Python在不同列表中同时出现的两项 [英] Python co-occurrence of two items in different lists

问题描述

我有一个列表列表，如:

I have a list of lists like :

names =  [['cat', 'fish'], ['cat'], ['fish', 'dog', 'cat'],
 ['cat', 'bird', 'fish'], ['fish', 'bird']]

我想计算在整个列表中在一起提到的每对名字的次数，输出结果将是:

I want to count number of times that each pair of names mentioned together in the whole list and the output would be like:

{ ['cat', 'fish']: 3, ['cat', 'dog']: 1,['cat','bird']:1
 ['fish','dog'] : 1, ['fish','bird']:2} 

我尝试过:

from collections import Counter
from collections import defaultdict

co_occurences = defaultdict(Counter)
for tags in names:
    for key in tags:
        co_occurences[key].update(tags)

print co_occurences

，但它不计算主列表中的共发生次数.

but it doesn't count co=occurrences in the main list.

推荐答案

您可以在python中使用按位与，并通过将列表列表转换为集合列表来进行比较

You could use the bitwise AND in python and compare them by converting the list of lists to a list of sets

>>> set(['cat','dog']) & set(['cat','dog','monkey','horse','fish'])
set(['dog', 'cat'])

您可以使用此属性并获得所需的计数.

You could use this property and achieve the count you've wanted.

def listOccurences(item, names):
    # item is the list that you want to check, eg. ['cat','fish']
    # names contain the list of list you have.
    set_of_items = set(item) # set(['cat','fish'])
    count = 0
    for value in names:
        if set_of_items & set(value) == set_of_items:
            count+=1
    return count

names =  [['cat', 'fish'], ['cat'], ['fish', 'dog', 'cat'],['cat', 'bird', 'fish'], ['fish', 'bird']]
# Now for each of your possibilities which you can generate
# Chain flattens the list, set removes duplicates, and combinations generates all possible pairs.
permuted_values = list(itertools.combinations(set(itertools.chain.from_iterable(names)), 2))
d = {}
for v in permuted_values:
    d[str(v)] = listOccurences(v, names)
# The key in the dict being a list cannot be possible unless it's converted to a string.
print(d)
# {"['fish', 'dog']": 1, "['cat', 'dog']": 1, "['cat', 'fish']": 3, "['cat', 'bird']": 1, "['fish', 'bird']": 2}

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