Zip items from two lists that have the same file name?
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
add a comment |
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
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
add a comment |
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
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
python pandas
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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.
your code returns.
– Ev. Kounis
Nov 20 at 7:53
infiles = defaultdict(list)
. What is thelist
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
|
show 4 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
your code returns.
– Ev. Kounis
Nov 20 at 7:53
infiles = defaultdict(list)
. What is thelist
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
|
show 4 more comments
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.
your code returns.
– Ev. Kounis
Nov 20 at 7:53
infiles = defaultdict(list)
. What is thelist
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
|
show 4 more comments
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.
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.
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
infiles = defaultdict(list)
. What is thelist
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
|
show 4 more comments
your code returns.
– Ev. Kounis
Nov 20 at 7:53
infiles = defaultdict(list)
. What is thelist
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
|
show 4 more comments
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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