Zip items from two lists that have the same file name?












2














I have two list: this:



list1(has way more items)



['C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp']


and this:



list2(has way more items)



['C:\Users\user\Desktop\programs\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\AWE\AWE.shp', #THIS IS EXTRA
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp']


How to ensure that the pairs will match with the corresponding same name on the other list after the zip?



Maybe we match with their previous folder? Like:



if list1[0].split('\')[-2] == list2[0].split('\')[-2]:
final = [(f,s) for f,s in zip(list1,list2)]
final


wanted final output :



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),etc..]









share|improve this question






















  • Do the two lists have the same length?
    – Martin Thoma
    Nov 20 at 7:51










  • No they don't. it needs also condition for that.
    – user10671234
    Nov 20 at 7:57
















2














I have two list: this:



list1(has way more items)



['C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp']


and this:



list2(has way more items)



['C:\Users\user\Desktop\programs\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\AWE\AWE.shp', #THIS IS EXTRA
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp']


How to ensure that the pairs will match with the corresponding same name on the other list after the zip?



Maybe we match with their previous folder? Like:



if list1[0].split('\')[-2] == list2[0].split('\')[-2]:
final = [(f,s) for f,s in zip(list1,list2)]
final


wanted final output :



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),etc..]









share|improve this question






















  • Do the two lists have the same length?
    – Martin Thoma
    Nov 20 at 7:51










  • No they don't. it needs also condition for that.
    – user10671234
    Nov 20 at 7:57














2












2








2







I have two list: this:



list1(has way more items)



['C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp']


and this:



list2(has way more items)



['C:\Users\user\Desktop\programs\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\AWE\AWE.shp', #THIS IS EXTRA
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp']


How to ensure that the pairs will match with the corresponding same name on the other list after the zip?



Maybe we match with their previous folder? Like:



if list1[0].split('\')[-2] == list2[0].split('\')[-2]:
final = [(f,s) for f,s in zip(list1,list2)]
final


wanted final output :



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),etc..]









share|improve this question













I have two list: this:



list1(has way more items)



['C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp']


and this:



list2(has way more items)



['C:\Users\user\Desktop\programs\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\AWE\AWE.shp', #THIS IS EXTRA
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp']


How to ensure that the pairs will match with the corresponding same name on the other list after the zip?



Maybe we match with their previous folder? Like:



if list1[0].split('\')[-2] == list2[0].split('\')[-2]:
final = [(f,s) for f,s in zip(list1,list2)]
final


wanted final output :



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),etc..]






python pandas






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 20 at 7:36









user10671234

376




376












  • Do the two lists have the same length?
    – Martin Thoma
    Nov 20 at 7:51










  • No they don't. it needs also condition for that.
    – user10671234
    Nov 20 at 7:57


















  • Do the two lists have the same length?
    – Martin Thoma
    Nov 20 at 7:51










  • No they don't. it needs also condition for that.
    – user10671234
    Nov 20 at 7:57
















Do the two lists have the same length?
– Martin Thoma
Nov 20 at 7:51




Do the two lists have the same length?
– Martin Thoma
Nov 20 at 7:51












No they don't. it needs also condition for that.
– user10671234
Nov 20 at 7:57




No they don't. it needs also condition for that.
– user10671234
Nov 20 at 7:57












1 Answer
1






active

oldest

votes


















2














I would just group the files with a collections.defaultdict(), then output the pairs of length 2 in a separate list.



Demo:



from os.path import basename
from collections import defaultdict
from pprint import pprint

f1 = [
"C:\Users\user\Desktop\prog1\merge\AST\AST.shp",
"C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp",
]

f2 = [
"C:\Users\user\Desktop\programs\merge\AST\AST.shp",
"C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\programs\merge\AWE\AWE.shp", # THIS IS EXTRA
"C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp",
]

files = defaultdict(list)
for path in f1 + f2:
filename = path.split('\')[-1]
files[filename].append(path)

pairs = [tuple(v) for k, v in files.items() if len(v) == 2]
pprint(pairs)


Output:



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp',
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp')]


Note: Using os.path.basename() to extract the filename from Windows paths will only work on Windows. It will simply do nothing on Unix enviorments.






share|improve this answer























  • your code returns .
    – Ev. Kounis
    Nov 20 at 7:53










  • in files = defaultdict(list). What is the list variable?
    – user10671234
    Nov 20 at 7:54






  • 1




    @RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
    – Ev. Kounis
    Nov 20 at 7:56








  • 1




    I think it is great.
    – user10671234
    Nov 20 at 8:25






  • 1




    @RoadRunner On retrospect, it does make sense. +1
    – Ev. Kounis
    Nov 20 at 8:45











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














I would just group the files with a collections.defaultdict(), then output the pairs of length 2 in a separate list.



Demo:



from os.path import basename
from collections import defaultdict
from pprint import pprint

f1 = [
"C:\Users\user\Desktop\prog1\merge\AST\AST.shp",
"C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp",
]

f2 = [
"C:\Users\user\Desktop\programs\merge\AST\AST.shp",
"C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\programs\merge\AWE\AWE.shp", # THIS IS EXTRA
"C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp",
]

files = defaultdict(list)
for path in f1 + f2:
filename = path.split('\')[-1]
files[filename].append(path)

pairs = [tuple(v) for k, v in files.items() if len(v) == 2]
pprint(pairs)


Output:



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp',
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp')]


Note: Using os.path.basename() to extract the filename from Windows paths will only work on Windows. It will simply do nothing on Unix enviorments.






share|improve this answer























  • your code returns .
    – Ev. Kounis
    Nov 20 at 7:53










  • in files = defaultdict(list). What is the list variable?
    – user10671234
    Nov 20 at 7:54






  • 1




    @RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
    – Ev. Kounis
    Nov 20 at 7:56








  • 1




    I think it is great.
    – user10671234
    Nov 20 at 8:25






  • 1




    @RoadRunner On retrospect, it does make sense. +1
    – Ev. Kounis
    Nov 20 at 8:45
















2














I would just group the files with a collections.defaultdict(), then output the pairs of length 2 in a separate list.



