Question

我想从一个列表（基础）的元素中创建一个列表，但仅考虑另一列表（其他）的元素。

基本列表为：

base = [['A', 'B'], ['B'], ['A', 'B', 'C', 'D', 'E'], ['B'], ['A', 'B', 'C'], ['A'], ['B', 'C'], ['A', 'B'], ['C', 'A', 'B'], ['A'], ['B', 'C']]

另一个列表是：

other = ['A', 'B']

预期结果是：

expected = [['A', 'B'], ['B'], ['A', 'B'], ['B'], ['A', 'B'], ['A'], ['B'], ['A', 'B'], ['A', 'B'], ['A'], ['B']]

是否有一种pythonic的方法？

Answer 1

赞！

base = [['A', 'B'], ['B'], ['A', 'B', 'C', 'D', 'E'], ['B'], ['A', 'B', 'C'], ['A'], ['B', 'C'], ['A', 'B'], ['C', 'A', 'B'], ['A'], ['B', 'C']]
other = ['A', 'B']
expected = [[i for i in l if i in other] for l in base]
print(expected)

输出：

[['A', 'B'],
 ['B'],
 ['A', 'B'],
 ['B'],
 ['A', 'B'],
 ['A'],
 ['B'],
 ['A', 'B'],
 ['A', 'B'],
 ['A'],
 ['B']]

破门而入： expected = [[i for i in l if i in other] for l in base]等同于：

expected = []

for l in base: # For every list in base
    lst = [] # There will be a list in expected
    for i in l: # For every element in l
        if i in other: # If the element is in base
            lst.append(i) # Append the element to the list that will go into expected
    expected.append(lst) # Append lst to expected

Answer 2

使用列表理解：

result = [[x for x in arr if x in other] for arr in base]

此处可以应用一种快速优化。此处可以使用Set数据结构存储other进行O（1）查找。

因此，改进版本：

otherSet = set(other)
result = [[x for x in arr if x in otherSet] for arr in base]

Answer 3

您可以使用map和filter：

list(map(lambda x: list(filter(lambda f: f in other, x)), base))

[['A', 'B'],
 ['B'],
 ['A', 'B'],
 ['B'],
 ['A', 'B'],
 ['A'],
 ['B'],
 ['A', 'B'],
 ['A', 'B'],
 ['A'],
 ['B']]

仅考虑其他列表的元素来创建列表

3 个答案:

赞！