Python实现列表的列表中空行与空列的移除（非Pandas方案）

第一步：剔除全为空元素的子列表 #

直接用列表推导式筛选掉所有元素都是None的子列表：

# 过滤掉所有元素都是None的子列表
filtered_lol = [sublist for sublist in lol if not all(x is None for x in sublist)]

拿示例输入来说，里面的[None, None, None]会被直接去掉，得到[["a", None, "c"], ["g", None, "i"], ["j", None, None]]。

第二步：移除所有子列表中该索引均为空的列 #

先找出那些在所有过滤后的子列表里，对应位置全是None的索引，然后只保留剩下的索引对应的元素：

# 找出需要保留的索引：排除所有子列表该位置都是None的索引
keep_indices = [idx for idx in range(len(filtered_lol[0])) if not all(sublist[idx] is None for sublist in filtered_lol)]
# 提取每个子列表中符合要求的元素
result = [[sublist[idx] for idx in keep_indices] for sublist in filtered_lol]

示例里过滤后的子列表中，索引1的元素全是None，所以这个索引会被排除，最终得到目标结果[["a", "c"], ["g", "i"], ["j", None]]。

完整代码示例 #

可以把两步拆分成更易读的版本：

# 示例输入
lol = [["a", None, "c"], [None, None, None], ["g", None, "i"], ["j", None, None]]

# 分步实现
filtered = [sublist for sublist in lol if not all(x is None for x in sublist)]
keep_indices = [i for i in range(len(filtered[0])) if not all(s[i] is None for s in filtered)]
final_result = [[s[i] for i in keep_indices] for s in filtered]

print(final_result)  # 输出: [['a', 'c'], ['g', 'i'], ['j', None]]

内容的提问来源于stack exchange，提问作者Moses Bakst