“Pd.cut in Pandas” Ответ

Разрешение DataFrame на основе диапазона

test = pd.DataFrame({'days': [0,20,30,31,45,60]})

test['range1'] = pd.cut(test.days, [0,30,60], include_lowest=True)
#30 value is in [30, 60) group
test['range2'] = pd.cut(test.days, [0,30,60], right=False)
#30 value is in (0, 30] group
test['range3'] = pd.cut(test.days, [0,30,60])
print (test)
   days          range1    range2    range3
0     0  (-0.001, 30.0]   [0, 30)       NaN
1    20  (-0.001, 30.0]   [0, 30)   (0, 30]
2    30  (-0.001, 30.0]  [30, 60)   (0, 30]
3    31    (30.0, 60.0]  [30, 60)  (30, 60]
4    45    (30.0, 60.0]  [30, 60)  (30, 60]
5    60    (30.0, 60.0]       NaN  (30, 60]
Gifted Gecko

Pd.cut in Pandas

>>> pd.qcut(range(5), 3, labels=["good", "medium", "bad"])
... 
[good, good, medium, bad, bad]
Categories (3, object): [good < medium < bad]
Famous Flamingo

DataFrame Cut

import pandas as pd

#设置切分区域
listBins = [0, 10, 20, 30, 40, 50, 60, 1000000]

#设置切分后对应标签
listLabels = ['0_10','11_20','21_30','31_40','41_50','51_60','61及以上']

#利用pd.cut进行数据离散化切分
"""
pandas.cut(x,bins,right=True,labels=None,retbins=False,precision=3,include_lowest=False)
x:需要切分的数据
bins:切分区域
right : 是否包含右端点默认True，包含
labels:对应标签，用标记来代替返回的bins，若不在该序列中，则返回NaN
retbins:是否返回间距bins
precision:精度
include_lowest:是否包含左端点，默认False，不包含
"""
df['fenzu'] = pd.cut(df['data'], bins=listBins, labels=listLabels, include_lowest=True)
Red Team

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