pd.cut 可以同时使用区间范围和标签吗？

Question

I'm fiddling around with something like this.

bins = [0, .25, .5, .75, 1, 1.25, 1.5, 1.75, 2]
labels = ['0', '.25', '.5', '.75', '1', '1.25', '1.5', '1.75', '2']
dataset['RatingScore'] = pd.cut(dataset['Rating'], bins, labels)

What I am actually getting is a range, like this: (0.75, 1.0]

I would like to get results like this: .75 or 1 or 1.25

Is it possible to get a specific number and NOT a range? Thanks.

Andy, your code runs, and it gives me actual numbers, rather than ranges, but I'm seeing a lot of gaps too.

Answer 1

You pass labels to the 3rd parameter of pd.cut . The third parameter of pd.cut is right=... . It accepts True/False as values. labels is non-empty list, so it is considered as True . Therefore, pd.cut executes as there is no label. You need to use keyword parameter to correctly specify list labels as labels for pd.cut . Another thing, number of bins must be one item more than labels . You need to add np.inf to the right of list bins

s = pd.Series([0.2, 0.6, 0.1, 0.9, 2])
bins = [0, 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2, np.inf]
labels = ['0', '.25', '.5', '.75', '1', '1.25', '1.5', '1.75', '2']

s_cat = pd.cut(s, bins=bins, labels=labels)

Out[1165]:
0       0
1      .5
2       0
3     .75
4    1.75
dtype: category
Categories (9, object): [0 < .25 < .5 < .75 ... 1.25 < 1.5 < 1.75 < 2]

Answer 2

If you don't add infinity to the bins you'll have as possible output float ( np.nan ) or interval let says you want to take the right interval you could try as follow

import pandas as pd
import numpy as np

def fun(x):
    if isinstance(x, float) is True:
        return np.nan
    else:
        return x.right

df = pd.DataFrame({"Rating":[.1* i for i in range(10)]})
bins = [0, .25, .5, .75, 1, 1.25, 1.5, 1.75, 2]
df["RatingScore"] = pd.cut(df['Rating'], bins)

df["RatingScore"].apply(fun)

0     NaN
1    0.25
2    0.25
3    0.50
4    0.50
5    0.50
6    0.75
7    0.75
8    1.00
9    1.00