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I got data=.... where I've done splitting so text.split("=") and chose the second list element.
But with my new data in this way it deletes even the data= and the "{" "}" Elements
Any suggestions?
You should choose the second element through the end of the list, then rejoin them.
text = "=".join(text.split("=")[1:])
Or you could do a string replacement:
text = text.replace("data=data=", "data=")
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Below is my list,
['pending/', 'pending/2021-08-01/', 'pending/2021-06-01/', 'pending/2021-06-18/']
And I need to sort the list and filter it to a below format. Please suggest a quicker way to achieve it
['pending/2021-06-01/', 'pending/2021-06-18/', 'pending/2021-08-01/']
When your format is fixed and always starts with "pending" you can use the normal sorted function and count the / in a list comprehension.
>>> values = ['pending/', 'pending/2021-08-01/', 'pending/2021-06-01/', 'pending/2021-06-18/']
>>> sorted(x for x in values if x.count('/') == 2)
['pending/2021-06-01/', 'pending/2021-06-18/', 'pending/2021-08-01/']
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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 am looking for a way to automatically extract dates from a string, but following each other without a delimiter
For example my string is: \n-\n24-04-201923-04-201922-04-201921-04-201920-04-201919-04-201918-04-2019
How can I get this output:
24-04-2019
23-04-2019
22-04-2019
21-04-2019
20-04-2019
19-04-2019
18-04-2019
Any help would be appreciated!
Given that they're all of equal length, you can just clear the \n's then use textwrap:
import textwrap
print(textwrap.wrap(my_string, 10))
You can remove \n's using strip():
my_string = my_string.strip()
You can use this code also.
string='\n-\n24-04-201923-04-201922-04-201921-04-201920-04-201919-04-201918-04-2019'
newStr=string[3:]
for char in range(0,len(newStr),10):
print newStr[char:char+10]
Here's the output
24-04-2019
23-04-2019
22-04-2019
21-04-2019
20-04-2019
19-04-2019
18-04-2019
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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 am trying to convert a tuple:
('Cobra',)
to a string, which when printed yields:
Cobra
#Assuming you have a list of tuples
sample = [('cobra',),('Cat',),('Dog',),('hello',),('Cobra',)]
#For each tuple in the list, Get the first element of each tuple
x = [i[0] for i in sample]