Pandas groupby how to compute counts in ranges

We can use pd.cut to bin the values into ranges, then we can groupby these ranges, and finally call count to count the values now binned into these ranges:

np.random.seed(0)
df = pd.DataFrame({"a": np.random.random_integers(1, high=100, size=100)})
ranges = [0,10,20,30,40,50,60,70,80,90,100]
df.groupby(pd.cut(df.a, ranges)).count()

            a
a            
(0, 10]    11
(10, 20]   10
(20, 30]    8
(30, 40]   13
(40, 50]   11
(50, 60]    9
(60, 70]   10
(70, 80]   11
(80, 90]   13
(90, 100]   4

Surprised I haven't seen this yet, so without further ado, here is

.value_counts(bins=N)

Computing bins with pd.cut followed by a groupBy is a 2-step process. value_counts allows you a shortcut using the bins argument:

# Uses Ed Chum's setup. Cross check our answers match!
np.random.seed(0)
df = pd.DataFrame({"a": np.random.random_integers(1, high=100, size=100)})

df['a'].value_counts(bins=10, sort=False)

(0.9, 10.9]      11
(10.9, 20.8]     10
(20.8, 30.7]      8
(30.7, 40.6]     13
(40.6, 50.5]     11
(50.5, 60.4]      9
(60.4, 70.3]     10
(70.3, 80.2]     11
(80.2, 90.1]     13
(90.1, 100.0]     4
Name: a, dtype: int64

This creates 10 evenly-spaced right-closed intervals and bincounts your data. sort=False will be required to avoid value_counts ordering the result in decreasing order of count.

---

Binning by Unequal Ranges

For this, you can pass a list to bins argument:

bins = [0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
df['a'].value_counts(bins=bins, sort=False) 

(-0.001, 10.0]    11
(10.0, 20.0]      10
(20.0, 30.0]       8
(30.0, 40.0]      13
(40.0, 50.0]      11
(50.0, 60.0]       9
(60.0, 70.0]      10
(70.0, 80.0]      11
(80.0, 90.0]      13
(90.0, 100.0]      4
Name: a, dtype: int64