Question

我想在两个数组之间创建一个映射。但是在python中，这样做导致最后一个元素被选中的映射。

array_1 = [0,0,0,1,2,3]
array_2 = [4,4,5,6,8,7]
mapping = dict(zip(array_1, array_2))
print(mapping)

映射导致{0: 5, 1: 6, 2: 8, 3: 7}

在这种情况下，如何为键4选择最常见的元素0。

Answer 1

您可以创建一个包含键和键值列表的字典。然后，您可以遍历此字典中的值列表，并使用Counter.most_common

将值更新为列表中最频繁出现的项目

from collections import defaultdict, Counter

array_1 = [0,0,0,1,2,3]
array_2 = [4,4,5,6,8,7]

mapping = defaultdict(list)

#Create the mapping with a list of values
for key, value in zip(array_1, array_2):
    mapping[key].append(value)

print(mapping)
#defaultdict(<class 'list'>, {0: [4, 4, 5], 1: [6], 2: [8], 3: [7]})

res = defaultdict(int)

#Iterate over mapping and chose the most frequent element in the list, and make it the value
for key, value in mapping.items():
    #The most frequent element will be the first element of Counter.most_common
    res[key] = Counter(value).most_common(1)[0][0]

print(dict(res))

输出将为

{0: 4, 1: 6, 2: 8, 3: 7}

Answer 2

您可以使用Counter计算所有映射的频率，然后按键和频率对这些映射进行排序：

from collections import Counter

array_1 = [0,0,0,1,2,3]
array_2 = [4,4,5,6,8,7]
c = Counter(zip(array_1, array_2))
dict(i for i, _ in sorted(c.items(), key=lambda x: (x[0], x[1]), reverse=True))
# {3: 7, 2: 8, 1: 6, 0: 4}

创建字典时如何选择最常出现的元素？

2 个答案: