pandas - 如何将 lambda 函数正确应用到 Pandas 数据框列中

Question

我有一个 Pandas 数据框，sample ，其中一列名为 PR我正在应用 lambda 函数，如下所示:

sample['PR'] = sample['PR'].apply(lambda x: NaN if x < 90)

然后我收到以下语法错误消息:

sample['PR'] = sample['PR'].apply(lambda x: NaN if x < 90)
                                                         ^
SyntaxError: invalid syntax

我究竟做错了什么？

Answer 1

您需要 mask :

sample['PR'] = sample['PR'].mask(sample['PR'] < 90, np.nan)

另一个解决方案 loc 和 boolean indexing :

sample.loc[sample['PR'] < 90, 'PR'] = np.nan

样本:

import pandas as pd
import numpy as np

sample = pd.DataFrame({'PR':[10,100,40] })
print (sample)
    PR
0   10
1  100
2   40

sample['PR'] = sample['PR'].mask(sample['PR'] < 90, np.nan)
print (sample)
      PR
0    NaN
1  100.0
2    NaN

sample.loc[sample['PR'] < 90, 'PR'] = np.nan
print (sample)
      PR
0    NaN
1  100.0
2    NaN

编辑:

使用 apply 的解决方案:

sample['PR'] = sample['PR'].apply(lambda x: np.nan if x < 90 else x)

计时 len(df)=300k :

sample = pd.concat([sample]*100000).reset_index(drop=True)

In [853]: %timeit sample['PR'].apply(lambda x: np.nan if x < 90 else x)
10 loops, best of 3: 102 ms per loop

In [854]: %timeit sample['PR'].mask(sample['PR'] < 90, np.nan)
The slowest run took 4.28 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 3.71 ms per loop