I have a large nested list and each list within the nested list contains a list of numbers that are formatted as floats. However every individual list in the nested list is the same except for a few exceptions. I want to extract the numbers that are common to all of the lists in the nested list. A simple example of my problem is shown below:
nested_list = [[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0]]
In the following case I would want to extract the following:
common_vals = [2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0]
I tried to use set intersections to solve this but since I wasn't able to get this to work on all of the elements of the nested list.
You can use reduce and set.intersection:
>>> reduce(set.intersection, map(set, nested_list))
set([2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0])
Use itertools.imap for memory efficient solution.
Timing Comparisons:
>>> lis = [[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0]]
>>> %timeit set.intersection(*map(set, lis))
100000 loops, best of 3: 12.5 us per loop
>>> %timeit set.intersection(*(set(e) for e in lis))
10000 loops, best of 3: 14.4 us per loop
>>> %timeit reduce(set.intersection, map(set, lis))
10000 loops, best of 3: 12.8 us per loop
>>> %timeit reduce(set.intersection, imap(set, lis))
100000 loops, best of 3: 13.1 us per loop
>>> %timeit set.intersection(set(lis[0]), *islice(lis, 1, None))
100000 loops, best of 3: 10.6 us per loop
>>> lis = [[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0]]*1000
>>> %timeit set.intersection(*map(set, lis))
10 loops, best of 3: 16.4 ms per loop
>>> %timeit set.intersection(*(set(e) for e in lis))
10 loops, best of 3: 15.8 ms per loop
>>> %timeit reduce(set.intersection, map(set, lis))
100 loops, best of 3: 16.3 ms per loop
>>> %timeit reduce(set.intersection, imap(set, lis))
10 loops, best of 3: 13.8 ms per loop
>>> %timeit set.intersection(set(lis[0]), *islice(lis, 1, None))
100 loops, best of 3: 8.4 ms per loop
>>> lis = [[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0], [2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0]]*10**5
>>> %timeit set.intersection(*map(set, lis))
1 loops, best of 3: 1.92 s per loop
>>> %timeit set.intersection(*(set(e) for e in lis))
1 loops, best of 3: 2.17 s per loop
>>> %timeit reduce(set.intersection, map(set, lis))
1 loops, best of 3: 2.14 s per loop
>>> %timeit reduce(set.intersection, imap(set, lis))
1 loops, best of 3: 1.52 s per loop
>>> %timeit set.intersection(set(lis[0]), *islice(lis, 1, None))
1 loops, best of 3: 913 ms per loop
Conclusion:
Steven Rumbalski's solution is clearly the best one in terms of efficiency.
Ashwini Chaudhary's solution is elegant, but could be quite inefficient for large inputs because it creates many intermediate sets. If your nested_list is large do this:
>>> set.intersection(set(nested_list[0]), *itertools.islice(nested_list, 1, None))
set([2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0])
Try this, it's the simplest solution:
set.intersection(*map(set, nested_list))
Or if you prefer to use generator expressions, which should be a more efficient solution in terms of memory usage:
set.intersection(*(set(e) for e in nested_list))
Count occurences of each element in sets of lists occuring in nested_list, if occurence equals number of lements in nested_list, it is common to all. You do not need the set conversion if elements of nested_list do not have numbers repeated in them
nested_list = [[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0],
[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0,11.0,12.0,13.0,14.0,15.0]]
from collections import Counter
result = [val for val,cnt in Counter([x for t in nested_list for x in set(t)]).items() if cnt == len(nested_list)]
print result
# [2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0]
来源：https://stackoverflow.com/questions/17350775/how-to-find-elements-that-are-common-to-all-lists-in-a-nested-list