如何根据行值从熊猫数据框中删除一行

Question

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想使用 Pandas 方法删除与计算阈值关联的行。 我这样做的尝试是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，我想删除与变量threshold具有相同值的行。 任何帮助将不胜感激。

Answer 1

我认为这应该有效：

S = S[S['A']!=threshold]

Answer 2

我想删除相当于阈值的行。 该算法的重点是从数据集中删除唯一值。 这可以通过

s = s[s['A'] != threshold]

它是这样使用的

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})
        s = s[s['A'] != threshold]
        print(I)

    print(I)
    # 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(maxInformationGain, s)

Answer 3

如果您希望删除等于阈值的行而不是保留不等于阈值的行，请使用drop ：

s.drop(s[s['A'] == threshold)].index, inplace=True)