Shuffle DataFrame rows

The idiomatic way to do this with Pandas is to use the .sample method of your data frame to sample all rows without replacement:

df.sample(frac=1)

The frac keyword argument specifies the fraction of rows to return in the random sample, so frac=1 means to return all rows (in random order).


Note: If you wish to shuffle your dataframe in-place and reset the index, you could do e.g.

df = df.sample(frac=1).reset_index(drop=True)

Here, specifying drop=True prevents .reset_index from creating a column containing the old index entries.

Follow-up note: Although it may not look like the above operation is in-place, python/pandas is smart enough not to do another malloc for the shuffled object. That is, even though the reference object has changed (by which I mean id(df_old) is not the same as id(df_new)), the underlying C object is still the same. To show that this is indeed the case, you could run a simple memory profiler:

$ python3 -m memory_profiler .\test.py
Filename: .\test.py

Line #    Mem usage    Increment   Line Contents
================================================
     5     68.5 MiB     68.5 MiB   @profile
     6                             def shuffle():
     7    847.8 MiB    779.3 MiB       df = pd.DataFrame(np.random.randn(100, 1000000))
     8    847.9 MiB      0.1 MiB       df = df.sample(frac=1).reset_index(drop=True)


You can simply use sklearn for this

from sklearn.utils import shuffle
df = shuffle(df)

TL;DR: np.random.shuffle(ndarray) can do the job.
So, in your case

np.random.shuffle(DataFrame.values)

DataFrame, under the hood, uses NumPy ndarray as a data holder. (You can check from DataFrame source code)

So if you use np.random.shuffle(), it would shuffle the array along the first axis of a multi-dimensional array. But the index of the DataFrame remains unshuffled.

Though, there are some points to consider.

  • function returns none. In case you want to keep a copy of the original object, you have to do so before you pass to the function.
  • sklearn.utils.shuffle(), as user tj89 suggested, can designate random_state along with another option to control output. You may want that for dev purposes.
  • sklearn.utils.shuffle() is faster. But WILL SHUFFLE the axis info(index, column) of the DataFrame along with the ndarray it contains.

Benchmark result

between sklearn.utils.shuffle() and np.random.shuffle().

ndarray

nd = sklearn.utils.shuffle(nd)

0.10793248389381915 sec. 8x faster

np.random.shuffle(nd)

0.8897626010002568 sec

DataFrame

df = sklearn.utils.shuffle(df)

0.3183923360193148 sec. 3x faster

np.random.shuffle(df.values)

0.9357550159329548 sec

Conclusion: If it is okay to axis info(index, column) to be shuffled along with ndarray, use sklearn.utils.shuffle(). Otherwise, use np.random.shuffle()

used code

import timeit
setup = '''
import numpy as np
import pandas as pd
import sklearn
nd = np.random.random((1000, 100))
df = pd.DataFrame(nd)
'''

timeit.timeit('nd = sklearn.utils.shuffle(nd)', setup=setup, number=1000)
timeit.timeit('np.random.shuffle(nd)', setup=setup, number=1000)
timeit.timeit('df = sklearn.utils.shuffle(df)', setup=setup, number=1000)
timeit.timeit('np.random.shuffle(df.values)', setup=setup, number=1000)

pythonbenchmarking


You can shuffle the rows of a data frame by indexing with a shuffled index. For this, you can eg use np.random.permutation (but np.random.choice is also a possibility):

In [12]: df = pd.read_csv(StringIO(s), sep="\s+")

In [13]: df
Out[13]: 
    Col1  Col2  Col3  Type
0      1     2     3     1
1      4     5     6     1
20     7     8     9     2
21    10    11    12     2
45    13    14    15     3
46    16    17    18     3

In [14]: df.iloc[np.random.permutation(len(df))]
Out[14]: 
    Col1  Col2  Col3  Type
46    16    17    18     3
45    13    14    15     3
20     7     8     9     2
0      1     2     3     1
1      4     5     6     1
21    10    11    12     2

If you want to keep the index numbered from 1, 2, .., n as in your example, you can simply reset the index: df_shuffled.reset_index(drop=True)