Reputation: 725
I have a python dataframe
with 1.5 million rows and 8 columns. I want combine few columns and create a new column. I know how to do this but wanted to know which one is faster and efficient. I am reproducing my code here
import pandas as pd
import numpy as np
df=pd.Dataframe(columns=['A','B','C'],data=[[1,2,3],[4,5,6],[7,8,9]])
Now here is what I want to achieve
df['D']=0.5*df['A']+0.3*df['B']+0.2*df['C']
The other alternative is to use the apply functionality
of pandas
df['D']=df.apply(lambda row: 0.5*row['A']+0.3*row['B']+0.2*row['C'])
I wanted to know which method takes less time when we have 1.5 millon rows and have to combine 8 columns
Upvotes: 3
Views: 3489
Reputation: 294508
Using @jezrael's setup
df=pd.DataFrame(columns=['A','B','C'],data=[[1,2,3],[4,5,6],[7,8,9]])
df = pd.concat([df]*30000).reset_index(drop=True)
Far more efficient to use a dot
product.
np.array([[.5, .3, .2]]).dot(df.values.T).T
Upvotes: 3
Reputation: 863291
First method is faster, because is vectorized:
df=pd.DataFrame(columns=['A','B','C'],data=[[1,2,3],[4,5,6],[7,8,9]])
print (df)
#[30000 rows x 3 columns]
df = pd.concat([df]*10000).reset_index(drop=True)
df['D1']=0.5*df['A']+0.3*df['B']+0.2*df['C']
#similar timings with mul function
#df['D1']=df['A'].mul(0.5)+df['B'].mul(0.3)+df['C'].mul(0.2)
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
Another testing in 1.5M
size DataFrame
, apply
method is very slow:
#[1500000 rows x 6 columns]
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
Upvotes: 3