在每行中应用逻辑比应用lambda更快吗?

使用 np.where()

df["Valindex"] = np.where(df["RB"].eq(0), df["Valindex0"], df["Valindex1"])

np.select()

conditions = [df["RB"].eq(0), df["RB"].eq(1)]
labels = [df["Valindex0"], df["Valindex1"]]
df["Valindex"] = np.select(conditions, labels)

产量 df

    contract    Valindex0   RB  Valindex1   Valindex
2   A00118      51          0   50          51
3   A00118      42          1   47          47
4   A00118      44          1   47          47

你可以试试 Series.where

df["Valindex"] = df["Valindex0"].where(df["RB"].eq(0), df["Valindex1"])


  contract  Valindex0  RB  Valindex1  Valindex
2   A00118         51   0         50        51
3   A00118         42   1         47        47
4   A00118         44   1         47        47

让我们试试看

df['Valindex'] = df['Valindex0']

df.loc[df.RB==1,'Valindex'] = df['Valindex1']

运行一组方法并在1000行上对每个方法计时,似乎Anurag建议的方法3效果最好。每次跑步的时间会有所不同,但排名不会改变。

#Method 1 (original) - 0.0168s
df['Valindex'] = df.apply(lambda df: df["Valindex" + str(df["RB"])], axis=1)

#Method 2 - 0.0015s
df['Valindex'] = (df['RB'] == 0 ) * df['Valindex0'] + df['RB'] * df['Valindex1']

#Method 3 - 0.0009s
df["Valindex"] = np.where(df["RB"].eq(0), df["Valindex0"], df["Valindex1"])

#Method 4 - 0.0014s
df["Valindex"] = df["Valindex0"].where(df["RB"].eq(0), df["Valindex1"])

#Method 5 - 0.0022s
df['Valindex'] = df['Valindex0']
df.loc[df.RB==1,'Valindex'] = df['Valindex1']