I'd like to extract a list containing more than one identical element.
Specifically
li = [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [2, 3, 4], [1, 2, 3], [2, 3, 4], [5, 6, 7]]
li contains two [1,2,3], three [2,3,4], and one [3,4,5][4,5,6][5,6,7] and
I would like to extract a list containing two or more identical elements and obtain results similar to li2 below.
li2 = [[1,2,3], [2,3,4]]
Thank you for your cooperation.
python
This is an example of Python 3 running
li = [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [2, 3, 4], [1, 2, 3], [2, 3, 4], [5, 6, 7]]
from collections import Counter
c=Counter(map(tuple,li))
Finally, if the li2 element doesn't have to be a list:
li2=list(filter(lambdax:c[x]>1,c))
If it has to be a list:
li2=list(map(list,filter(lambdax:c[x]>1,c)))
If you also want to keep the order:
li2=sorted(list(map(list,filter(lambdax:c[x]>1,c))),key=li.index)
Method 1
how to use itertools.groupby.When the same elements are sorted and aligned, they are grouped together.The order of results is not maintained.
from ittertools import groupby
print [k for k, gin groupby(sorted(li)) iflen(list(g))>=2]
Method 2
Using collections.Counter, the order of results is not maintained.
import collections
li = [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [2, 3, 4], [1, 2, 3], [2, 3, 4], [5, 6, 7]]
li2=map(lambda(k,v):list(k),#Tuple only and return to list
filter(lambda(x,v): Extract 2 or more as a result of counting v>=2,#Counter
collections.Counter(
Make it a hashable tuple to count in map(tuple,li)#Counter
).items()
)
)
print li2
Run Results
[2,3,4], [1,2,3]]
I have already answered, but I feel that List Compression is more refreshing than map or lambda.
from collections import Counter
li = [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [2, 3, 4], [1, 2, 3], [2, 3, 4], [5, 6, 7]]
c=Counter(tuple(items) for items in li)
more_than_one = [ items for items, count in c. most_common() if count > 1 ]
more_than_one:
[(2,3,4), (1,2,3)]
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