如果单个列具有NaN，如果存在NaN，如何更改整行

Question

我有一个这样的DataFrame

                    gauge       satellite
1979-06-23 18:00:00 6.700000    2.484378
1979-06-27 03:00:00 NaN         8.891460
1979-06-27 06:00:00 1.833333    4.053460
1979-06-27 09:00:00 NaN         2.876649
1979-07-31 18:00:00 6.066667    1.438324

我想要这样一个数据帧

                    gauge       satellite
1979-06-23 18:00:00 6.700000    2.484378
1979-06-27 03:00:00 NaN         NaN
1979-06-27 06:00:00 1.833333    4.053460
1979-06-27 09:00:00 NaN         NaN
1979-07-31 18:00:00 6.066667    1.438324

Answer 1

我将做什么reindex

df.dropna().reindex(df.index)

Answer 2

mask：

df.mask(df.gauge.isna())

---

                        gauge  satellite
1979-06-23 18:00:00  6.700000   2.484378
1979-06-27 03:00:00       NaN        NaN
1979-06-27 06:00:00  1.833333   4.053460
1979-06-27 09:00:00       NaN        NaN
1979-07-31 18:00:00  6.066667   1.438324

Answer 3

使用np.where添加nan

import numpy as np
df['satellite'] = np.where(df['gauge'].isnull(),np.nan,df['satellite'])

第二个解决方案

使用.loc和isnull

df.loc[df['guage'].isnull(),'satellite'] = np.nan

Answer 4

您可以使用np.where：

df['satellite'] = np.where(df['gauge'].isna(), np.NaN, df['satellite'])
df['gauge'] = np.where(df['satellite'].isna(), np.NaN, df['gauge'])

Answer 5

您需要确定行中是否有np.nan。 .any(1)为您提供了连续的遮罩。

df.loc[df.isna().any(1)] = np.nan

输出：

                        gauge       satellite
1979-06-23  18:00:00    6.700000    2.484378
1979-06-27  03:00:00    NaN         NaN
1979-06-27  06:00:00    1.833333    4.053460
1979-06-27  09:00:00    NaN         NaN
1979-07-31  18:00:00    6.066667    1.438324