I would like to convert the following string
"For "The" Win","Way "To" Go"
to
"For ""The"" Win","Way ""To"" Go"
The straightforward regex would be
str2 = re.sub(r'(?<!,|^)"(?=\w)|(?<=\w)"(?!,|$)', '""', str1,flags=re.MULTILINE)
i.e., Double the quotes that are
Followed by a letter but not preceded by a comma or the beginning of line
Preceded by a letter but not followed by a comma or the end of line
The problem is I am using python and it's regex engine does not allow using the OR operator in the lookbehind construct. I get the error
sre_constants.error: look-behind requires fixed-width pattern
What I am looking for is a regex that will replace the '"' around 'The' and 'To' with '""'.
I can use the following regex (An answer provided to another question)
\b\s*"(?!,|[ \t]*$)
but that consumes the space just before the 'The' and 'To' and I get the below
"For""The"" Win","Way""To"" Go"
Is there a workaround so that I can double the quotes around 'The' and 'To' without consuming the spaces just before them?
Instead of saying not preceded by comma or the line start, say preceded by a non-comma character:
r'(?<=[^,])"(?=\w)|(?<=\w)"(?!,|$)'
Looks to me like you don't need to bother with anchors.
If there is a character before the quote, you know it's not at the beginning of the string.
If that character is not a newline, you're not at the beginning of a line.
If the character is not a comma, you're not at the beginning of a field.
So you don't need to use anchors, just do a positive lookbehind/lookahead for a single character:
result = re.sub(r'(?<=[^",\r\n])"(?=[^,"\r\n])', '""', subject)
I threw in the " on the chance that there might be some quotes that are already escaped. But realistically, if that's the case you're probably screwed anyway. ;)
re.sub(r'\b(\s*)"(?!,|[ \t]*$)', r'\1""', s)
Most direct workaround whenever you encounter this issue: explode the look-behind into two look-behinds.
str2 = re.sub(r'(?<!,)(?<!^)"(?=\w)|(?<=\w)"(?!,|$)', '""', str1,flags=re.MULTILINE)
(don't name your strings str)
str2 = re.sub('(?<=[^,])"(?=\w)'
'|'
'(?<=\w)"(?!,|$)',
'""', ss,
flags=re.MULTILINE)
I always wonder why people use raw strings for regex patterns when it isn't needed.
Note I changed your str which is the name of a builtin class to ss
.
For `"fun" :
str2 = re.sub('"'
'('
'(?<=[^,]")(?=\w)'
'|'
'(?<=\w")(?!,|$)'
')',
'""', ss,
flags=re.MULTILINE)
or also
str2 = re.sub('(?<=[^,]")(?=\w)'
'|'
'(?<=\w")(?!,|$)',
'"', ss,
flags=re.MULTILINE)
Related
I need to truncate string by special characters '-', '(', '/' with one leading whitespace, i.e. ' -', ' (', ' /'.
how to do that?
patterns=r'[-/()]'
try:
return row.split(re.findall(patterns, row)[0], 1)[0]
except:
return row
the above code picked up all special characters but without the leading space.
patterns=r'[s-/()]'
this one does not work.
Try this pattern
patterns=r'^\s[-/()]'
or remove ^ depending on your needs.
It looks like you want to get a part of the string before the first occurrence of \s[-(/] pattern.
Use
return re.sub(r'\s[-(/].*', '', row)
This code will return a part of row string without all chars after the first occurrence of a whitespace (\s) followed with -, ( or / ([-(/]).
See the regex demo.
Please try this pattern patterns = r'\s+-|\s\/|\s\(|\s\)'
I want a regex search to end when it reaches ". ", but not when it reaches "."; I'm aware of using [^...] to exclude single characters, and have been using this to stop my search when it reaches a certain character. This does not work with strings though, as [^. ] stops when it reaches either character. Say I've got the code
import re
def main():
my_string = "The value of the float is 2.5. The int's value is 2.\n"
re.search("[^.]*", my_string)
main()
Which gives a match object with the string
"The value of the float is 2"
How can I change this so that it only stops after the string ". "?
