v "b0b"
doesn't really work here. What version of grep do you use?– Arkadiusz Drabczyk
12 mins ago
add a comment |
grep
's -w
is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0
in 0.02
it had asserted the negation logic to remove the line.
Using sed
is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
add a comment |
grep
's -w
is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0
in 0.02
it had asserted the negation logic to remove the line.
Using sed
is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
add a comment |
grep
's -w
is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0
in 0.02
it had asserted the negation logic to remove the line.
Using sed
is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
grep
's -w
is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0
in 0.02
it had asserted the negation logic to remove the line.
Using sed
is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
answered 4 mins ago
InianInian
4,4851025
4,4851025
add a comment |
add a comment |
When the lines you want to delete only contain a 0
followed by the next line you can select those lines by issuing the following command:
grep "0$"
This will only print the occurrences of 0
that are at the end of a line.
I hope this helps!
add a comment |
When the lines you want to delete only contain a 0
followed by the next line you can select those lines by issuing the following command:
grep "0$"
This will only print the occurrences of 0
that are at the end of a line.
I hope this helps!
add a comment |
When the lines you want to delete only contain a 0
followed by the next line you can select those lines by issuing the following command:
grep "0$"
This will only print the occurrences of 0
that are at the end of a line.
I hope this helps!
When the lines you want to delete only contain a 0
followed by the next line you can select those lines by issuing the following command:
grep "0$"
This will only print the occurrences of 0
that are at the end of a line.
I hope this helps!
answered 3 mins ago
majesticLSDmajesticLSD
423
423
add a comment |
add a comment |
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