[英]Find frequency of pair of numbers in 2D array in python
我想在2D數組中找到對的頻率。 樣本輸入如下:
list_of_items = [[12,14,18],[12,19,54,89,105],[ 14, 19],[54, 88 ,105,178]]
預期產出如下:
(12,14):1
(12,18):1
(12,19):1
(12,54):1
(12,88):0
.
.
.
(54,105):2
.
.
我嘗試過以下代碼,但我認為這不是最佳解決方案:
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
您可以使用itertools中的組合並使用集合中的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})
必須枚舉每一對以進行計數,並且此方法使您只能枚舉每對。 應該是最好的時間復雜性。
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