Question

我想在2D数组中找到对的频率。样本输入如下：

list_of_items = [[12,14,18],[12,19,54,89,105],[ 14, 19],[54, 88 ,105,178]]

预期输出如下：

(12,14):1
(12,18):1
(12,19):1
(12,54):1
(12,88):0
.
.
.
(54,105):2
.
.

我尝试过以下代码，但我认为这不是最佳解决方案：

number_set = [ 12, 14, 18,19,54,88,89 , 105, 178]

def get_frequency_of_pairs(list_of_items, number_set):
    x=1
    combination_list = []
    result = {}
    for i in number_set:
       for j in range(x,len(number_set)):
           combination_list = combination_list +[(i,number_set[j])]
       x = x+1
    for t in combination_list:
        result[t]=0
    for t in combination_list:
       for items in list_of_items:
           if( set(t).issubset(items) ):
              result[t]=result[t]+1
    return result

Answer 1

您可以使用itertools中的组合并使用集合中的Counter，如下所示：

counts = collections.Counter()
list_of_items = [[12,14,18], [12,19,54,89,105], [14,19], [54,88,105,178]]
for sublist in list_of_items:
    counts.update(itertools.combinations(sublist, 2))

print counts
Counter({(54, 105): 2, (88, 105): 1, (54, 89): 1, (19, 105): 1, (12, 14): 1, (14, 19): 1, (14, 18): 1, (12, 89): 1, (12, 19): 1, (89, 105): 1, (12, 18): 1, (19, 89): 1, (19, 54): 1, (105, 178): 1, (88, 178): 1, (54, 178): 1, (12, 105): 1, (12, 54): 1, (54, 88): 1})

必须枚举每一对以进行计数，并且此方法使您只能枚举每对。应该是最好的时间复杂性。

在python中查找2D数组中的数字对的频率

1 个答案: