如何从pandas value_counts（）中获取第一个值

Question

我正在编写一个程序，通过熵离散化对一组属性进行离散化。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

我的程序面临的具体问题是确定数据集中的类数。这发生在numberOfClasses = s['Class'].value_counts()。我想使用pandas方法返回不同类的数量。在这个例子中，只有两个。但是我回来了

Number of classes: 2    5
1    4

从print语句中

import pandas as pd
import numpy as np
import entropy_based_binning as ebb
from math import log2

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):

    informationGain = {}
    # while(uniqueValue(s)):
    # Step 1: pick a threshold
    threshold = 6

    # 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(informationGain)

    # # Step 5: calculate the max information gain
    # minInformationGain = min(informationGain)

    # # 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):
    # calculate the number of classes in s
    numberOfClasses = s['Class'].value_counts()
    print(f'Number of classes: {numberOfClasses}')
    # TODO calculate pi for each class.
    # calculate the frequency of class_i in S1
    p1 = 2/4
    p2 = 3/4
    ent = -(p1*log2(p2)) - (p2*log2(p2))

    return ent 

main()

理想情况下，我想打印Number of classes: 2。通过这种方式，我可以在类上循环，并从数据集中计算属性A的频率。我已经查阅了pandas文档，但是我被价值观所束缚。任何帮助都将不胜感激

Answer 1

也许可以试试：

number_of_classes = len(s['Class'].unique())

它将返回列Class中唯一类的数量

甚至更短：

s['Class'].nunique()