Build a dataframe based on one repetetive row

user_18601215 0
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I have a row and I want to build a dataframe based on this row. For example, I have a row with 5 value. I want to build a df with 4 rows and 5 column. here is the desired output for me:

df = pd.DataFrame()
df['a']=[1]
df['b']=[2]
df['c']=[3]
df['d']=[1]
df['e']=[2]

Output:

      a     b     c     d     e
0     1     2     3     1     2
1     1     2     3     1     2
2     1     2     3     1     2
3     1     2     3     1     2

Can you help me with it?

pandas dataframe python
2 Answers

You can repeat the index:

out = (df.loc[df.index.repeat(4)]
         .reset_index(drop=True)
      )

Output:

   a  b  c  d  e
0  1  2  3  1  2
1  1  2  3  1  2
2  1  2  3  1  2
3  1  2  3  1  2

Used input:

d = {'a': [1], 'b': [2], 'c': [3], 'd': [1], 'e': [2]}
df = pd.DataFrame(d)

rows and columns

Using repeat:

df.loc[df.index.repeat(4), df.columns.repeat(2)].reset_index(drop=True)

Output:

   a  a  b  b  c  c  d  d  e  e
0  1  1  2  2  3  3  1  1  2  2
1  1  1  2  2  3  3  1  1  2  2
2  1  1  2  2  3  3  1  1  2  2
3  1  1  2  2  3  3  1  1  2  2

Using numpy.tile for a different order:

import numpy as np
df.loc[df.index.repeat(4), np.tile(df.columns, 2)].reset_index(drop=True)

Output:

   a  b  c  d  e  a  b  c  d  e
0  1  2  3  1  2  1  2  3  1  2
1  1  2  3  1  2  1  2  3  1  2
2  1  2  3  1  2  1  2  3  1  2
3  1  2  3  1  2  1  2  3  1  2
user_163434640

Use pandas.concat

n = 4 
res = pd.concat([df]*n, ignore_index=True)

Example:

>>> df = pd.DataFrame([[1, 2, 3, 1, 2]], columns=list('abcde'))
>>> df

   a  b  c  d  e
0  1  2  3  1  2

>>> pd.concat([df]*4, ignore_index=True)

   a  b  c  d  e
0  1  2  3  1  2
1  1  2  3  1  2
2  1  2  3  1  2
3  1  2  3  1  2
user_171206920
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