我没有得到二进制分类问题的预期输出。
问题是使用二元分类将乳腺癌标记为: - 良性,或 - 恶性
它没有提供所需的输出。
首先有一个函数可以加载返回测试和训练形状数据的数据集:
x_train is of shape: (30, 381),
y_train is of shape: (1, 381),
x_test is of shape: (30, 188),
y_test is of shape: (1, 188).
然后有一个逻辑回归分类器的类,它预测输出。
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split
from sklearn.metrics import accuracy_score
import numpy as np
def load_dataset():
cancer_data = load_breast_cancer()
x_train, x_test, y_train, y_test = train_test_split(cancer_data.data, cancer_data.target, test_size=0.33)
x_train = x_train.T
x_test = x_test.T
y_train = y_train.reshape(1, (len(y_train)))
y_test = y_test.reshape(1, (len(y_test)))
m = x_train.shape[1]
return x_train, x_test, y_train, y_test, m
class Neural_Network():
def __init__(self):
np.random.seed(1)
self.weights = np.random.rand(30, 1) * 0.01
self.bias = np.zeros(shape=(1, 1))
def sigmoid(self, x):
return 1 / (1 + np.exp(-x))
def train(self, x_train, y_train, iterations, m, learning_rate=0.5):
for i in range(iterations):
z = np.dot(self.weights.T, x_train) + self.bias
a = self.sigmoid(z)
cost = (-1 / m) * np.sum(y_train * np.log(a) + (1 - y_train) * np.log(1 - a))
if (i % 500 == 0):
print("Cost after iteration %i: %f" % (i, cost))
dw = (1 / m) * np.dot(x_train, (a - y_train).T)
db = (1 / m) * np.sum(a - y_train)
self.weights = self.weights - learning_rate * dw
self.bias = self.bias - learning_rate * db
def predict(self, inputs):
m = inputs.shape[1]
y_predicted = np.zeros((1, m))
z = np.dot(self.weights.T, inputs) + self.bias
a = self.sigmoid(z)
for i in range(a.shape[1]):
y_predicted[0, i] = 1 if a[0, i] > 0.5 else 0
return y_predicted
if __name__ == "__main__":
'''
step-1 : Loading data set
x_train is of shape: (30, 381)
y_train is of shape: (1, 381)
x_test is of shape: (30, 188)
y_test is of shape: (1, 188)
'''
x_train, x_test, y_train, y_test, m = load_dataset()
neuralNet = Neural_Network()
'''
step-2 : Train the network
'''
neuralNet.train(x_train, y_train,10000,m)
y_predicted = neuralNet.predict(x_test)
print("Accuracy on test data: ")
print(accuracy_score(y_test, y_predicted)*100)
提供此输出的程序:
C:\Python36\python.exe C:/Users/LENOVO/PycharmProjects/MarkDmo001/Numpy.py
Cost after iteration 0: 5.263853
C:/Users/LENOVO/PycharmProjects/MarkDmo001/logisticReg.py:25: RuntimeWarning: overflow encountered in exp
return 1 / (1 + np.exp(-x))
C:/Users/LENOVO/PycharmProjects/MarkDmo001/logisticReg.py:33: RuntimeWarning: divide by zero encountered in log
cost = (-1 / m) * np.sum(y_train * np.log(a) + (1 - y_train) * np.log(1 - a))
C:/Users/LENOVO/PycharmProjects/MarkDmo001/logisticReg.py:33: RuntimeWarning: invalid value encountered in multiply
cost = (-1 / m) * np.sum(y_train * np.log(a) + (1 - y_train) * np.log(1 - a))
Cost after iteration 500: nan
Cost after iteration 1000: nan
Cost after iteration 1500: nan
Cost after iteration 2000: nan
Cost after iteration 2500: nan
Cost after iteration 3000: nan
Cost after iteration 3500: nan
Cost after iteration 4000: nan
Cost after iteration 4500: nan
Cost after iteration 5000: nan
Cost after iteration 5500: nan
Cost after iteration 6000: nan
Cost after iteration 6500: nan
Cost after iteration 7000: nan
Cost after iteration 7500: nan
Cost after iteration 8000: nan
Cost after iteration 8500: nan
Cost after iteration 9000: nan
Cost after iteration 9500: nan
Accuracy:
0.0
答案 0 :(得分:2)
问题是爆炸渐变。您需要将输入标准化为[0, 1]
。
如果你在训练数据中查看特征3和特征23,你会看到大于3000的值。在这些值与你的初始权重相乘后,它们仍然位于[0, 30]
范围内。因此,在第一次迭代中,z
向量仅包含值为50的正数。因此,a
向量(sigmoid的输出)如下所示:
[0.9994797 0.99853904 0.99358676 0.99999973 0.98392862 0.99983016 0.99818802 ...]
因此,在第一步中,您的模型始终以高置信度预测1。但这并不总是正确的,并且模型输出导致大梯度的高概率,当您查看dw
的最高值时可以看到。就我而言,
dw[3]
是388 dw[23]
是571 ,其他值位于[0, 55]
。因此,您可以清楚地看到这些特征中的大输入如何导致爆炸梯度。由于梯度下降现在朝着相反方向迈出了太大的一步,下一步中的权重不在[0, 0.01]
中,而在[-285, 0.002]
中,这只会使事情变得更糟。在下一次迭代中,z
包含大约-100万的值,这会导致sigmoid函数溢出。
[0, 1]
[-0.01, 0.01]
中使用权重,以便它们大致相互抵消。否则,z
中的值仍然会与您拥有的要素数呈线性关系。至于规范化输入,您可以使用sklearn' MinMaxScaler
:
x_train, x_test, y_train, y_test, m = load_dataset()
scaler = MinMaxScaler()
x_train_normalized = scaler.fit_transform(x_train.T).T
neuralNet = Neural_Network()
'''
step-2 : Train the network
'''
neuralNet.train(x_train_normalized, y_train,10000,m)
# Use the same transformation on the test inputs as on the training inputs
x_test_normalized = scaler.transform(x_test.T).T
y_predicted = neuralNet.predict(x_test_normalized)
.T
是因为sklearn希望训练输入的形状为(num_samples, num_features)
,而x_train
和x_test
的形状为(num_features, num_samples)
。