如何計算邏輯回歸精度

Question

我是python機器學習和編碼的完全入門者，我受過從零開始進行邏輯回歸編碼的任務，以了解幕后發生的事情。 到目前為止，我已經對假設函數，成本函數和梯度下降進行了編碼，然后對邏輯回歸進行了編碼。 但是，在為打印精度進行編碼時，我得到的輸出很低（0.69），它不會隨着迭代次數的增加或學習率的改變而變化。 我的問題是，下面的我的准確性代碼是否有問題？ 任何指向正確方向的幫助將不勝感激

X = data[['radius_mean', 'texture_mean', 'perimeter_mean',
   'area_mean', 'smoothness_mean', 'compactness_mean', 'concavity_mean',
   'concave points_mean', 'symmetry_mean', 'fractal_dimension_mean',
   'radius_se', 'texture_se', 'perimeter_se', 'area_se', 'smoothness_se',
   'compactness_se', 'concavity_se', 'concave points_se', 'symmetry_se',
   'fractal_dimension_se', 'radius_worst', 'texture_worst',
   'perimeter_worst', 'area_worst', 'smoothness_worst',
   'compactness_worst', 'concavity_worst', 'concave points_worst',
   'symmetry_worst', 'fractal_dimension_worst']]
X = np.array(X)
X = min_max_scaler.fit_transform(X)
Y = data["diagnosis"].map({'M':1,'B':0})
Y = np.array(Y)

X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.25)

X = data["diagnosis"].map(lambda x: float(x))

def Sigmoid(z):
    if z < 0:
        return 1 - 1/(1 + math.exp(z))
    else:
        return 1/(1 + math.exp(-z))

def Hypothesis(theta, x):
    z = 0
    for i in range(len(theta)):
        z += x[i]*theta[i]
    return Sigmoid(z)

def Cost_Function(X,Y,theta,m):
    sumOfErrors = 0
    for i in range(m):
        xi = X[i]
        hi = Hypothesis(theta,xi)
        error = Y[i] * math.log(hi if  hi >0 else 1)
        if Y[i] == 1:
            error = Y[i] * math.log(hi if  hi >0 else 1)
        elif Y[i] == 0:
            error = (1-Y[i]) * math.log(1-hi  if  1-hi >0 else 1)
        sumOfErrors += error

    constant = -1/m
    J = constant * sumOfErrors
    #print ('cost is: ', J ) 
    return J

def Cost_Function_Derivative(X,Y,theta,j,m,alpha):
    sumErrors = 0
    for i in range(m):
        xi = X[i]
        xij = xi[j]
        hi = Hypothesis(theta,X[i])
        error = (hi - Y[i])*xij
        sumErrors += error
    m = len(Y)
    constant = float(alpha)/float(m)
    J = constant * sumErrors
    return J

def Gradient_Descent(X,Y,theta,m,alpha):
    new_theta = []
    constant = alpha/m
    for j in range(len(theta)):
        CFDerivative = Cost_Function_Derivative(X,Y,theta,j,m,alpha)
        new_theta_value = theta[j] - CFDerivative
        new_theta.append(new_theta_value)
    return new_theta


def Accuracy(theta):
    correct = 0
    length = len(X_test, Hypothesis(X,theta))
    for i in range(length):
        prediction = round(Hypothesis(X[i],theta))
        answer = Y[i]
    if prediction == answer.all():
            correct += 1
    my_accuracy = (correct / length)*100
    print ('LR Accuracy %: ', my_accuracy)



def Logistic_Regression(X,Y,alpha,theta,num_iters):
    theta = np.zeros(X.shape[1])
    m = len(Y)
    for x in range(num_iters):
        new_theta = Gradient_Descent(X,Y,theta,m,alpha)
        theta = new_theta
        if x % 100 == 0:
            Cost_Function(X,Y,theta,m)
            print ('theta: ', theta)    
            print ('cost: ', Cost_Function(X,Y,theta,m))
    Accuracy(theta)

initial_theta = [0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]  
alpha = 0.0001
iterations = 1000
Logistic_Regression(X,Y,alpha,initial_theta,iterations)

這是使用威斯康星州乳腺癌數據集（ https://www.kaggle.com/uciml/breast-cancer-wisconsin-data ）中的數據，其中我權衡了30個特征-盡管將特征更改為已知相關的特征也不會改變我的准確性。

