基于python 等频分箱qcut问题的解决

import pandas as pd
 
def pct_rank_qcut(series, n):
  '''
  series：要分箱的列
  n：箱子数
  '''
  edages = pd.series([i/n for i in range(n)] # 转换成百分比
  func = lambda x: (edages >= x).argmax() #函数：(edages >= x)返回fasle/true列表中第一次出现true的索引值
  return series.rank(pct=1).astype(float).apply(func) #series.rank(pct=1)每个值对应的百分位数，最终返回对应的组数;rank()函数传入的数据类型若为object，结果会有问题，因此进行了astype

import numpy as np
import pandas as pd
 
# Discretization: Equal Width #
# Datas: Sample * Feature
def Discretization_EqualWidth(K, Datas, FeatureNumber):
  DisDatas = np.zeros_like(Datas)
  for i in range(FeatureNumber):
    DisOneFeature = pd.cut(Datas[:, i], K, labels=range(1, K+1))
    DisDatas[:, i] = DisOneFeature
  return DisDatas

import numpy as np
import pandas as pd
 
# Discretization: Equal Frequency #
# vector: single feature
def Rank_qcut(vector, K):
  quantile = np.array([float(i) / K for i in range(K + 1)]) # Quantile: K+1 values
  funBounder = lambda x: (quantile >= x).argmax()
  return vector.rank(pct=True).apply(funBounder)
 
# Discretization: Equal Frequency #
# Datas: Sample * Feature
def Discretization_EqualFrequency(K, Datas, FeatureNumber):
  DisDatas = np.zeros_like(Datas)
  w = [float(i) / K for i in range(K + 1)]
  for i in range(FeatureNumber):
    DisOneFeature = Rank_qcut(pd.Series(Datas[:, i]), K)
    #print(DisOneFeature)
    DisDatas[:, i] = DisOneFeature
  return DisDatas