python - 使用条件语句从 pandas df 列中减去标量给出 ValueError: The truth value of a Series is ambiguous

Question

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commission从该列中的每个条目中减去标量。

例如，df_trades.loc[:, 'CASH']打印出来

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，我想要结果：

2011-01-10   -2558.0000
2011-01-11       0.0000
2011-01-12       0.0000
2011-01-13   -2582.0000

Answer 1

采用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)

或者df.mask

df['CASH'] = df['CASH'].mask(df['CASH'] != 0 ,df['CASH']+commission)

日期
2011-01-10 -2558.0
2011-01-11 0.0
2011-01-12 0.0
2011-01-13 -2582.0
名称：现金，数据类型：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

Answer 2

这应该这样做：

df.loc[df['CASH'] != 0, 'CASH'] -= 1