Remove specific words containing dots from a string
$begingroup$
Hi I got words that contain one of the following tokens: CO. PRO. IST.
How to clean my words from these tokens? Do you have any regex pattern?
for example the words :
convolutional NN in data science--> extract nothing
co. Neural N --> extract co
co NN --> extract nothing
Thanks!
r rstudio regex dplyr
asked 18 hours ago
3nomis3nomis
1827
$endgroup$
add a comment |
$begingroup$
Hi I got words that contain one of the following tokens: CO. PRO. IST.
How to clean my words from these tokens? Do you have any regex pattern?
for example the words :
convolutional NN in data science--> extract nothing
co. Neural N --> extract co
co NN --> extract nothing
Thanks!
r rstudio regex dplyr
asked 18 hours ago
3nomis3nomis
1827
$endgroup$
add a comment |
$begingroup$
Hi I got words that contain one of the following tokens: CO. PRO. IST.
How to clean my words from these tokens? Do you have any regex pattern?
for example the words :
convolutional NN in data science--> extract nothing
co. Neural N --> extract co
co NN --> extract nothing
Thanks!
r rstudio regex dplyr
asked 18 hours ago
3nomis3nomis
1827
$endgroup$
Hi I got words that contain one of the following tokens: CO. PRO. IST.
How to clean my words from these tokens? Do you have any regex pattern?
for example the words :
convolutional NN in data science--> extract nothing
co. Neural N --> extract co
co NN --> extract nothing
Thanks!
r rstudio regex dplyr
r rstudio regex dplyr
asked 18 hours ago
3nomis3nomis
1827
asked 18 hours ago
3nomis3nomis
1827
asked 18 hours ago
3nomis3nomis
1827
asked 18 hours ago
3nomis3nomis
1827
asked 18 hours ago
3nomis3nomis
1827
1827
add a comment |
add a comment |
1 Answer
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$begingroup$
you can use lookahead with str_extract from the stringr package.
e.g.
stringr::str_extract(string = "co.asdkhsda", pattern = "co(?=\.)")
# extracts 'co' only when followed by a literal '.'
stringr::str_extract("convolutional NN in data science", pattern = "co(?=\.)")
# extracts nothing, returns NA
The regex pattern
co(?=\.)
looks for 'co', followed by '\.'
The '\'-part is used to escape the special meaning of the '.'
The dot usually matches any character, and therefore need to be escaped.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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1 Answer
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$begingroup$
you can use lookahead with str_extract from the stringr package.
e.g.
stringr::str_extract(string = "co.asdkhsda", pattern = "co(?=\.)")
# extracts 'co' only when followed by a literal '.'
stringr::str_extract("convolutional NN in data science", pattern = "co(?=\.)")
# extracts nothing, returns NA
The regex pattern
co(?=\.)
looks for 'co', followed by '\.'
The '\'-part is used to escape the special meaning of the '.'
The dot usually matches any character, and therefore need to be escaped.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
you can use lookahead with str_extract from the stringr package.
e.g.
stringr::str_extract(string = "co.asdkhsda", pattern = "co(?=\.)")
# extracts 'co' only when followed by a literal '.'
stringr::str_extract("convolutional NN in data science", pattern = "co(?=\.)")
# extracts nothing, returns NA
The regex pattern
co(?=\.)
looks for 'co', followed by '\.'
The '\'-part is used to escape the special meaning of the '.'
The dot usually matches any character, and therefore need to be escaped.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
you can use lookahead with str_extract from the stringr package.
e.g.
stringr::str_extract(string = "co.asdkhsda", pattern = "co(?=\.)")
# extracts 'co' only when followed by a literal '.'
stringr::str_extract("convolutional NN in data science", pattern = "co(?=\.)")
# extracts nothing, returns NA
The regex pattern
co(?=\.)
looks for 'co', followed by '\.'
The '\'-part is used to escape the special meaning of the '.'
The dot usually matches any character, and therefore need to be escaped.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
you can use lookahead with str_extract from the stringr package.
e.g.
stringr::str_extract(string = "co.asdkhsda", pattern = "co(?=\.)")
# extracts 'co' only when followed by a literal '.'
stringr::str_extract("convolutional NN in data science", pattern = "co(?=\.)")
# extracts nothing, returns NA
The regex pattern
co(?=\.)
looks for 'co', followed by '\.'
The '\'-part is used to escape the special meaning of the '.'
The dot usually matches any character, and therefore need to be escaped.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 15 hours ago
JaggeJagge
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 15 hours ago
JaggeJagge
1
answered 15 hours ago
JaggeJagge
1
1
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jagge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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