计算Python中的列表列表中列表的出现

Question

I have a list that looks something like this:

co_list = [[387, 875, 125, 822], [397, 994, 135, 941], [397, 994, 135, 941], [397, 994, 135, 941], [397, 994, 135, 941], [1766, 696, 1504, 643]. . . ]

I need to count the number of identical co-ordinates lists and return the count, 4 in this case.

So far I have tried:

def most_common(lst):
    lst = list(lst)
    return max(set(lst), key=lst.count)

for each in kk :
    print most_common(each) 

Using which I get the most occurring element in each list. But my intention is to get a list if it's occurrence is more than 3.

Expected Output:

(element, count) = ([397, 994, 135, 941], 4) 

Any help would be appreciated. Thanks.

Answer 1

You can use collections.Counter for that task:

from collections import Counter

co_list = [[387, 875, 125, 822], [397, 994, 135, 941], [397, 994, 135, 941], [397, 994, 135, 941], [397, 994, 135, 941], [1766, 696, 1504, 643]]

common_list, appearances = Counter([tuple(x) for x in co_list]).most_common(1)[0]  # Note 1
if appearances > 3:
    print((list(common_list), appearances))  # ([397, 994, 135, 941], 4)
else:
    print('No list appears more than 3 times!')

1) The inner list s are converted to tuple s because Counter builds a dict and list s being not hashable cannot be used as key s.

Answer 2

from collections import Counter

def get_most_common_x_list(origin_list, x):

    counter = Counter(tuple(item) for item in origin_list)

    for item, count in most_common_list = counter.most_common():

        if count > x:

            yield list(item), count

        else:

            break