Replace all .png to nothing in python [closed] - python

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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 '

Related

formating string and passing argument at runtime [closed]

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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}}')

Python Regular Expression getting sub string [closed]

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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']

python regex with multiple separators [closed]

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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']

Regular expression operations in Python [closed]

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How to get what I want? For example, I have a string like this
'RC00001 C00003_C00004RC00087 C00756_C01545RC01045 C06756_C03485'
I want to get
'RC00001 C00003_C00004','RC00087 C00756_C01545','RC01045 C06756_C03485'
What should I do? I have tried many times, but I failed. Please help me! Thank you!

answer=[]
a="RC00001 C00003_C00004RC00087 C00756_C01545RC01045 C06756_C03485"
b = a.split("RC")
for i in b[1:]:
answer.append("RC%s" % (i))
print(answer)
This will output:
['RC00001 C00003_C00004', 'RC00087 C00756_C01545', 'RC01045 C06756_C03485']

If you want to achieve this using regex, you could try the following
import re
input_str = 'RC00001 C00003_C00004RC00087 C00756_C01545RC01045 C06756_C03485'
pattern = '(RC[\d+]+\s+C[\d]+_C[\d]+)'
print(re.findall(pattern, input_str))
# output
# [('RC00001 C00003_C00004', 'RC00087 C00756_C01545', 'RC01045 C06756_C03485')]
provided the format is always RC{numbers} C{numbers}

Regex replacing part of link [closed]

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I need to write a regex, that will traverse through my file and replace each :
href='#!/api/API.
with
href='http://www.domain.com/API. . I currently have this kind of code, that will be run from console :
def replace(file, pattern, substring):
my_file = open(file)
for line in my_file:
my_file.write(line.replace(pattern, substring))
Will sending the given strings work in this case ?

So you want to replace #!/api with http://www.domain.com?
You can just do:
line.replace('#!/api','http://www.domain.com')
No need for regex.

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