user248237dfsf user248237dfsf - 3 months ago 61
Python Question

shuffling/permutating a DataFrame in pandas

What's a simple and efficient way to shuffle a dataframe in pandas, by rows or by columns? I.e. how to write a function

shuffle(df, n, axis=0)
that takes a dataframe, a number of shuffles
n
, and an axis (
axis=0
is rows,
axis=1
is columns) and returns a copy of the dataframe that has been shuffled
n
times.

Edit: key is to do this without destroying the row/column labels of the dataframe. If you just shuffle
df.index
that loses all that information. I want the resulting
df
to be the same as the original except with the order of rows or order of columns different.

Edit2: My question was unclear. When I say shuffle the rows, I mean shuffle each row independently. So if you have two columns
a
and
b
, I want each row shuffled on its own, so that you don't have the same associations between
a
and
b
as you do if you just re-order each row as a whole. Something like:

for 1...n:
for each col in df: shuffle column
return new_df


But hopefully more efficient than naive looping. This does not work for me:

def shuffle(df, n, axis=0):
shuffled_df = df.copy()
for k in range(n):
shuffled_df.apply(np.random.shuffle(shuffled_df.values),axis=axis)
return shuffled_df

df = pandas.DataFrame({'A':range(10), 'B':range(10)})
shuffle(df, 5)

Answer
In [16]: def shuffle(df, n=1, axis=0):     
    ...:     df = df.copy()
    ...:     for _ in range(n):
    ...:         df.apply(np.random.shuffle, axis=axis)
    ...:     return df
    ...:     

In [17]: df = pd.DataFrame({'A':range(10), 'B':range(10)})

In [18]: shuffle(df)

In [19]: df
Out[19]: 
   A  B
0  8  5
1  1  7
2  7  3
3  6  2
4  3  4
5  0  1
6  9  0
7  4  6
8  2  8
9  5  9