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card_travel2 - 3 yrslocation_onKolkata
From above string I want to get 2 - 3 yrs which regular expression can I use?
Hope you are looking something like this,
import re
pattern = r'\d+\s+-\s+\d+\s+yrs'
text = 'card_travel2 - 3 yrslocation_onKolkata'
re.findall(pattern, text)
output
['2 - 3 yrs']
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eg-In this instead of using 21 (a value), I want to use a variable to generalize it
print("{:-^21}".format(".|."*(2*(i+1)-1)))
I want to use something like this
print("{:-^M}".format(".|."*(2*(i+1)-1)))
That can easily enough be done. For example:
M = 40
i = 3
print("{val:-^{width}}".format(width=M, val=".|."*(2*(i+1)-1)))
Outputs:
---------.|..|..|..|..|..|..|.----------
You could also do it with f-strings (note the outer ' because " is used on the inner expression):
print(f'{".|."*(2*(i+1)-1):-^{M}}')
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import re
s="fig shown abcd.png referring 12254383.png"
p=re.sub("\(.*?).png\", '', s)
print(p)
Output expected:
fig shown referring
Please help to remove *.png
Here's a sample of the regex that works:
https://regex101.com/r/HJLzTo/1
pat = re.compile(r'[\w]+\.png')
pat.sub('', "fig shown abcd.png referring 12254383.png")
Result:
'fig shown referring '
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I am trying to separate all the images from the following string.
how can I get a list of images that start with "comp1/img_" and are either split by a "," or a ";"
/*jsonp*/jsonresp({"img_set":"comp1/img_23434;comp1/img_3243r43r,comp1/img_o43nfjr;comp1/img_wjfno43,comp1/img_nrejfner;comp1/img_jrenckerjv,comp1/img_23434k;comp1/img_rkfnk4n"},"fknreff\",");
so I would end up with a list like...
comp1/img_23434
comp1/img_3243r43r
comp1/img_o43nfjr
comp1/img_wjfno43
comp1/img_nrejfner
comp1/img_jrenckerjv
comp1/img_23434k
comp1/img_rkfnk4n
any help would be appreciated.
thanks
You can do this:
>>> data = '/*jsonp*/jsonresp({"img_set":"comp1/img_23434;comp1/img_3243r43r,comp1/img_o43nfjr;comp1/img_wjfno43,comp1/img_nrejfner;comp1/img_jrenckerjv,comp1/img_23434k;comp1/img_rkfnk4n"},"fknreff\",");'
>>> import re
>>> re.findall(r'comp1/img_[^;,"]+', data)
['comp1/img_23434', 'comp1/img_3243r43r', 'comp1/img_o43nfjr', 'comp1/img_wjfno43', 'comp1/img_nrejfner', 'comp1/img_jrenckerjv', 'comp1/img_23434k', 'comp1/img_rkfnk4n']
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I have a list output:
['Go497f9te(40RAAC34)\n','G0THDU433(40RAAC33)\n']
and I want to clean it up in order to output:
[40RAAC34,40RAAC33]
If you have a string:
'hello (world)'
and want the text between the brackets, you can either use a regex:
import re
re.findall('\((.*?)\)', s)[0]
#'world'
or, if you are sure that there is only one set of brackets (i.e. no leading ) chars) then you can just use slicing:
s[s.index('(')+1:s.index(')')]
#'world'
So then you just need to throw this into a list-comprehension or similar.
l = ['Go497f9te(40RAAC34)\n','G0THDU433(40RAAC33)\n']
[s[s.index('(')+1:s.index(')')] for s in l]
#['40RAAC34', '40RAAC33']
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I have a string
<b>Status : Active<br>Code : C1654<br><br>Shop <a class="top"<b>Shop A</b></a></b>
And I want get Active , C1654 and Shop A.
How to do the same thing in Python?
use python re module
(you didn't explain the pattern you wanted to follow so I can just give an example that will work for the above string):
import re
results = []
reg = '.*?>.*?: (.+?)<br'
my_str = '<b>Status : Active<br>Code : C1654<br><br>Shop <a class="top"<b>Shop A</b></a></b>'
results+=re.findall(reg,my_str)
reg2 = '<a.*?<b>(.*?)</b>'
results += re.findall(reg2,my_str)
print result
>>>['Active', 'C1654', 'Shop A']
I hope I helped