如何在Python中转置不规则（锯齿状）二维数组？

你的需求完全可行，由于NumPy的transpose函数仅适用于规则形状的数组，而原生zip会以最短子数组长度截断，所以需要结合Python原生工具处理：

方法一：手动按列收集元素 #

先将NumPy对象数组转为Python列表，再遍历每一列索引收集有效元素：

import numpy as np

seats = np.array([
                   [1, 2, 3, 4, 5], 
                  [1, 2, 3, 4], 
                  [1, 2, 3], 
                  [1, 2, 3, 4],
                  [1, 2],
                  [1, 2]
                 ], dtype = object)

# 转换为Python列表便于处理
seats_list = seats.tolist()
# 获取最长子数组的长度，确定转置后的列数
max_col = max(len(row) for row in seats_list)

transposed_list = []
for col_idx in range(max_col):
    # 收集所有行中存在该列索引的元素
    current_col = [row[col_idx] for row in seats_list if col_idx < len(row)]
    transposed_list.append(current_col)

# 转换回NumPy对象数组
transposed_seats = np.array(transposed_list, dtype=object)

方法二：利用itertools.zip_longest简化代码 #

借助zip_longest对齐所有子数组，再过滤填充的空值：

import numpy as np
from itertools import zip_longest

seats = np.array([
                   [1, 2, 3, 4, 5], 
                  [1, 2, 3, 4], 
                  [1, 2, 3], 
                  [1, 2, 3, 4],
                  [1, 2],
                  [1, 2]
                 ], dtype = object)

seats_list = seats.tolist()
# 用zip_longest对齐所有行，填充None
transposed_list = []
for group in zip_longest(*seats_list, fillvalue=None):
    # 过滤掉填充的None，保留有效元素
    current_col = [val for val in group if val is not None]
    transposed_list.append(current_col)

transposed_seats = np.array(transposed_list, dtype=object)

两种方法最终生成的transposed_seats都完全符合你期望的格式：

array([[1, 1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2, 2],
       [3, 3, 3, 3],
       [4, 4, 4],
       [5]], dtype=object)

内容的提问来源于stack exchange，提问作者Wunderkind