查找符合条件的列表的最后一个元素

Question

我有矩阵（列表的类型列表），我只想 select 符合我的条件的最后一行。具体来说，我有这个矩阵：

matrix = [
         [[0], whatever0],
         [[1], whatever1],
         [[1, 0], whatever10],
         [[1, 1], whatever11],
         [[1, 2], whatever12],
         [[1, 3], whatever13],
         [[1, 4], whatever14],
         ]

并且我只想 select 最后一行 i 使得 len(matrix[i][0]) == 1，所以答案应该 return [[1]，whatever1 ].

还没有找到如何回答这个具体问题。谢谢。

Answer 1

你可以这样做：

matrix = [
    [[0], 'whatever0'],
    [[1], 'whatever1'],
    [[1, 0], 'whatever10'],
    [[1, 1], 'whatever11'],
    [[1, 2], 'whatever12'],
    [[1, 3], 'whatever13'],
    [[1, 4], 'whatever14'],
]

result = next((e for e in reversed(matrix) if len(e[0]) == 1), None)
print(result)

输出

[[1], 'whatever1']

说明

首先使用 reversed and return only the elements that have len(e[0]) == 1. Then with next 向后遍历列表只得到第一个结果。

Answer 2

简单地向后迭代将是最有效的：

for i in range(len(matrix), -1, -1):
    if len(matrix[i][0]) == 1:
        print(matrix[i])
        break
else:
    print('No matching row found')

Answer 3

检查this

>>> matrix = [
...          [[0], "whatever0"],
...          [[1], "whatever1"],
...          [[1, 1], "whatever11"],
...          [[1, 2], "whatever12"],
...          [[1, 3], "whatever13"],
...          [[1, 4], "whatever14"]
...          ]

res=list(filter(lambda x: x[0][0]>0, matrix))[-1]

res 包含每个矩阵项的第一项的第一项的最后一个元素（粗体）

...[[0], "whatever0"],

...[[1],"whatever1"],

... [[1, 1], "whatever11"]

...[[1, 2]**, "whatever12"],

...[[1, 3]**, "whatever13"],

... [[1, 4]**, "whatever14"]

查找符合条件的列表的最后一个元素

Find last element of a list that fits a critera

python

select

element

list