Question

我有以下包含整数和nan值的列表：

x = [8, 8, 3, 3, 2, 9, nan, nan, nan, 4, 1, 9, 4, 8, 2, nan, nan, nan, nan, 2, 2, 1, 1, 1, 2, nan, nan, nan, nan, 4, 1, 9, 5, 8, 3, 8, 8, 8, 3, 4, 2, nan]

我想加入整数以成为一个数字，并将nan值保持在其位置。另外，每个新数字都应包含6位数字。

新列表应如下所示：

x = [883329, nan, nan, nan, 419482, nan, nan, nan, nan, 221112, nan, nan, nan, nan, 419583, 888342, nan]

我尝试了以下代码，但这不是我想要的

y =''.join(str(n) for n in x)
k=list(map(''.join, zip(*[iter(y)]*6))) 
k = [883329, nannan, nan419, 482nan, nannan, nan221, 112nan, nannan, nan419, 583888, 342nan]

关于如何解决此问题的任何建议？

Answer 1

您可以使用groupby和grouper配方执行以下操作：

from numpy import nan, isnan
from itertools import groupby, zip_longest

x = [8, 8, 3, 3, 2, 9, nan, nan, nan, 4, 1, 9, 4, 8, 2, nan, nan, nan, nan, 2, 2, 1, 1, 1, 2, nan, nan, nan, nan, 4, 1,
     9, 5, 8, 3, 8, 8, 8, 3, 4, 2, nan]


def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(fillvalue=fillvalue, *args)


result = []
for k, v in groupby(x, key=isnan):
    if k:
        result.extend(list(v))
    else:
        result.extend(int(''.join(g)) for g in grouper(6, map(str, v)))

print(result)

输出

[883329, nan, nan, nan, 419482, nan, nan, nan, nan, 221112, nan, nan, nan, nan, 419583, 888342, nan]

仅连接包含NaN值的列表中的整数[Python]

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