Bonus question, is there any way to tell regex to stop whenever it reaches one of multiple strings? Using the above code as an example, if I wanted the search to end when it found the string ". " or the string ".\n", how would I go about it? Thanks!
To match from the start of a string till the . followed with whitespace, use
^(.*?)\.\s
If you want to only require a space or newline after a dot, use either of (the second is best if you have single chars only, use alternation if there are multicharacter alternatives)
^(.*?)\.(?: |\n)
^(.*?)\.[ \n]
See the regex demo.
Details
^ - start of a string
(.*?) - Capturing group 1: any 0+ chars other than linebreak chars, as few as possible
\. - a literal . char
\s - a whitespace char
(?: |\n) / [ \n] - a non-capturing group matching either a space or (|) a newline.
Python demo:
import re
my_string = "The value of the float is 2.5. The int's value is 2.\n"
m = re.search("^(.*?)\.\s", my_string) # Try to find a match
if m: # If there is a match
print(m.group(1)) # Show Group 1 value
NOTE If there can be line breaks in the input, pass re.S or re.DOTALL flag:
m = re.search("^(.*?)\.\s", my_string, re.DOTALL)
Besides classic approach explained by Wiktor, also splitting may be interesting solution in this case.
>>> my_string
"The value of the float is 2.5. The int's value is 2.\n"
>>> re.split('\. |\.\n', my_string)
['The value of the float is 2.5', "The int's value is 2", '']
If you want to include periods at the end of the sentence, you can do something like this:
['{}.'.format(sentence) for sentence in re.split('\. |\.\n', my_string) if sentence]
To handle multiple empty spaces between the sentences:
>>> str2 = "The value of the float is 2.5. The int's value is 2.\n\n "
>>> ['{}.'.format(sentence)
for sentence in re.split('\. \s*|\.\n\s*', str2)
if sentence
]
['The value of the float is 2.5.', "The int's value is 2."]
I'm trying to add \n after the quotation mark (") and space.
The closest that I could find is re.sub however it remove certain characters.
line = 'Type: "SecurityIncident" RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"'
q = re.sub(r'[\d\w]" ', '\n', line)
print(q)
Output:
Type: "SecurityInciden\nRowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F2\n
Looking for a solution without any character being remove.
Your attempted regex [\d\w]" is almost fine but has some little short comings. You don't need to write \d with \w in a character set as that is redundant as \w already contains \d within it. Since \w alone is enough to represent an alphabet or digit or underscore, hence no need to enclose it in character set [] hence you can just write \w and your updated regex becomes \w".
But now if you match this regex and substitute it with \n it will match a literal alphabet t then " and a space and it will be replaced by \n which is why you are getting this output,
SecurityInciden\nRowID
You need to capture the matched string in group1 and while substituting, you need to use it while substituting so that doesn't get replaced hence you should use \1\n as replacement instead of just \n
Try this updated regex,
(\w" )
And replace it by \1\n
Demo1
If you notice, there is an extra space at the end of line in the first line and if you don't want that space there, you can take that space out of those capturing parenthesis and use this regex,
(\w")
^ space here
Demo2
Here is a sample python code,
import re
line = 'Type: "SecurityIncident" RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"'
q = re.sub(r'(\w") ', r'\1\n', line)
print(q)
Output,
Type: "SecurityIncident"
RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"
Try this:
import re
line = 'Type: "SecurityIncident" RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"'
pattern = re.compile('(\w+): (".+?"\s?)', re.IGNORECASE)
q = re.sub(pattern, r'\g<1>: \g<2>\n', line)
print(repr(q))
It should give you following resutls:
Type: "SecurityIncident" \nRowID:
"FB013B06-B04C-4FEB-A5A5-3B858F910F29"\n
In your regex you are removing the t from incident because you are matching it and not using it in the replacement.
Another option to get your result might be to split on a double quote followed by a whitespace when preceded with a word character using a positive lookbehind.
Then join the result back together using a newline.
(?<=\w)"
Regex demo | Python demo
For example:
import re
line = 'Type: "SecurityIncident" RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"'
print("\n".join(re.split(r'(?<=\w)" ', line)))
Result
Type: "SecurityIncident
RowID: "FB013B06-B04C-4FEB-A5A5-3B858F910F29"
I am learning Regular Expressions, so apologies for a simple question.
