# How to shuffle only a fraction of a column in a Pandas dataframe?

I would like to shuffle a fraction (for example 40%) of the values of a specific column in a Pandas dataframe.

How would you do it? Is there a simple idiomatic way to do that, maybe using np.random, or sklearn.utils.shuffle?

I have searched and only found answers related to shuffling the whole column, or shuffling complete rows in the df, but none related to shuffling only a fraction of a column.

I have actually managed to do it, apparently, but I get a warning, so I figure even if in this simple example it seems to work, that is probably not the way to do it.

Here's what I've done:

import pandas as pd
import numpy as np

df = pd.DataFrame({'i':range(20),
'L':[chr(97+i) for i in range(20)]
})

df['L2'] = df['L']

df.T

    0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19
i   0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19
L   a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t
L2  a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t


For now, L2 is simply a copy of column L. I keep L as the original, and I want to shuffle L2, so I can visually compare both. The i column is simply a dummy column. It's there to show that I want to keep all my columns intact, except for a fraction of L2 that I want to shuffle.

n_rows=len(df)
n_shuffle=int(n_rows*0.4)
n_rows, n_shuffle

(20, 8)

pick_rows=np.random.permutation(list(range(n_rows)))[0:n_shuffle]
pick_rows

array([ 3,  0, 11, 16, 14,  4,  8, 12])

shuffled_values=np.random.permutation(df['L2'][pick_rows])
shuffled_values

array(['l', 'e', 'd', 'q', 'o', 'i', 'm', 'a'], dtype=object)

df['L2'][pick_rows]=shuffled_values


I get this warning:

C:\Users\adumont\.conda\envs\fastai-cpu\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
"""Entry point for launching an IPython kernel.

df.T


I get the following, which is what I expected (40% of the values of L2 are now shuffled):

    0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19
i   0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19
L   a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t
L2  e   b   c   l   i   f   g   h   m   j   k   d   a   n   o   p   q   r   s   t


You can see the notebook here (it's rendered better on nbviewer than here): https://nbviewer.jupyter.org/gist/adumont/bc2bac1b6cf7ba547e7ba6a19c01adb6

Thanks in advance.

## 1 Answer

I don't think there is any idiomatic way of doing this since it's quite unusual operation, normally the whole row or column should be shuffled. What you are doing looks like a good approach.

The error SettingWithCopyWarning you get is a common warning that you could be operating on a copy of the original data and not a view (the origianl). For more information I would recommend checking the answers here: https://stackoverflow.com/questions/20625582/how-to-deal-with-settingwithcopywarning-in-pandas.

To avoid the error and make the code more compact you could do it as follows:

import random

fraction = 0.4
n_rows = len(df)
n_shuffle=int(n_rows*fraction)

pick_rows = random.sample(range(1, n_rows), n_shuffle)

df.loc[pick_rows, 'L2'] = np.random.permutation(df.loc[pick_rows, 'L2'])


Note that the use of loc here will make sure that no copy is created and everything is done on the origianl dataframe (i.e. this will not give the SettingWithCopyWarning warning).