用 NaN 值迭代连接 pandas 中的列答案

【问题标题】：Iteratively concatenate columns in pandas with NaN values用 NaN 值迭代连接 pandas 中的列
【发布时间】：2017-10-17 11:17:53
【问题描述】：

我有一个pandas.DataFrame 数据框：

import pandas as pd

df = pd.DataFrame({"x": ["hello there you can go home now", "why should she care", "please sort me appropriately"], 
    "y": [np.nan, "finally we were able to go home", "but what about meeeeeeeeeee"],
    "z": ["", "alright we are going home now", "ok fine shut up already"]})

cols = ["x", "y", "z"]

我想迭代地连接这些列，而不是像这样写：

df["concat"] = df["x"].str.cat(df["y"], sep = " ").str.cat(df["z"], sep = " ")

我知道将三列放在一起似乎微不足道，但我实际上有 30 列。所以，我想做类似的事情：

df["concat"] = df[cols[0]]
for i in range(1, len(cols)):
    df["concat"] = df["concat"].str.cat(df[cols[i]], sep = " ")

现在，最初的 df["concat"] = df[cols[0]] 行工作正常，但是位置 df.loc[1, "y"] 中的 NaN 值会混淆连接。最终，由于这个空值，整个1st 行在df["concat"] 中最终成为NaN。我怎样才能解决这个问题？我需要指定pd.Series.str.cat 的选项吗？

【问题讨论】：

标签： python pandas

【解决方案1】：

选项 1

pd.Series(df.fillna('').values.tolist()).str.join(' ')

0                    hello there you can go home now  
1    why should she care finally we were able to go...
2    please sort me appropriately but what about me...
dtype: object

选项 2

df.fillna('').add(' ').sum(1).str.strip()

0                      hello there you can go home now
1    why should she care finally we were able to go...
2    please sort me appropriately but what about me...
dtype: object

【讨论】：

【解决方案2】：

选项 3

In [3061]: df.apply(lambda x: x.str.cat(sep=''), axis=1)
Out[3061]:
0                      hello there you can go home now
1    why should she carefinally we were able to go ...
2    please sort me appropriatelybut what about mee...
dtype: object

【讨论】：