Python Pandas 在 dataframe 中复制行

小开

最佳答案

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

小开

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

小开

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.

小开

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)

小开

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)

小开

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...