问题描述
my_list_of_lists =
[['sparrow','sparrow','junco','jay','robin'],['sparrow','robin','robin']]
我想对所有列表的每个位置进行成对比较,如下所示:
#1 with 2
['sparrow','robin']
['sparrow','robin']
#1 with 3
['sparrow','robin']
#2 with 3
['sparrow','robin']
因此1与2的配对:
pairs =[('sparrow','sparrow'),('sparrow','junco'),('junco',('jay','robin'),('robin','robin')]
我想在每次成对比较中获得对的计数和频率:
pairs =[('sparrow','sparrow') ('junco','robin')]
sparrowsparrow_counts = 2
juncosparrow_counts = 2
jayrobin_counts = 1
robinrobin = 1
frequency_of_combos = [('sparrow','sparrow'):.333,'junco'):.333,'robin'):.167,'robin'): .167]
我尝试压缩,但最终将所有列表(不是成对的)压缩到元组中,其余的我都感到困惑。
我认为它与How to calculate counts and frequencies for pairs in list of lists?有点相关,但是我不知道如何将其应用于我的数据。
解决方法
压缩两个列表,然后过滤掉不匹配的对,并使用collections.Counter对它们进行计数:
from collections import Counter
a = ['sparrow','sparrow','junco','jay','robin']
b = ['sparrow','robin','robin']
c = Counter([ i for i in zip(a,b) if i[0] == i[1]])
print(c)
Counter({('sparrow','sparrow'): 2,('robin','robin'): 1})
您似乎已经弄清楚了频率部分,但这应该可以清除zip和Counter的使用。