如何比较和统计两个列表中出现的次数

Question

我有两个不同的列表，大小不同(列表A的大小在1000范围内，列表B的大小在10000范围内(。

A=[[0, 0, 0],
[0, 0, 1],
[0, 0, 2],
[0, 0, 3],
[0, 0, 4],
[0, 0, 5],
[0, 1, 0],
[0, 1, 1],
[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 1, 5],
[0, 1, 6],
[0, 1, 7],
[0, 1, 8],
[0, 1, 9],
[0, 2, 0],
[0, 2, 1],
[0, 2, 2]] 
B=[[1, 1, 2],
[0, 0, 2],
[0, 0, 1],
[4, 2, 2],
[3, 1, 2],
[1, 0, 1],
[1, 1, 2],
[0, 1, 2],
[0, 0, 0],
[2, 2, 3],
[1, 2, 1],
[0, 2, 1],
[0, 2, 0],
[0, 2, 1],
[0, 1, 3],
[0, 0, 0],
[1, 2, 5],
[0, 4, 3],
[0, 1, 3]]

我需要比较列表和列表B，找出A的每个元素在B中出现的次数。例如，我需要找出[0,0,0](A的第一个元素(在B中发生的次数。谢谢你的帮助。

Answer 1

这应该有效：

A = [[0, 0, 0],
[0, 0, 1],
[0, 0, 2],
[0, 0, 3],
[0, 0, 4],
[0, 0, 5],
[0, 1, 0],
[0, 1, 1],
[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 1, 5],
[0, 1, 6],
[0, 1, 7],
[0, 1, 8],
[0, 1, 9],
[0, 2, 0],
[0, 2, 1],
[0, 2, 2]] 
B = [[1, 1, 2],
[0, 0, 2],
[0, 0, 1],
[4, 2, 2],
[3, 1, 2],
[1, 0, 1],
[1, 1, 2],
[0, 1, 2],
[0, 0, 0],
[2, 2, 3],
[1, 2, 1],
[0, 2, 1],
[0, 2, 0],
[0, 2, 1],
[0, 1, 3],
[0, 0, 0],
[1, 2, 5],
[0, 4, 3],
[0, 1, 3]]
nums = []
for i in A:
for j in B:
if i in B:
nums.append(str(i))
nums_freq = {}
nums = list(dict.fromkeys(nums))
for i in nums:
count = 0
for j in B:
if i == str(j):
if i in nums_freq.keys():
nums_freq[i] += 1
else:
nums_freq[i] = 1

num_freq:的值

{'[0, 0, 0]': 2,
'[0, 0, 1]': 1,
'[0, 0, 2]': 1,
'[0, 1, 2]': 1,
'[0, 1, 3]': 2,
'[0, 2, 0]': 1,
'[0, 2, 1]': 2}

Answer 2

>>> import operator
>>> for e in A:
...     print(e, 'appearing in :',  operator.countOf(B, e))
...
[0, 0, 0] appearing in : 2
[0, 0, 1] appearing in : 1
[0, 0, 2] appearing in : 1
[0, 0, 3] appearing in : 0
[0, 0, 4] appearing in : 0
[0, 0, 5] appearing in : 0
[0, 1, 0] appearing in : 0
[0, 1, 1] appearing in : 0
[0, 1, 2] appearing in : 1
[0, 1, 3] appearing in : 2
[0, 1, 4] appearing in : 0
[0, 1, 5] appearing in : 0
[0, 1, 6] appearing in : 0
[0, 1, 7] appearing in : 0
[0, 1, 8] appearing in : 0
[0, 1, 9] appearing in : 0
[0, 2, 0] appearing in : 1
[0, 2, 1] appearing in : 2
[0, 2, 2] appearing in : 0

Answer 3

我可以建议两种不同的方法：

使用计数器：

from collections import Counter
cb = Counter(tuple(b) for b in B)
list((a, cb[tuple(a)]) for a in A))

使用嵌套理解：

list((a, sum(all(ia == ib for ia, ib in zip(a, b)) for b in B)) for a in A)

---

实际上，您应该考虑将内部集合作为元组而不是列表，因为元组支持元素相等，并且是可哈希的。

如果A和B是元组的列表，那么它将简单到：

Counter(a for a in A for b in B if a == b)