Question

我有一个pandas数据框sample，其中一个名为PR的列应用了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

我做错了什么？

Answer 1

您需要mask：

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

loc和boolean indexing的另一种解决方案：

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

样品：

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：

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

Answer 2

您需要在lambda函数中添加else，因为您要告诉您在满足条件（此处x <90）的情况下该怎么做，但您没有告诉在不满足条件的情况下该怎么做。 / p>

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

如何将lambda函数正确应用到pandas数据框列中

2 个答案: