第一种方法更快,因为它是矢量化的:
df=pd.DataFrame(columns=['A','B','C'],data=[[1,2,3],[4,5,6],[7,8,9]])
print (df)
df = pd.concat([df]*10000).reset_index(drop=True)
df['D1']=0.5*df['A']+0.3*df['B']+0.2*df['C']
df['D']=df.apply(lambda row: 0.5*row['A']+0.3*row['B']+0.2*row['C'], axis=1)
print (df)
In [54]: %timeit df['D2']=df['A'].mul(0.5)+df['B'].mul(0.3)+df['C'].mul(0.2)
The slowest run took 10.84 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 950 µs per loop
In [55]: %timeit df['D1']=0.5*df['A']+0.3*df['B']+0.2*df['C']
The slowest run took 4.76 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.2 ms per loop
In [56]: %timeit df['D']=df.apply(lambda row: 0.5*row['A']+0.3*row['B']+0.2*row['C'], axis=1)
1 loop, best of 3: 928 ms per loop
在另一个大小为1.5M
的DataFrame
测试中,apply
方法非常慢:
df = pd.concat([df]*500000).reset_index(drop=True)
In [62]: %timeit df['D2']=df['A'].mul(0.5)+df['B'].mul(0.3)+df['C'].mul(0.2)
10 loops, best of 3: 34.8 ms per loop
In [63]: %timeit df['D1']=0.5*df['A']+0.3*df['B']+0.2*df['C']
10 loops, best of 3: 31.5 ms per loop
In [64]: %timeit df['D']=df.apply(lambda row: 0.5*row['A']+0.3*row['B']+0.2*row['C'], axis=1)
1 loop, best of 3: 47.3 s per loop