通过准备NaN使所有子列表的长度相同

首先,使用 max() . 然后,对于每个子列表,使用一个切片分配将该列表的内容替换为正确的 NaN 然后是原始列表。

import random, math

n = 5
lists = [list(range(0,random.randint(1,5))) for x in range(n)]

# get the maximum length
maxlen = len(max(lists, key=len))

# pad left of each sublist with NaN to make it as long as the longest
for sublist in lists:
    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist

import random
import numpy as np

n = 5
a = [list(range(0,random.randint(1,5))) for x in range(n)]
for c in a:
    while len(c) < n:
        c.insert(0, np.nan)
print(a)

[南,南,南,南,0],[南,南,南,南,0],[南,南,0,1],[南,南,0,1,2],[南,南,南,0,1]]

这是我认为你正在寻找的一个例子。

这不是最常见的蟒蛇密码,但我认为它可以做到:

longest_list_length = 0
for sublist in output:
  if len(sublist) > longest_list_length:
    longest_list_length = len(sublist)

for sublist in output:
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
  sublist = nans_to_prepend + sublist

如果你只关心生成 np.nan 在随机创建期间填充,这个简单的一行程序就足够了:

[[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]

# [[nan, 0, 1, 2, 3], 
#  [nan, nan, 0, 1, 2], 
#  [nan, nan, nan, 0, 1], 
#  [nan, 0, 1, 2, 3],
#  [0, 1, 2, 3, 4]]

你当然可以换 5 具有 n 确定你的最大长度。