This question already has answers here:
Escaping regex string
(4 answers)
Closed 3 years ago.
ı am trying to stemmize words in tex of dataframe
data is a dataframe , karma is text column , zargan is the dict of word and root of word
for a in range(1,100000):
for j in data.KARMA[a].split():
pattern = r'\b'+j+r'\b'
data.KARMA[a] = re.sub(pattern, str(zargan.get(j,j)),data.KARMA[a])
print(data.KARMA[1])
I want to change the word and root in the texts
Looks like j contains some regular expression special character like *. If you want it to be interpreted as literal text, you can say
pattern = r'\b'+re.escape(j)+r'\b'
and possibly the same for r if it should similarly be coerced into a literal string.
Related
This question already has answers here:
What do ^ and $ mean in a regular expression?
(2 answers)
Closed 2 years ago.
I've got a problem with carets and dollar signs in Python.
I want to find every word which starts with a number and ends with a letter
Here is what I've tried already:
import re
text = "Cell: 415kkk -555- 9999ll Work: 212-555jjj -0000"
phoneNumRegex = re.compile(r'^\d+\w+$')
print(phoneNumRegex.findall(text))
Result is an empty list:
[]
The result I want:
415kkk, 9999ll, 555jjj
Where is the problem?
Problems with your regex:
^...$ means you only want full matches over the whole string - get rid of that.
r'\w+' means "any word character" which means letters + numbers (case ignorant) plus underscore '_'. So this would match '5555' for '555' via
r'\d+' and another '5' as '\w+' hence add it to the result.
You need
import re
text = "Cell: 415kkk -555- 9999ll Work: 212-555jjj -0000"
phoneNumRegex = re.compile(r'\b\d+[a-zA-Z]+\b')
print(phoneNumRegex.findall(text))
instead:
['415kkk', '9999ll', '555jjj']
The '\b' are word boundaries so you do not match 'abcd1111' inside '_§$abcd1111+§$'.
Readup:
re-syntax
regex101.com - Regextester website that can handle python syntax
This question already has answers here:
How to match a whole word with a regular expression?
(4 answers)
Closed 3 years ago.
I want to replace only specific word in one string. However, some other words have that word inside but I don't want them to be changed.
For example, for the below string I only want to replace x with y in z string. how to do that?
x = "the"
y = "a"
z = "This is the thermometer"
import re
pattern=r'\bthe\b' # \b - start and end of the word
repl='a'
string = 'This is the thermometer'
string=re.sub(pattern, repl, string)
In your case you can use re.sub(x, y, z).
You can read the documentation here for more information.
This question already has answers here:
How to extract numbers from a string in Python?
(19 answers)
Closed 3 years ago.
So, I have a string "AB256+74POL". I want to extract the numbers only into a list say num = [256,74]. How to do this in python?
I have tried string.split('+') and followed by iterating over the two parts and adding the characters which satisfy isdigit(). But is there an easier way to that?
import re
a = 'AB256+74POL'
array = re.findall(r'[0-9]+', a)
"".join([c if c.isdigit() else " " for c in mystring]).split()
Explanation
Strings are iterable in python. So we iterate on each character in the string, and replace non digits with spaces, then split the result to get all sequences of digits in a list.
This question already has answers here:
How can I tell if a string repeats itself in Python?
(13 answers)
Closed 3 years ago.
I need to split a string by using repeated characters.
For example:
My string is "howhowhow"
I need output as 'how,how,how'.
I cant use 'how' directly in my reg exp. because my input varies. I should check the string whether it is repeating the character and need to split that characters.
import re
string = "howhowhow"
print(','.join(re.findall(re.search(r"(.+?)\1", string).group(1), string)))
OUTPUT
howhowhow -> how,how,how
howhowhowhow -> how,how,how,how
testhowhowhow -> how,how,how # not clearly defined by OP
The pattern is non-greedy so that howhowhowhow doesn't map to howhow,howhow which is also legitimate. Remove the ? if you prefer the longest match.
lengthofRepeatedChar = 3
str1 = 'howhowhow'
HowmanyTimesRepeated = int(len(str1)/lengthofRepeatedChar)
((str1[:lengthofRepeatedChar]+',')*HowmanyTimesRepeated)[:-1]
'how,how,how'
Works When u know the length of repeated characters
This question already has answers here:
Replace all the occurrences of specific words
(4 answers)
Find substring in string but only if whole words?
(8 answers)
Closed 6 years ago.
Want to replace a certain words in a string but keep getting the followinf result:
String: "This is my sentence."
User types in what they want to replace: "is"
User types what they want to replace word with: "was"
New string: "Thwas was my sentence."
How can I make sure it only replaces the word "is" instead of any string of the characters it finds?
Code function:
import string
def replace(word, new_word):
new_file = string.replace(word, new_word[1])
return new_file
Any help is much appreciated, thank you!
using regular expression word boundary:
import re
print(re.sub(r"\bis\b","was","This is my sentence"))
Better than a mere split because works with punctuation as well:
print(re.sub(r"\bis\b","was","This is, of course, my sentence"))
gives:
This was, of course, my sentence
Note: don't skip the r prefix, or your regex would be corrupt: \b would be interpreted as backspace.
A simple but not so all-round solution (as given by Jean-Francios Fabre) without using regular expressions.
' '.join(x if x != word else new_word for x in string.split())