I want to select the words that have a '-' (minus sign) in it but not at the beginning and not at the end of the word
I tried (using findall):
r'\b-\b'
for
str = 'word semi-column peace'
but, of course got only:
['-']
Thank you!
What you actually want to do is a regex like this:
\w+-\w+
What this means is find a alphanumeric character at least once as indicated by the utilization of '+', then find a '-', following by another alphanumeric character at least once, again, as indicated by the '+' again.
str is a built in name, better not to use it for naming
st = 'word semi-column peace'
# \w+ word - \w+ word after -
print(re.findall(r"\b\w+-\w+\b",st))
['semi-column']
a '-' (minus sign) in it but not at the beginning and not at the end of the word
Since "-" is not a word character, you can't use word boundaries (\b) to prevent a match from words with hyphens at the beggining or end. A string like "-not-wanted-" will match both \b\w+-\w+\b and \w+-\w+.
We need to add an extra condition before and after the word:
Before: (?<![-\w]) not preceded by either a hyphen nor a word character.
After: (?![-\w]) not followed by either a hyphen nor a word character.
Also, a word may have more than 1 hyphen in it, and we need to allow it. What we can do here is repeat the last part of the word ("hyphen and word characters") once or more:
\w+(?:-\w+)+ matches:
\w+ one or more word characters
(?:-\w+)+ a hyphen and one or more word characters, and also allows this last part to repeat.
Regex:
(?<![-\w])\w+(?:-\w+)+(?![-\w])
regex101 demo
Code:
import re
pattern = re.compile(r'(?<![-\w])\w+(?:-\w+)+(?![-\w])')
text = "-abc word semi-column peace -not-wanted- one-word dont-match- multi-hyphenated-word"
result = re.findall(pattern, text)
ideone demo
You can also use the following regex:
>>> st = "word semi-column peace"
>>> print re.findall(r"\S+\-\S+", st)
['semi-column']
You can try something like this: Centering on the hyphen, I match until there is a white space in either direction from the hyphen I also make check to see if the words are surrounded by hyphens (e.g -test-cats-) and if they are I make sure not to include them. The regular expression should also work with findall.
st = 'word semi-column peace'
m = re.search(r'([^ | ^-]+-[^ | ^-]+)', st)
if m:
print m.group(1)
I'm having trouble matching a string with regexp (I'm not that experienced with regexp). I have a string which contains a forward slash after each word and a tag. An example:
led/O by/O Timothy/PERSON R./PERSON Geithner/PERSON ,/O the/O president/O of/O the/O New/ORGANIZATION
In those strings, I am only interested in all strings that precede /PERSON. Here's the regexp pattern that I came up with:
(\w)*\/PERSON
And my code:
match = re.findall(r'(\w)*\/PERSON', string)
Basically, I am matching any word that comes before /PERSON. The output:
>>> reg
['Timothy', '', 'Geithner']
My problem is that the second match, matched to an empty string as for R./PERSON, the dot is not a word character. I changed my regexp to:
match = re.findall(r'(\w|.*?)\/PERSON', string)
But the match now is:
['led/O by/O Timothy', ' R.', ' Geithner']
It is taking everything prior to the first /PERSON which includes led/O by/O instead of just matching Timothy. Could someone please help me on how to do this matching, while including a full stop as an abbreviation? Or at least, not have an empty string match?
Thanks,
Match everything but a space character ([^ ]*). You also need the star (*) inside the capture:
match = re.findall(r'([^ ]*)\/PERSON', string)
Firstly, (\w|.) matches "a word character, or any character" (dot matches any character which is why you're getting those spaces).
Escaping this with a backslash will do the trick: (\w|\.)
Second, as #Ionut Hulub points out you may want to use + instead of * to ensure you match something but Regular Expressions work on the principle of "leftmost, longest" so it'll always try to match the longest part that it can before the slash.
If you want to match any non-whitespace character you can use \S instead of (\w|\.), which may actually be what you want.