如何将列表列转换为非嵌套列表? [英] How to convert list column to a non-nested list?

问题描述

当列元素为列表时，如何将列转换为非嵌套列表?

How to convert a column to a non-nested list while the column elements are list?

例如，该列就像

column
[1, 2, 3]
[1, 2]

我想最后关注

[1,2,3,1,2]

但是现在有了column.tolist()，我会得到

But now with column.tolist(), I will get

[[1,2,3],[1,2]]

感谢您的帮助.我的目的是找到最简单(优雅)和最有效的方法来执行此操作.现在，我使用@jezrael方法.

Thanks for help. My intention is to find the most simple (elegant) and efficient method to do this. Now I use @jezrael method.

from itertools import chain
output = list(chain.from_iterable(df[column])

@piRSquared提供了最简单的方法，但速度可能较慢.

The simplest method is provided by @piRSquared, but maybe slower.

output = df[column].values.sum()

推荐答案

您可以使用 numpy.concatenate :

You can use numpy.concatenate:

print (np.concatenate(df['column'].values).tolist())
[1, 2, 3, 1, 2]

或者:

from  itertools import chain
print (list(chain.from_iterable(df['column'])))
[1, 2, 3, 1, 2]

另一种解决方案，谢谢 juanpa .arrivillaga :

Another solution, thanks juanpa.arrivillaga:

print ([item for sublist in df['column'] for item in sublist])
[1, 2, 3, 1, 2]

时间:

df = pd.DataFrame({'column':[[1,2,3], [1,2]]})
df = pd.concat([df]*10000).reset_index(drop=True)
print (df)

In [77]: %timeit (np.concatenate(df['column'].values).tolist())
10 loops, best of 3: 22.7 ms per loop

In [78]: %timeit (list(chain.from_iterable(df['column'])))
1000 loops, best of 3: 1.44 ms per loop

In [79]: %timeit ([item for sublist in df['column'] for item in sublist])
100 loops, best of 3: 2.31 ms per loop

In [80]: %timeit df.column.sum()
1 loop, best of 3: 1.34 s per loop

这篇关于如何将列表列转换为非嵌套列表?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！