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pythonpandasdataframegroup-bypivot-table
Salvador Dali
Salvador Dali

Reputation: 222581

Groupby value counts on the dataframe pandas

I have the following dataframe:

df = pd.DataFrame([
    (1, 1, 'term1'),
    (1, 2, 'term2'),
    (1, 1, 'term1'),
    (1, 1, 'term2'),
    (2, 2, 'term3'),
    (2, 3, 'term1'),
    (2, 2, 'term1')
], columns=['id', 'group', 'term'])

I want to group it by id and group and calculate the number of each term for this id-group pair.

So in the end I want to get something like this:

enter image description here

Anyway I can achieve this without looping?

Upvotes: 103

Views: 205707

Answers (6)

ali bakhtiari
ali bakhtiari

Reputation: 1097

Another alternative:

df.assign(count=1).groupby(['id', 'group','term']).sum().unstack(fill_value=0).xs("count", 1)

term      term1  term2  term3
id group                     
1  1          2      1      0
   2          0      1      0
2  2          1      0      1
   3          1      0      0

Upvotes: 0

Barth
Barth

Reputation: 91

If you want to use value_counts you can use it on a given series, and resort to the following:

df.groupby(["id", "group"])["term"].value_counts().unstack(fill_value=0)

or in an equivalent fashion, using the .agg method:

df.groupby(["id", "group"]).agg({"term": "value_counts"}).unstack(fill_value=0)

Another option is to directly use value_counts on the DataFrame itself without resorting to groupby:

df.value_counts().unstack(fill_value=0)

Upvotes: 8

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210852

using pivot_table() method:

In [22]: df.pivot_table(index=['id','group'], columns='term', aggfunc='size', fill_value=0)
Out[22]:
term      term1  term2  term3
id group
1  1          2      1      0
   2          0      1      0
2  2          1      0      1
   3          1      0      0

Timing against 700K rows DF:

In [24]: df = pd.concat([df] * 10**5, ignore_index=True)

In [25]: df.shape
Out[25]: (700000, 3)

In [3]: %timeit df.groupby(['id', 'group', 'term'])['term'].size().unstack(fill_value=0)
1 loop, best of 3: 226 ms per loop

In [4]: %timeit df.pivot_table(index=['id','group'], columns='term', aggfunc='size', fill_value=0)
1 loop, best of 3: 236 ms per loop

In [5]: %timeit pd.crosstab([df.id, df.group], df.term)
1 loop, best of 3: 355 ms per loop

In [6]: %timeit df.groupby(['id','group','term'])['term'].size().unstack().fillna(0).astype(int)
1 loop, best of 3: 232 ms per loop

In [7]: %timeit df.groupby(['id', 'group', 'term']).size().unstack(fill_value=0)
1 loop, best of 3: 231 ms per loop

Timing against 7M rows DF:

In [9]: df = pd.concat([df] * 10, ignore_index=True)

In [10]: df.shape
Out[10]: (7000000, 3)

In [11]: %timeit df.groupby(['id', 'group', 'term'])['term'].size().unstack(fill_value=0)
1 loop, best of 3: 2.27 s per loop

In [12]: %timeit df.pivot_table(index=['id','group'], columns='term', aggfunc='size', fill_value=0)
1 loop, best of 3: 2.3 s per loop

In [13]: %timeit pd.crosstab([df.id, df.group], df.term)
1 loop, best of 3: 3.37 s per loop

In [14]: %timeit df.groupby(['id','group','term'])['term'].size().unstack().fillna(0).astype(int)
1 loop, best of 3: 2.28 s per loop

In [15]: %timeit df.groupby(['id', 'group', 'term']).size().unstack(fill_value=0)
1 loop, best of 3: 1.89 s per loop

Upvotes: 28

piRSquared
piRSquared

Reputation: 294318

I use groupby and size

df.groupby(['id', 'group', 'term']).size().unstack(fill_value=0)

enter image description here

Timing

enter image description here

1,000,000 rows

df = pd.DataFrame(dict(id=np.random.choice(100, 1000000),
                       group=np.random.choice(20, 1000000),
                       term=np.random.choice(10, 1000000)))

enter image description here

Upvotes: 174

jezrael
jezrael

Reputation: 862741

You can use crosstab:

print (pd.crosstab([df.id, df.group], df.term))
term      term1  term2  term3
id group                     
1  1          2      1      0
   2          0      1      0
2  2          1      0      1
   3          1      0      0

Another solution with groupby with aggregating size, reshaping by unstack:

df.groupby(['id', 'group', 'term'])['term'].size().unstack(fill_value=0)

term      term1  term2  term3
id group                     
1  1          2      1      0
   2          0      1      0
2  2          1      0      1
   3          1      0      0

Timings:

df = pd.concat([df]*10000).reset_index(drop=True)

In [48]: %timeit (df.groupby(['id', 'group', 'term']).size().unstack(fill_value=0))
100 loops, best of 3: 12.4 ms per loop

In [49]: %timeit (df.groupby(['id', 'group', 'term'])['term'].size().unstack(fill_value=0))
100 loops, best of 3: 12.2 ms per loop

Upvotes: 17

A.Kot
A.Kot

Reputation: 7903

Instead of remembering lengthy solutions, how about the one that pandas has built in for you:

df.groupby(['id', 'group', 'term']).count()

Upvotes: 26

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