CODEWITHSUNDEEP
pythonnumpypandassubsampling
Nishant
Nishant

Reputation: 55866

Subsample pandas dataframe

I have a DataFrame loaded from a .tsv file. I wanted to generate some exploratory plots. The problem is that the data set is large (~1 million rows), so there are too many points on the plot to see a trend. Plus, it is taking a while to plot.

I wanted to sub-sample 10000 randomly distributed rows. This should be reproducible so the same sequence of random numbers is generated in each run.

This: Sample two pandas dataframes the same way seems to be on the right track, but I cannot guarantee the subsample size.

Upvotes: 17

Views: 29721

Answers (3)

joris
joris

Reputation: 139172

You can select random elements from the index with np.random.choice. Eg to select 5 random rows:

df = pd.DataFrame(np.random.rand(10))

df.loc[np.random.choice(df.index, 5, replace=False)]

This function is new in 1.7. If you want a solution with an older numpy, you can shuffle the data and taken the first elements of that:

df.loc[np.random.permutation(df.index)[:5]]

In this way your DataFrame is not sorted anymore, but if this is needed for plotting (for example, a line plot), you can simply do .sort() afterwards.

Upvotes: 19

Alex Coventry
Alex Coventry

Reputation: 70847

These days, one can simply use the sample method on a DataFrame:

>>> help(df.sample)
Help on method sample in module pandas.core.generic:

sample(self, n=None, frac=None, replace=False, weights=None, random_state=None, axis=None) method of pandas.core.frame.DataFrame instance
    Returns a random sample of items from an axis of object.

Replicability can be achieved by using the random_state keyword:

>>> len(set(df.sample(n=1, random_state=np.random.RandomState(0)).iterations.values[0] for _ in xrange(1000)))
1
>>> len(set(df.sample(n=1).iterations.values[0] for _ in xrange(1000)))
40

Upvotes: 14

Andy Hayden
Andy Hayden

Reputation: 375485

Unfortunately np.random.choice appears to be quite slow for small samples (less than 10% of all rows), you may be better off using plain ol' sample:

from random import sample
df.loc[sample(df.index, 1000)]

For large DataFrame (a million rows), we see small samples:

In [11]: %timeit df.loc[sample(df.index, 10)]
1000 loops, best of 3: 1.19 ms per loop

In [12]: %timeit df.loc[np.random.choice(df.index, 10, replace=False)]
1 loops, best of 3: 1.36 s per loop

In [13]: %timeit df.loc[np.random.permutation(df.index)[:10]]
1 loops, best of 3: 1.38 s per loop

In [21]: %timeit df.loc[sample(df.index, 1000)]
10 loops, best of 3: 14.5 ms per loop

In [22]: %timeit df.loc[np.random.choice(df.index, 1000, replace=False)]
1 loops, best of 3: 1.28 s per loop    

In [23]: %timeit df.loc[np.random.permutation(df.index)[:1000]]
1 loops, best of 3: 1.3  s per loop

But around 10% it gets about the same:

In [31]: %timeit df.loc[sample(df.index, 100000)]
1 loops, best of 3: 1.63 s per loop

In [32]: %timeit df.loc[np.random.choice(df.index, 100000, replace=False)]
1 loops, best of 3: 1.36 s per loop

In [33]: %timeit df.loc[np.random.permutation(df.index)[:100000]]
1 loops, best of 3: 1.4 s per loop

and if you are sampling everything (don't use sample!):

In [41]: %timeit df.loc[sample(df.index, 1000000)]
1 loops, best of 3: 10 s per loop

Note: both numpy.random and random accept a seed, to reproduce randomly generated output.

As @joris points out in the comments, choice (without replacement) is actually sugar for permutation so it's no suprise it's constant time and slower for smaller samples...

Upvotes: 18

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