本文介绍了使用条件语句从pandas df列中减去标量会导致ValueError:Series的真值不明确的处理方法,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
我正在尝试执行:
if df_trades.loc[:, 'CASH'] != 0: df_trades.loc[:, 'CASH'] -= commission
,然后我得到错误. df_trades.loc[:, 'CASH']
是一列浮点数.我想从该列的每个条目中减去标量commission
.
and then I get the error. df_trades.loc[:, 'CASH']
is a column of floats. I want to subtract the scalar commission
from each entry in that column.
例如,df_trades.loc[:, 'CASH']
打印出
2011-01-10 -2557.0000
2011-01-11 0.0000
2011-01-12 0.0000
2011-01-13 -2581.0000
如果commission
是1
,我需要结果:
2011-01-10 -2558.0000
2011-01-11 0.0000
2011-01-12 0.0000
2011-01-13 -2582.0000
推荐答案
使用np.where
commission = -1
df['CASH'] = np.where(df['CASH'] != 0, df['CASH'] + commission , df['CASH'])
或df.where
即
df['CASH'] = df['CASH'].where(df['CASH'] == 0,df['CASH']+commission)
或df.mask
df['CASH'] = df['CASH'].mask(df['CASH'] != 0 ,df['CASH']+commission)
Date
2011-01-10 -2558.0
2011-01-11 0.0
2011-01-12 0.0
2011-01-13 -2582.0
Name: CASH, dtype: float64
%%timeit
commission = -1
df['CASH'] = np.where(df['CASH'] != 0, df['CASH'] + commission , df['CASH'])
1000 loops, best of 3: 750 µs per loop
%%timeit
df['CASH'].mask(df['CASH'] != 0 ,df['CASH']+commission)
1000 loops, best of 3: 1.45 ms per loop
%%timeit
df['CASH'].where(df['CASH'] == 0,df['CASH']+commission)
1000 loops, best of 3: 1.55 ms per loop
%%timeit
df.loc[df['CASH'] != 0, 'CASH'] += commission
100 loops, best of 3: 2.37 ms per loop
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