Answer 1

Python為我們提供了這個scikit-learn庫，使我們的工作更加輕松，這對我來說很有效：

from sklearn.metrics import accuracy_score

y_pred = log.predict(x_test)

score =accuracy_score(y_test,y_pred)

Answer 2

我不確定您如何得出alpha值為0.0001 ，但我認為它太低了。 將代碼與癌症數據一起使用表明，每次迭代的成本都在降低-這只是冰山一角。

當我將其提高到0.5時，我仍然得到了降低的成本，但是在一個更合理的水平上。 經過1000次迭代后，它報告：

cost:  0.23668000993020666

修復了Accuracy函數后，我在數據測試段上獲得了92％的收益。

您已經安裝了Numpy，如X = np.array(X) 。 您應該真正考慮將其用於操作。 這樣的工作將會快幾個數量級 。 這是向量化版本，可立即提供結果，而無需等待：

import math
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from sklearn.preprocessing import MinMaxScaler
from sklearn.model_selection import train_test_split

df = pd.read_csv("cancerdata.csv")
X = df.values[:,2:-1].astype('float64')
X = (X - np.mean(X, axis =0)) /  np.std(X, axis = 0)

## Add a bias column to the data
X = np.hstack([np.ones((X.shape[0], 1)),X])
X = MinMaxScaler().fit_transform(X)
Y = df["diagnosis"].map({'M':1,'B':0})
Y = np.array(Y)
X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.25)


def Sigmoid(z):
    return 1/(1 + np.exp(-z))

def Hypothesis(theta, x):   
    return Sigmoid(x @ theta) 

def Cost_Function(X,Y,theta,m):
    hi = Hypothesis(theta, X)
    _y = Y.reshape(-1, 1)
    J = 1/float(m) * np.sum(-_y * np.log(hi) - (1-_y) * np.log(1-hi))
    return J

def Cost_Function_Derivative(X,Y,theta,m,alpha):
    hi = Hypothesis(theta,X)
    _y = Y.reshape(-1, 1)
    J = alpha/float(m) * X.T @ (hi - _y)
    return J

def Gradient_Descent(X,Y,theta,m,alpha):
    new_theta = theta - Cost_Function_Derivative(X,Y,theta,m,alpha)
    return new_theta

def Accuracy(theta):
    correct = 0
    length = len(X_test)
    prediction = (Hypothesis(theta, X_test) > 0.5)
    _y = Y_test.reshape(-1, 1)
    correct = prediction == _y
    my_accuracy = (np.sum(correct) / length)*100
    print ('LR Accuracy %: ', my_accuracy)

def Logistic_Regression(X,Y,alpha,theta,num_iters):
    m = len(Y)
    for x in range(num_iters):
        new_theta = Gradient_Descent(X,Y,theta,m,alpha)
        theta = new_theta
        if x % 100 == 0:
            #print ('theta: ', theta)    
            print ('cost: ', Cost_Function(X,Y,theta,m))
    Accuracy(theta)

ep = .012

initial_theta = np.random.rand(X_train.shape[1],1) * 2 * ep - ep
alpha = 0.5
iterations = 2000
Logistic_Regression(X_train,Y_train,alpha,initial_theta,iterations)

我想我可能有不同版本的scikit，因為我更改了MinMaxScaler行以使其工作。 結果是，我可以眨眼間進行10K次迭代，將模型應用於測試集的結果的准確性約為97％。

Answer 3

准確性是最直觀的性能指標之一，它只是正確預測的觀測值與總觀測值的比率。 更高的精度意味着模型的性能更好。

Accuracy = TP+TN/TP+FP+FN+TN

TP = True positives
TN = True negatives
FN = False negatives
TN = True negatives

在使用准確性度量時，誤報和誤報的成本應相近。 更好的指標是F1分數，由

F1-score = 2*(Recall*Precision)/Recall+Precision where,

Precision = TP/TP+FP
Recall = TP/TP+FN

在這里閱讀更多

https://zh.wikipedia.org/wiki/Precision_and_recall

使用python進行機器學習的好處在於，像scikit-learn這樣的重要模塊是開源的，因此您始終可以查看實際的代碼。 請使用以下鏈接訪問scikit學習指標源代碼，該源代碼將使您了解scikit-learn在執行此操作時如何計算准確性得分

from sklearn.metrics import accuracy_score
accuracy_score(y_true, y_pred)

https://github.com/scikit-learn/scikit-learn/tree/master/sklearn/metrics