如何在Python中基于两个条件更改列的值 | 如何在Python中基于两个条件更改列的值

本文介绍了如何在Python中基于两个条件更改列的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个数据集，其中包含游戏时间和事件时间.

I have a dataset where I have the time in a game and the time of an event.

  EVENT     GAME
  0:34      0:43
  NaN       0:23
  2:34      3:43
  NaN       4:50

我要替换GAME<的EVENT栏中的NaN；按GAME列中的值0.24.

I want to replace the NaN in the EVENT column where GAME < 0.24 by the value in the GAME column.

 df['EVENT'][(df['GAME'] < '0:24') & (df['EVENT'] == 'NaN')] = df['GAME']

我已经尝试过了，但是没用.抱歉，如果很明显.我是Python的新手.

I have tried this but it dosen't work. Sorry if it is obvious. I am new to Python.

推荐答案

您可以使用 isnull 用于检查NaN:

df.loc[(df['GAME'] < '0:24') & (df['EVENT'].isnull()), 'EVENT'] = df['GAME']
print (df)

  EVENT  GAME
0  0:34  0:43
1  0:23  0:23
2  2:34  3:43
3   NaN  4:50

使用 mask 的另一种解决方案:

Another solution with mask:

mask = (df['GAME'] < '0:24') & (df['EVENT'].isnull())
df['EVENT'] = df['EVENT'].mask(mask, df['GAME'])
print (df)
  EVENT  GAME
0  0:34  0:43
1  0:23  0:23
2  2:34  3:43
3   NaN  4:50

或 numpy.where :

df['EVENT'] = np.where(mask, df['GAME'], df['EVENT'])
print (df)
  EVENT  GAME
0  0:34  0:43
1  0:23  0:23
2  2:34  3:43
3   NaN  4:50

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