Python Pandas replicate rows in dataframe

Question

If the dataframe looks like:

Store,Dept,Date,Weekly_Sales,IsHoliday
1,1,2010-02-05,24924.5,FALSE
1,1,2010-02-12,46039.49,TRUE
1,1,2010-02-19,41595.55,FALSE
1,1,2010-02-26,19403.54,FALSE
1,1,2010-03-05,21827.9,FALSE
1,1,2010-03-12,21043.39,FALSE
1,1,2010-03-19,22136.64,FALSE
1,1,2010-03-26,26229.21,FALSE
1,1,2010-04-02,57258.43,FALSE

And I wanna duplicate rows with IsHoliday equal to TRUE, I can do:

is_hol = df['IsHoliday'] == True
df_try = df[is_hol]
df=df.append(df_try*10)

But is there a better way to do this as I need to duplicate holiday rows 5 times, and I have to append 5 times if using the above way.

https://stackoverflow.com/questions/50788508/replicating-rows-in-pandas/50788670#50788670 — U13-Forward, Jan 01 '21 at 09:55

Karl D. · Accepted Answer · 2014-06-04T07:08:56.490

You can put df_try inside a list and then do what you have in mind:

>>> df.append([df_try]*5,ignore_index=True)

    Store  Dept       Date  Weekly_Sales IsHoliday
0       1     1 2010-02-05      24924.50     False
1       1     1 2010-02-12      46039.49      True
2       1     1 2010-02-19      41595.55     False
3       1     1 2010-02-26      19403.54     False
4       1     1 2010-03-05      21827.90     False
5       1     1 2010-03-12      21043.39     False
6       1     1 2010-03-19      22136.64     False
7       1     1 2010-03-26      26229.21     False
8       1     1 2010-04-02      57258.43     False
9       1     1 2010-02-12      46039.49      True
10      1     1 2010-02-12      46039.49      True
11      1     1 2010-02-12      46039.49      True
12      1     1 2010-02-12      46039.49      True
13      1     1 2010-02-12      46039.49      True

score 49 · Answer 2 · answered Dec 29 '16 at 22:58

Other way is using concat() function:

import pandas as pd

In [603]: df = pd.DataFrame({'col1':list("abc"),'col2':range(3)},index = range(3))

In [604]: df
Out[604]: 
  col1  col2
0    a     0
1    b     1
2    c     2

In [605]: pd.concat([df]*3, ignore_index=True) # Ignores the index
Out[605]: 
  col1  col2
0    a     0
1    b     1
2    c     2
3    a     0
4    b     1
5    c     2
6    a     0
7    b     1
8    c     2

In [606]: pd.concat([df]*3)
Out[606]: 
  col1  col2
0    a     0
1    b     1
2    c     2
0    a     0
1    b     1
2    c     2
0    a     0
1    b     1
2    c     2

This should be preferred over accepted answer, not that `append()` got deprecated. — gsamaras, Apr 08 '23 at 13:30

score 29 · Answer 3 · answered May 23 '18 at 14:23

29

This is an old question, but since it still comes up at the top of my results in Google, here's another way.

import pandas as pd
import numpy as np

df = pd.DataFrame({'col1':list("abc"),'col2':range(3)},index = range(3))

Say you want to replicate the rows where col1="b".

reps = [3 if val=="b" else 1 for val in df.col1]
df.loc[np.repeat(df.index.values, reps)]

You could replace the 3 if val=="b" else 1 in the list interpretation with another function that could return 3 if val=="b" or 4 if val=="c" and so on, so it's pretty flexible.

answered May 23 '18 at 14:23

snooze_bear

589
6
14

1

This will also replicate the index values as well, correct? – RufusVS Sep 18 '18 at 16:59
I don't know if I'm doing it wrong but this is very slow for me. – grofte Jul 25 '19 at 10:54
This more elegant way is also pretty fast for me compared with using .append(..). My use case is just to replicate a dataframe with a single row 1000x of times. – kawingkelvin Nov 18 '19 at 22:31
Not scalable: try doing `for val in df.col1` on 100 million line dataframe.. :( – Marco Nov 16 '22 at 16:13

grofte · Answer 4 · 2019-07-25T11:59:53.490

Appending and concatenating is usually slow in Pandas so I recommend just making a new list of the rows and turning that into a dataframe (unless appending a single row or concatenating a few dataframes).

import pandas as pd

df = pd.DataFrame([
[1,1,'2010-02-05',24924.5,False],
[1,1,'2010-02-12',46039.49,True],
[1,1,'2010-02-19',41595.55,False],
[1,1,'2010-02-26',19403.54,False],
[1,1,'2010-03-05',21827.9,False],
[1,1,'2010-03-12',21043.39,False],
[1,1,'2010-03-19',22136.64,False],
[1,1,'2010-03-26',26229.21,False],
[1,1,'2010-04-02',57258.43,False]
], columns=['Store','Dept','Date','Weekly_Sales','IsHoliday'])

temp_df = []
for row in df.itertuples(index=False):
    if row.IsHoliday:
        temp_df.extend([list(row)]*5)
    else:
        temp_df.append(list(row))

df = pd.DataFrame(temp_df, columns=df.columns)

For the record, it is probably faster to do an outer join but the code gets quite convoluted for a small gain. — grofte, Jul 25 '19 at 12:15

score 5 · Answer 5 · answered Jun 23 '21 at 06:34

5

You can do it in one line:

df.append([df[df['IsHoliday'] == True]] * 5, ignore_index=True)

or

df.append([df[df['IsHoliday']]] * 5, ignore_index=True)

answered Jun 23 '21 at 06:34

Mykola Zotko

15,583
3
71
73

buddemat · Answer 6 · 2021-10-03T14:24:20.557

Another alternative to append() is to first replace the values of a column by a list of entries and then explode() (either using ignore_index=True or not, depending on what you want):

df['IsHoliday'] = df['IsHoliday'].apply(lambda x: 5*[x] if (x == True) else x)

df.explode('IsHoliday', ignore_index=True)

The nice thing about this one is that you can already use the list in the apply() call to build copies of rows with modified values in a column, in case you wanted to do that later anyways...

Python Pandas replicate rows in dataframe

6 Answers6

Linked

Related