Demo:



from os.path import basename
from collections import defaultdict
from pprint import pprint

f1 = [
"C:\Users\user\Desktop\prog1\merge\AST\AST.shp",
"C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp",
]

f2 = [
"C:\Users\user\Desktop\programs\merge\AST\AST.shp",
"C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\programs\merge\AWE\AWE.shp", # THIS IS EXTRA
"C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp",
]

files = defaultdict(list)
for path in f1 + f2:
filename = path.split('\')[-1]
files[filename].append(path)

pairs = [tuple(v) for k, v in files.items() if len(v) == 2]
pprint(pairs)


Output:



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp',
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp')]


Note: Using os.path.basename() to extract the filename from Windows paths will only work on Windows. It will simply do nothing on Unix enviorments.






share|improve this answer























  • your code returns .
    – Ev. Kounis
    Nov 20 at 7:53










  • in files = defaultdict(list). What is the list variable?
    – user10671234
    Nov 20 at 7:54






  • 1




    @RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
    – Ev. Kounis
    Nov 20 at 7:56








  • 1




    I think it is great.
    – user10671234
    Nov 20 at 8:25






  • 1




    @RoadRunner On retrospect, it does make sense. +1
    – Ev. Kounis
    Nov 20 at 8:45














2












2








2






I would just group the files with a collections.defaultdict(), then output the pairs of length 2 in a separate list.



Demo:



from os.path import basename
from collections import defaultdict
from pprint import pprint

f1 = [
"C:\Users\user\Desktop\prog1\merge\AST\AST.shp",
"C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp",
]

f2 = [
"C:\Users\user\Desktop\programs\merge\AST\AST.shp",
"C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\programs\merge\AWE\AWE.shp", # THIS IS EXTRA
"C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp",
]

files = defaultdict(list)
for path in f1 + f2:
filename = path.split('\')[-1]
files[filename].append(path)

pairs = [tuple(v) for k, v in files.items() if len(v) == 2]
pprint(pairs)


Output:



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp',
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp')]


Note: Using os.path.basename() to extract the filename from Windows paths will only work on Windows. It will simply do nothing on Unix enviorments.






share|improve this answer














I would just group the files with a collections.defaultdict(), then output the pairs of length 2 in a separate list.



Demo:



from os.path import basename
from collections import defaultdict
from pprint import pprint

f1 = [
"C:\Users\user\Desktop\prog1\merge\AST\AST.shp",
"C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp",
]

f2 = [
"C:\Users\user\Desktop\programs\merge\AST\AST.shp",
"C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp",
"C:\Users\user\Desktop\programs\merge\AWE\AWE.shp", # THIS IS EXTRA
"C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp",
]

files = defaultdict(list)
for path in f1 + f2:
filename = path.split('\')[-1]
files[filename].append(path)

pairs = [tuple(v) for k, v in files.items() if len(v) == 2]
pprint(pairs)


Output:



[('C:\Users\user\Desktop\prog1\merge\AST\AST.shp',
'C:\Users\user\Desktop\programs\merge\AST\AST.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTI\ASTI.shp',
'C:\Users\user\Desktop\programs\merge\ASTI\ASTI.shp'),
('C:\Users\user\Desktop\prog1\merge\ASTO\ASTO.shp',
'C:\Users\user\Desktop\programs\merge\ASTO\ASTO.shp')]


Note: Using os.path.basename() to extract the filename from Windows paths will only work on Windows. It will simply do nothing on Unix enviorments.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 20 at 8:21

























answered Nov 20 at 7:49









RoadRunner

10.3k31340




10.3k31340












  • your code returns .
    – Ev. Kounis
    Nov 20 at 7:53










  • in files = defaultdict(list). What is the list variable?
    – user10671234
    Nov 20 at 7:54






  • 1




    @RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
    – Ev. Kounis
    Nov 20 at 7:56








  • 1




    I think it is great.
    – user10671234
    Nov 20 at 8:25






  • 1




    @RoadRunner On retrospect, it does make sense. +1
    – Ev. Kounis
    Nov 20 at 8:45


















  • your code returns .
    – Ev. Kounis
    Nov 20 at 7:53










  • in files = defaultdict(list). What is the list variable?
    – user10671234
    Nov 20 at 7:54






  • 1




    @RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
    – Ev. Kounis
    Nov 20 at 7:56








  • 1




    I think it is great.
    – user10671234
    Nov 20 at 8:25






  • 1




    @RoadRunner On retrospect, it does make sense. +1
    – Ev. Kounis
    Nov 20 at 8:45
















your code returns .
– Ev. Kounis
Nov 20 at 7:53




your code returns .
– Ev. Kounis
Nov 20 at 7:53












in files = defaultdict(list). What is the list variable?
– user10671234
Nov 20 at 7:54




in files = defaultdict(list). What is the list variable?
– user10671234
Nov 20 at 7:54




1




1




@RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
– Ev. Kounis
Nov 20 at 7:56






@RoadRunner That is what I get. The online compiler might be running a different OS though. Hmm
– Ev. Kounis
Nov 20 at 7:56






1




1




I think it is great.
– user10671234
Nov 20 at 8:25




I think it is great.
– user10671234
Nov 20 at 8:25




1




1




@RoadRunner On retrospect, it does make sense. +1
– Ev. Kounis
Nov 20 at 8:45




@RoadRunner On retrospect, it does make sense. +1
– Ev. Kounis
Nov 20 at 8:45


















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