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)

---

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)

---

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

---

Following could be one of ways:

dataframe = dataframe.sample(frac=1, random_state=42).reset_index(drop=True)

where

frac=1 means all rows of a data frame

random_state=42 means keeping the same order in each execution

reset_index(drop=True) means reinitialize index for randomized dataframe

---

(I don't have enough reputation to comment this on the top post, so I hope someone else can do that for me.) There was a concern raised that the first method:

df.sample(frac=1)

It makes a deep copy or just changed the dataframe. I ran the following code:

print(hex(id(df)))
print(hex(id(df.sample(frac=1))))
print(hex(id(df.sample(frac=1).reset_index(drop=True))))

and my results were:

0x1f8a784d400
0x1f8b9d65e10
0x1f8b9d65b70

which means the method is not returning the same object, as was suggested in the last comment. So this method does indeed make a shuffled copy.

---

What is also useful, if you use it for Machine_learning and want to separate always the same data, you could use:

df.sample(n=len(df), random_state=42)

This makes sure, that you keep your random choice always replicable

---

Here is another way to do this:

df_shuffled = df.reindex(np.random.permutation(df.index))

---

shuffle the pandas data frame by taking a sample array in this case index and randomize its order then set the array as an index of data frame. Now sort the data frame according to index. Here goes your shuffled dataframe

import random
df = pd.DataFrame({"a":[1,2,3,4],"b":[5,6,7,8]})
index = [i for i in range(df.shape[0])]
random.shuffle(index)
df.set_index([index]).sort_index()

output

    a   b
0   2   6
1   1   5
2   3   7
3   4   8

Insert you data frame in the place of mine in above code .

---

Shuffle the DataFrame using sample() by passing the frac parameter. Save the shuffled DataFrame to a new variable.

new_variable = DataFrame.sample(frac=1)

---

Here is another way:

df['rnd'] = np.random.rand(len(df))
df = df.sort_values(by='rnd', inplace=True).drop('rnd', axis=1)

---