使用條件語句來減去大熊貓DF列標給ValueError異常：一個系列的真值是不確定

Question

我試圖執行：

if df_trades.loc[:, 'CASH'] != 0: df_trades.loc[:, 'CASH'] -= commission 

，然後我得到的錯誤。 df_trades.loc[:, 'CASH']是一列浮標。我想從該列中的每個條目中減去標量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，我想要的結果：

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

這應做到：

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