使用 lambda 以字符串开头时替换 DataFrame 列中的值 [英] Replace values in DataFrame column when they start with string using lambda

问题描述

我有一个数据框:

将pandas导入为pd将 numpy 导入为 npx = {'值':['测试'，'XXX123'，'XXX456'，'测试']}df = pd.DataFrame(x)

我想使用 lambda 用 np.nan 替换以 XXX 开头的值.

我尝试了很多关于替换、应用和映射的事情，我能做的最好的是 False, True, True, False.

以下有效，但我想知道更好的方法，我认为应用、替换和 lambda 可能是更好的方法.

df.Value.loc[df.Value.str.startswith('XXX', na=False)] = np.nan

解决方案

使用

I have a DataFrame:

import pandas as pd
import numpy as np
x = {'Value': ['Test', 'XXX123', 'XXX456', 'Test']}
df = pd.DataFrame(x)

I want to replace the values starting with XXX with np.nan using lambda.

I have tried many things with replace, apply and map and the best I have been able to do is False, True, True, False.

The below works, but I would like to know a better way to do it and I think the apply, replace and a lambda is probably a better way to do it.

df.Value.loc[df.Value.str.startswith('XXX', na=False)] = np.nan

解决方案

use the apply method

In [80]: x = {'Value': ['Test', 'XXX123', 'XXX456', 'Test']}
In [81]: df = pd.DataFrame(x)
In [82]: df.Value.apply(lambda x: np.nan if x.startswith('XXX') else x)
Out[82]:
0    Test
1     NaN
2     NaN
3    Test
Name: Value, dtype: object

Performance Comparision of apply, where, loc

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