Question

我需要在Python中用以下列表的值替换空值。如果列表中的所有项目都不为空，则其值将相同。

列表示例

[[1538140080000, None], [1538140140000, None], [1538140200000, None], [1538140260000, None], [1538140320000, None], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

我希望得到的输出是

[[1538140080000, 92881926.0], [1538140140000, 92881926.0], [1538140200000, 92881926.0], [1538140260000, 92881926.0], [1538140320000, 92881926.0], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

在列表中，（键，值）中的任何一个都将具有所需的值（除null之外），我希望将其替换为列表中的所有键。

Answer 1

let predicate = NSPredicate(format: " %K != %@ AND %K != nil", remoteAttributes.lineOwner, "")

输出

l = [[1538140080000, None], [1538140140000, None], [1538140200000, None], [1538140260000, None], [1538140320000, None], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

replacement = next(filter(None, list(zip(*l))[1]))

for sl in l:
     for i, x in enumerate(sl):
         if x is None:
             sl[i] = replacement
print(l)

Answer 2

列表理解可以轻松完成任务

l = [[q if q else 0 for q in i] for i in l]

输出

[[1538140080000, 0], [1538140140000, 0], [1538140200000, 0], [1538140260000, 0], [1538140320000, 0], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

已更新

var = [[i[q] for i in l if i[q]] for q in range(len(l[0]))]
l  = [[q if q else var[w][0] for w,q in enumerate(i)] for i in l]

输出

[[1538140080000, 92881926.0], [1538140140000, 92881926.0], [1538140200000, 92881926.0], [1538140260000, 92881926.0], [1538140320000, 92881926.0], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

Answer 3

如果您进行硬编码替换，则可能是单行。

此外，如果您允许懒惰求值，则可以消除list（...）调用。

replacement = 0
l = list( map(lambda (a,b): [a,replacement if b == None else b], l) )

Answer 4

如果您愿意使用第三方库，Pandas提供了矢量化方法：

import pandas as pd

L = [[1538140080000, None], [1538140140000, None], [1538140200000, None],
     [1538140260000, None], [1538140320000, None], [1538140380000, 92881926.0],
     [1538140440000, 92881926.0]]

res = pd.DataFrame(L).bfill().ffill().values.tolist()

print(res)

[[1538140080000.0, 92881926.0], [1538140140000.0, 92881926.0], [1538140200000.0, 92881926.0],
 [1538140260000.0, 92881926.0], [1538140320000.0, 92881926.0], [1538140380000.0, 92881926.0],
 [1538140440000.0, 92881926.0]]

Answer 5

获取不是None的值，因为它将永远只是一个值，我们可以使用set

将l中的项目重新定义为[i[0], *value]

value = set([y for x, y in l if y != None])
l = [[i[0], *value] for i in l]

[[1538140080000, 92881926.0], [1538140140000, 92881926.0], [1538140200000, 92881926.0], [1538140260000, 92881926.0], [1538140320000, 92881926.0], [1538140380000, 92881926.0], [1538140440000, 92881926.0]]

Python循环迭代，替换列表的时间序列中的空值

5 个答案: