如何将 lambda 函数正确应用到 Pandas 数据框列中
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How to properly apply a lambda function into a pandas data frame column
提问by Amani
I have a pandas data frame, sample, with one of the columns called PRto which am applying a lambda function as follows:
我有一个 Pandas 数据框，sample其中一列被调用PR，正在应用 lambda 函数，如下所示：
sample['PR'] = sample['PR'].apply(lambda x: NaN if x < 90)
I then get the following syntax error message:
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
What am I doing wrong?
我究竟做错了什么？
回答by jezrael
You need mask:
你需要mask：
sample['PR'] = sample['PR'].mask(sample['PR'] < 90, np.nan)
Another solution with locand boolean indexing:
使用loc和的另一种解决方案boolean indexing：
sample.loc[sample['PR'] < 90, 'PR'] = np.nan
Sample:
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
EDIT:
编辑：
Solution with apply:
解决方案apply：
sample['PR'] = sample['PR'].apply(lambda x: np.nan if x < 90 else x)
Timingslen(df)=300k:
时间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
回答by kali prasad deverasetti
You need to add elsein your lambda function. Because you are telling what to do in case your condition(here x < 90) is met, but you are not telling what to do in case the condition is not met.
您需要添加elselambda 函数。因为您是在告诉在满足条件（此处 x < 90）的情况下该做什么，但您没有告诉在不满足条件的情况下该做什么。
sample['PR'] = sample['PR'].apply(lambda x: 'NaN' if x < 90 else x)
