CODEWITHSUNDEEP
pythonpandasdataframerangebinning
Alex Man
Alex Man

Reputation: 477

Return the lowr or upper bound of a range after binning in Python

I convert the following df into bins using pd.cut in following: 

import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,100,size=(5, 4)), columns=list('ABCD'))
print(df)
newDF = pd.cut(df.A, 2, precision=0)
print(newDF)

A   B   C   D
0  83  43  99  85
1   6  57  44  45
2   5  72  10  53
3  24  50  23  18
4  75  25  96  27
0    (44.0, 83.0]
1     (5.0, 44.0]
2     (5.0, 44.0]
3     (5.0, 44.0]
4    (44.0, 83.0]

Is there any way to return the lower bound or upper bound of the range instead of the whole range? For example, from the above example:

0    44.0
1    5.0
2    5.0
3    5.0
4    44.0

Upvotes: 3

Views: 2655

Answers (3)

piRSquared
piRSquared

Reputation: 294228

This isn't too much different that @ansev's answer. However, I really wanted an IntervalDtype accessor for pd.Series objects so that this sort of thing would work.

# THIS IS NOT REAL!
# JUST AN EXAMPLE
# OF WHAT I WANT
newDF.astype(pd.IntervalDtype()).interval.left

So, in search for such a thing, I came across the same things as @ansev. I'd expect this to change in the future. I suspect they will add an IntervalDtype accessor (maybe).

I'll offer a simple list comprehension. What this offers is a simple solution without creating additional pandas objects.

cats = newDF.cat.categories
codes = newDF.cat.codes
pd.Series([cats[code].left for code in codes], newDF.index)

0    54.0
1    54.0
2    14.0
3    14.0
4    54.0
dtype: float64

Upvotes: 0

Quang Hoang
Quang Hoang

Reputation: 150735

For numerical values, if you pass a constant to bins, pd.cut will just cut for np.linspace(min,max, bins+1). So

bins = 2
interval_bins = np.linspace(df.A.min(), df.A.max(),bins+1)

lefts = interval_bins[:-1]
rights = inteval_bins[1:]

Upvotes: 3

ansev
ansev

Reputation: 30920

Use Series.map:

pd.cut(df.A, 2, precision=0).map(lambda x: x.left)

or pd.IntervalIndex

s = pd.cut(df.A, 2, precision=0)
pd.Series(data=pd.IntervalIndex(s).left, index = s.index)

#print(df)
#
#
#    A   B   C   D
#0  26  70  28   2
#1  49  42  56  28
#2  48  26  40  19
#3   3  50  17   3
#4  20  34  54  42
#
#
#pd.cut(df.A, 2, precision=0).map(lambda x: x.left)
#
#0     3.0
#1    26.0
#2    26.0
#3     3.0
#4     3.0
#Name: A, dtype: category
#Categories (2, float64): [3.0 < 26.0]

Upvotes: 4

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