我正在编写一个程序，通过熵离散化来离散化一组属性。目标是解析数据集

A,Class
5,1
12.5,1
11.5,2
8.6,2
7,1
6,1
5.9,2
1.5,2
9,2
7.8,1
2.1,1
13.5,2
12.45,2

进入

A,Class
1,1
3,1
3,2
2,2
2,1
2,1
1,2
1,2
3,2
2,1
1,1
3,2
3,2

我面临的具体问题是，我希望使用pandas方法删除与计算出的阈值关联的行，我尝试这样做的方法是s['A'].drop[s.iloc[0]]。

import pandas as pd
import numpy as np
import entropy_based_binning as ebb
from math import log2
from random import randrange, uniform

def main():
    df = pd.read_csv('S1.csv')
    s = df
    s = entropy_discretization(s)

# This method discretizes s A1
# If the information gain is 0, i.e the number of 
# distinct class is 1 or
# If min f/ max f < 0.5 and the number of distinct values is floor(n/2)
# Then that partition stops splitting.
def entropy_discretization(s):

    I = {}
    while(uniqueValue(s)):
        # Step 1: pick a threshold
        threshold = s['A'].iloc[0]

        # Step 2: Partititon the data set into two parttitions
        s1 = s[s['A'] < threshold]
        print("s1 after spitting")
        print(s1)
        print("******************")
        s2 = s[s['A'] >= threshold]
        print("s2 after spitting")
        print(s2)
        print("******************")
            
        # Step 3: calculate the information gain.
        informationGain = information_gain(s1,s2,s)
        print(f'Calculated information gain {informationGain}')

        I.update({'informationGain':informationGain,'threshold':threshold})
        print(I)
        s['A'].drop[s.iloc[0]]

    # Step 5: calculate the max information gain
    maxInformationGain = np.amax(informationGain)
    print(f'Calculated maximum information gain {maxInformationGain}')

    # Step 6: keep the partitions of S based on the value of threshold_i
    s = bestPartition(minInformationGain, s)

def uniqueValue(s):
    # are records in s the same? return true
    if s.nunique()['A'] == 1:
        return False
    # otherwise false 
    else:
        return True

def bestPartition(maxInformationGain):
    # determine be threshold_i
    threshold_i = 6

    return 

def information_gain(s1, s2, s):
    # calculate cardinality for s1
    cardinalityS1 = len(pd.Index(s1['A']).value_counts())
    print(f'The Cardinality of s1 is: {cardinalityS1}')
    # calculate cardinality for s2
    cardinalityS2 = len(pd.Index(s2['A']).value_counts())
    print(f'The Cardinality of s2 is: {cardinalityS2}')
    # calculate cardinality of s
    cardinalityS = len(pd.Index(s['A']).value_counts())
    print(f'The Cardinality of s is: {cardinalityS}')
    # calculate informationGain
    informationGain = (cardinalityS1/cardinalityS) * entropy(s1) + (cardinalityS2/cardinalityS) * entropy(s2)
    print(f'The total informationGain is: {informationGain}')
    return informationGain


def entropy(s):
    print("calculating the entropy for s")
    print("*****************************")
    print(s)
    print("*****************************")

    # initialize ent
    ent = 0

    # calculate the number of classes in s
    numberOfClasses = s['Class'].nunique()
    print(f'Number of classes for dataset: {numberOfClasses}')
    value_counts = s['Class'].value_counts()
    p = []
    for i in range(0,numberOfClasses):
        n = s['Class'].count()
        # calculate the frequency of class_i in S1
        print(f'p{i} {value_counts.iloc[i]}/{n}')
        f = value_counts.iloc[i]
        pi = f/n
        p.append(pi)
    
    print(p)

    for pi in p:
        ent += -pi*log2(pi)

    return ent 

main()

理想情况下，我希望删除与变量threshold具有相同值的行。