用另一个数据框的值替换熊猫数据框的多个值的最快方法

Question

I am trying to replace multiple rows of pandas dataframe, with values from another dataframe.

Supposed I have 10,000 rows of customer_id in my dataframe df1 and I want to replace these customer_id with 3,000 values from df2 .

For the sake of illustration, let's generate the dataframes (below).

Say these 10 rows in df1 represent 10,000 rows, and the 3 rows from df2 represent 3,000 values.

import numpy as np
import pandas as pd
np.random.seed(42)

# Create df1 with unique values
arr1 = np.arange(100,200,10)
np.random.shuffle(arr1)
df1 = pd.DataFrame(data=arr1, 
                   columns=['customer_id'])

# Create df2 for new unique_values
df2 = pd.DataFrame(data = [1800, 1100, 1500],
                   index = [180, 110, 150], # this is customer_id column on df1
                   columns = ['customer_id_new'])

I want to replace 180 with 1800, 110 with 1100, and 150 with 1500.

I know we can do below ...

# Replace multiple values
replace_values = {180 : 1800, 110 : 1100, 150 : 1500 }                                                                                          
df1_replaced = df1.replace({'customer_id': replace_values})

And it works fine if I only have a few lines...

But if I have thousands of lines that I need to replace, how could I do this without typing out what values I want to change one at a time?

EDIT: To clarify, I don't need to use replace . Anything that could replace those values in df1 from values in df2 in the fastest most efficient way is ok.

Answer 1

df1['customer_id'] = df1['customer_id'].replace(df2['customer_id_new'])

或者，您可以就地进行。

df1['customer_id'].replace(df2['customer_id_new'], inplace=True)

Answer 2

You can try this, using map with a pd.Series:

 df1['customer_id'] = df1['customer_id'].map(df2.squeeze()).fillna(df1['customer_id'])

or

df1['customer_id'] = df1['customer_id'].map(df2['customer_id_new']).fillna(df1['customer_id'])

Output:

   customer_id
0       1800.0
1       1100.0
2       1500.0
3        100.0
4        170.0
5        120.0
6        190.0
7        140.0
8        130.0
9        160.0

Answer 3

Going with your original method using replace , you can simplify it with to_dict to create your mapping dictionary without having to do it manually:

df1["customer_id"] = df1["customer_id"].replace(df2["customer_id_new"].to_dict())

>>> df1
   customer_id
0         1800
1         1100
2         1500
3          100
4          170
5          120
6          190
7          140
8          130
9          160

Answer 4

In my opinion, apart from trying out useful answers mentioned above, you may try parallelising your data-frame in-case you have multi-core processor.

For example:

import pandas as pd, numpy as np, seaborn as sns
from multiprocessing import Pool

num_partitions = 10 #number of partitions to split data-frame
num_cores = 4 #number of cores on your machine

iris = pd.DataFrame(sns.load_dataset('iris'))
def parallelize_dataframe(df, func):
   df_split = np.array_split(df, num_partitions)
   pool = Pool(num_cores)
   df = pd.concat(pool.map(func, df_split))
   pool.close()
   pool.join()
   return df

In place of 'func' parameter, you may pass your replace method. Please let me know if it helps. In case of any error, do comment.

Thanks!