Binning a column with pandas
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Chapters
00:00 Binning A Column With Pandas
00:19 Accepted Answer Score 357
01:14 Answer 2 Score 25
02:15 Answer 3 Score 1
02:33 Answer 4 Score 2
02:51 Thank you
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Full question
https://stackoverflow.com/questions/4527...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #pandas #numpy #dataframe #binning
#avk47
    Rise to the top 3% as a developer or hire one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Popsicle Puzzles
--
Chapters
00:00 Binning A Column With Pandas
00:19 Accepted Answer Score 357
01:14 Answer 2 Score 25
02:15 Answer 3 Score 1
02:33 Answer 4 Score 2
02:51 Thank you
--
Full question
https://stackoverflow.com/questions/4527...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #pandas #numpy #dataframe #binning
#avk47
ACCEPTED ANSWER
Score 357
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
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.
ANSWER 2
Score 25
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)
ANSWER 3
Score 2
Convenient and fast option using Numpy
np.digitize is a convenient and fast option:
import pandas as pd
import numpy as np
df = pd.DataFrame({'x': [1,2,3,4,5]})
df['y'] = np.digitize(df['x'], bins=[3,5]) # convert column to bin
print(df)
returns
   x  y
0  1  0
1  2  0
2  3  1
3  4  1
4  5  2
ANSWER 4
Score 1
We could also use np.select:
bins = [0, 1, 5, 10, 25, 50, 100]
df['groups'] = (np.select([df['percentage'].between(i, j, inclusive='right') 
                           for i,j in zip(bins, bins[1:])], 
                          [1, 2, 3, 4, 5, 6]))
Output:
   percentage  groups
0       46.50       5
1       44.20       5
2      100.00       6
3       42.12       5