I have the following string:
mystr = '(string_to_delete_20221012_11-36) keep this (string_to_delete_20221016_22-22) keep this (string_to_delete_20221017_20-55) keep this'
I wish to delete all the entries (string_to_deletexxxxxxxxxxxxxxx) (including the trailing space)
I sort of need pseudo code as follows:
If you find a string (string_to_delete then replace that string and the timestamp, closing parenthesis and trailing space with null e.g. delete the string (string_to_delete_20221012_11-36)
I would use a list comprehension but given that not all strings are contained inside parenthesis I cannot see what I could use to create the list via a string.split().
Is this somethng that needs regular expressions?
it seemed like a good place to put regex:
import re
pattern = r'\(string_to_delete_.*?\)\s*'
mystr = '(string_to_delete_20221012_11-36) keep this (string_to_delete_20221016_22-22) keep this (string_to_delete_20221017_20-55) keep this'
for match in re.findall(pattern, mystr):
mystr = mystr.replace(match, '', 1) # replace 1st occurence of matched str with empty string
print(mystr)
results with:
>> keep this keep this keep this
brief regex breakdown: \(string_to_delete_.*?\)\s*
\( look for left parenthesis - escape needed
match string string_to_delete_
.*? look for zero or more characters if any
\) match closing parenthesis
\s* include zero or more whitespaces after that
I was working on a certain problem where I have form new sub-strings from a main string.
For e.g.
in_string=ste5ts01,s02,s03
The expected output strings are ste5ts01, ste5ts02, ste5ts03
There could be comma(,) or forward-slash (/) as the separator and in this case the delimiters are the letter s and ,
The pattern I have created so far:
pattern = r"([^\s,/]+)(?<num>\d+)([,/])(?<num>\d+)(?:\2(?<num>\d+))*(?!\S)"
The issue is, I am not able to figure out how to give the letter 's' as one of the delimiters.
Any help will be much appreciated!
You might use an approach using the PyPi regex module and named capture groups which are available in the captures:
=(?<prefix>s\w+)(?<num>s\d+)(?:,(?<num>s\d+))+
Explanation
= Match literally
(?<prefix>s\w+) Match s and 1+ word chars in group prefix
(?<num>s\d+) Capture group num match s and 1+ digits
(?:,(?<num>s\d+))+ Repeat 1+ times matching , and capture s followed by 1+ digits in group num
Example
import regex as re
pattern = r"=(?<prefix>s\w+)(?<num>s\d+)(?:,(?<num>s\d+))+"
s="in_string=ste5ts01,s02,s03"
matches = re.finditer(pattern, s)
for _, m in enumerate(matches, start=1):
print(','.join([m.group("prefix") + c for c in m.captures("num")]))
Output
ste5ts01,ste5ts02,ste5ts03
Is there a way to replace the matched pattern substring using a single re.sub() line?.
What I would like to avoid is using a string replace method to the current re.sub() output.
Input = "/J&L/LK/Tac1_1/shareloc.pdf"
Current output using re.sub("[^0-9_]", "", input): "1_1"
Desired output in a single re.sub use: "1.1"
According to the documentation, re.sub is defined as
re.sub(pattern, repl, string, count=0, flags=0)
If repl is a function, it is called for every non-overlapping occurrence of pattern.
This said, if you pass a lambda function, you can remain the code in one line. Furthermore, remember that the matched characters can be accessed easier to an individual group by: x[0].
I removed _ from the regex to reach the desired output.
txt = "/J&L/LK/Tac1_1/shareloc.pdf"
x = re.sub("[^0-9]", lambda x: '.' if x[0] is '_' else '', txt)
print(x)
There is no way to use a string replacement pattern in Python re.sub to replace with two possible strings, as there is no conditional replacement construct support in Python re.sub. So, using a callable as the replacement argument or use other work-arounds.
It looks like you only expect one match of <DIGITS>_<DIGITS> in the input string. In this case, you can use
import re
text = "/J&L/LK/Tac1_1/shareloc.pdf"
print( re.sub(r'^.*?(\d+)_(\d+).*', r'\1.\2', text, flags=re.S) )
# => 1.1
See the Python demo. See the regex demo. Details:
^ - start of string
.*? - zero or more chars as few as possible
(\d+) - Group 1: one or more digits
_ - a _ char
(\d+) - Group 2: one or more digits
.* - zero or more chars as many as possible.
'Neighborhood,eattend10,eattend11,eattend12,eattend13,mattend10,mattend11,mattend12,mattend13,
hsattend10,hsattend11,hsattend12,hsattend13,eenrol11,eenrol12,eenrol13,menrol11,menrol12,
menrol13,hsenrol11,hsenrol12,hsenrol13,aastud10,aastud11,aastud12,aastud13,wstud10,wstud11,
wstud12,wstud13,hstud10,hstud11,hstud12,hstud13,abse10,abse11,abse12,abse13,absmd10,absmd11,
absmd12,absmd13,abshs10,abshs11,abshs12,abshs13,susp10,susp11,susp12,susp13,farms10,farms11,
farms12,farms13,sped10,sped11,sped12,sped13,ready11,ready12,ready13,math310,math311,math312,
math313,read310,read311,read312,read313,math510,math511,math512,math513,read510,read511,read512,
read513,math810,math811,math812,math813,read810,read811,read812,read813,hsaeng10,hsaeng11,
hsaeng12,hsaeng13,hsabio10,hsabio11,hsabio12,hsabio13,hsagov10,hsagov11,hsagov13,hsaalg10,
hsaalg11,hsaalg12,hsaalg13,drop10,drop11,drop12,drop13,compl10,compl11,compl12,compl13,
sclsw11,sclsw12,sclsw13,sclemp13\
I have this data set. I need to know how many drop words are there and print them.
Or similarly for any word like mattend and print those.
I tried using findall but I think that's not correct
I assume we can use re.search or re.match.
How can I do it in RegEx?
You can use len() on re.findall() to get the length of the returned list:
import re
with open('example.csv') as f:
data = f.read().strip()
print(len(re.findall('drop',data)))
I think re.findall should be correct.
From python re module documentation:
Search:
Scan through string looking for the first location where this regular expression produces a match, and return a corresponding match object.
Match:
If zero or more characters at the beginning of string match this regular expression, return a corresponding match object.
Findall:
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result.
I tried it on your example and it worked for me:
re.findall("drop", str)
If you want to see digits after it you can try something like:
re.findall("drop\d*", str)
If you want to count the words you can use:
len(re.findall("drop\d*", str))
I am trying to split the string 'id#namespace' by #.
There is this special case that the id has a format name#gmail which makes the string I am trying to split look like name#gmail#namespace.
Is there any way to achieve that only split by the last # which will give me name#gmail and namespace?
If it is always the last index. Use lastIndex to find the last index of the character and then use substring
Something like this
int idx = string.lastIndexOf("#");
String[] splitStrings = {string.substring(0, idx), string.substring(idx)};
You could match the last # using a negative lookahead (?! to assert that there are no more # following:
#(?!.*#)
Regex demo
System.out.println("name#gmail#namespace".split("#(?!.*#)")[0]); //name#gmail