Binning a column with Python Pandas

I have a data frame column with numeric values:

df['percentage'].head()
46.5
44.2
100.0
42.12

I want to see the column as bin counts:

bins = [0, 1, 5, 10, 25, 50, 100]

How can I get the result as bins with their value counts?

[0, 1] bin amount
[1, 5] etc
[5, 10] etc
...

You can use pandas.cut:

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = pd.cut(df['percentage'], bins)
print (df)
   percentage     binned
0       46.50   (25, 50]
1       44.20   (25, 50]
2      100.00  (50, 100]
3       42.12   (25, 50]

bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
df['binned'] = pd.cut(df['percentage'], bins=bins, labels=labels)
print (df)
   percentage binned
0       46.50      5
1       44.20      5
2      100.00      6
3       42.12      5

Or numpy.searchsorted:

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = np.searchsorted(bins, df['percentage'].values)
print (df)
   percentage  binned
0       46.50       5
1       44.20       5
2      100.00       6
3       42.12       5

...and then value_counts or groupby and aggregate size:

s = pd.cut(df['percentage'], bins=bins).value_counts()
print (s)
(25, 50]     3
(50, 100]    1
(10, 25]     0
(5, 10]      0
(1, 5]       0
(0, 1]       0
Name: percentage, dtype: int64

s = df.groupby(pd.cut(df['percentage'], bins=bins)).size()
print (s)
percentage
(0, 1]       0
(1, 5]       0
(5, 10]      0
(10, 25]     0
(25, 50]     3
(50, 100]    1
dtype: int64

By default cut returns categorical.

Series methods like Series.value_counts() will use all categories, even if some categories are not present in the data, operations in categorical.

Using the Numba module for speed up.

On big datasets (more than 500k), pd.cut can be quite slow for binning data.

I wrote my own function in Numba with just-in-time compilation, which is roughly six times faster:

from numba import njit

@njit
def cut(arr):
    bins = np.empty(arr.shape[0])
    for idx, x in enumerate(arr):
        if (x >= 0) & (x < 1):
            bins[idx] = 1
        elif (x >= 1) & (x < 5):
            bins[idx] = 2
        elif (x >= 5) & (x < 10):
            bins[idx] = 3
        elif (x >= 10) & (x < 25):
            bins[idx] = 4
        elif (x >= 25) & (x < 50):
            bins[idx] = 5
        elif (x >= 50) & (x < 100):
            bins[idx] = 6
        else:
            bins[idx] = 7

    return bins

cut(df['percentage'].to_numpy())

# array([5., 5., 7., 5.])

Optional: you can also map it to bins as strings:

a = cut(df['percentage'].to_numpy())

conversion_dict = {1: 'bin1',
                   2: 'bin2',
                   3: 'bin3',
                   4: 'bin4',
                   5: 'bin5',
                   6: 'bin6',
                   7: 'bin7'}

bins = list(map(conversion_dict.get, a))

# ['bin5', 'bin5', 'bin7', 'bin5']

Speed comparison:

# Create a dataframe of 8 million rows for testing
dfbig = pd.concat([df]*2000000, ignore_index=True)

dfbig.shape

# (8000000, 1)

%%timeit
cut(dfbig['percentage'].to_numpy())

# 38 ms ± 616 µs per loop (mean ± standard deviation of 7 runs, 10 loops each)

%%timeit
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
pd.cut(dfbig['percentage'], bins=bins, labels=labels)

# 215 ms ± 9.76 ms per loop (mean ± standard deviation of 7 runs, 10 loops each)