Using conditional statement to subtract scalar from pandas df column gives 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

然后我得到了错误。 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。

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

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

Answer 2

应该这样做：

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

Using conditional statement to subtract scalar from pandas df column gives ValueError: The truth value of a Series is ambiguous

Using conditional statement to subtract scalar from pandas df column gives ValueError: The truth value of a Series is ambiguous

python

scalar

truthiness

pandas