如何在嵌套列表 Python 中定位索引

Question

从输入导入列表

def find_peak(m: List[List[int]]) -> List[int]:

    """

    Given an non-empty elevation map m, returns the cell of the

    highest point in m.

Examples (note some spacing has been added for human readablity)

>>> m = [[1,2,3],

         [9,8,7],

         [5,4,6]]

>>> find_peak(m)

[1,0]

>>> m = [[6,2,3],

         [1,8,7],

         [5,4,9]]

>>> find_peak(m)

[2,2]

"""

max_location = []

for sublist in m:

    max_location.append(max(sublist))

max_location = max(max_location)

for sublist in m:

    if max_location in sublist:

        return (m.index(max_location),sublist.index(max_location))

这并没有真正起作用，因为它只是返回该数字不在列表中

Answer 1

您还可以充分利用enumerate. 首先找到具有最大编号的行（及其索引）。然后在该行中找到该数字及其索引。在这两种情况下，您都需要为max函数提供一个键，以便它考虑值（而不是索引）：def find_peak(m):    i, max_row = max(enumerate(m), key=lambda x: max(x[1]))    j, max_val = max(enumerate(max_row), key=lambda x: x[1])    return [i, j]输出print(find_peak([[1,2,3], [9,8,7], [5,4,6]]))# [1, 0]print(find_peak([[6,2,3], [1,8,7], [5,4,9]]))# [2, 2]

Answer 2

我认为当您考虑迭代索引而不是列表中的项目时它更容易：from itertools import productd = [[1, 2, 3], [7, 9, 8], [4, 5, 6]]# generate all indicesx_len = range(len(d))y_len = range(len(d[0]))indices = product(x_len, y_len)# select maximal item by indexkey = lambda x: d[x[0]][x[1]]max_index = max(indices, key=key)print(max_index)

Answer 3

您可以在展平的输入结构中找到最大值，然后返回指定先前找到的最大值位置的坐标：from typing import Listdef find_peak(m: List[List[int]]) -> List[int]:  _max = max([i for b in m for i in b])  return [[i, a] for i in range(len(m)) for a in range(len(m[0])) if m[i][a] == _max][0]print(list(map(find_peak, [[[1, 2, 3], [9, 8, 7], [5, 4, 6]], [[6, 2, 3], [1, 8, 7], [5, 4, 9]]])))输出：[[1, 0], [2, 2]]