Question

如何在没有“全局”的函数外部找到 x 的尺寸（.shape）

import numpy as np

def f():
    x=np.array(([1.2],[3,4],[5,6]),dtype=float)

我也想在这段代码中了解相同的东西，例如 self.o_error 的尺寸。一般来说，一个类中的任何东西

import numpy as np

# X = (hours studying, hours sleeping), y = score on test, xPredicted = 4 hours studying & 8 hours sleeping (input data for prediction)
X = np.array(([2, 9], [1, 5], [3, 6]), dtype=float)
y = np.array(([92], [86], [89]), dtype=float)
xPredicted = np.array(([4,8]), dtype=float)

# scale units
X = X/np.amax(X, axis=0) # maximum of X array
xPredicted = xPredicted/np.amax(xPredicted, axis=0) # maximum of xPredicted (our input data for the prediction)
y = y/100 # max test score is 100

class Neural_Network(object):
  def __init__(self):
  #parameters
    self.inputSize = 2
    self.outputSize = 1
    self.hiddenSize = 3

  #weights
    self.W1 = np.random.randn(self.inputSize, self.hiddenSize) # (3x2) weight matrix from input to hidden layer
    self.W2 = np.random.randn(self.hiddenSize, self.outputSize) # (3x1) weight matrix from hidden to output layer

  def forward(self, X):
    #forward propagation through our network
    self.z = np.dot(X, self.W1) # dot product of X (input) and first set of 3x2 weights
    self.z2 = self.sigmoid(self.z) # activation function
    self.z3 = np.dot(self.z2, self.W2) # dot product of hidden layer (z2) and second set of 3x1 weights
    o = self.sigmoid(self.z3) # final activation function
    return o

  def sigmoid(self, s):
    # activation function
    return 1/(1+np.exp(-s))

  def sigmoidPrime(self, s):
    #derivative of sigmoid
    return s * (1 - s)

  def backward(self, X, y, o):
    # backward propagate through the network
    self.o_error = y - o # error in output
    self.o_delta = self.o_error*self.sigmoidPrime(o) # applying derivative of sigmoid to error

    self.z2_error = self.o_delta.dot(self.W2.T) # z2 error: how much our hidden layer weights contributed to output error
    self.z2_delta = self.z2_error*self.sigmoidPrime(self.z2) # applying derivative of sigmoid to z2 error

    self.W1 += X.T.dot(self.z2_delta) # adjusting first set (input --> hidden) weights
    self.W2 += self.z2.T.dot(self.o_delta) # adjusting second set (hidden --> output) weights

Answer 1

您可以在__init__函数中初始化变量，然后就可以在函数外部甚至在类外部访问变量。

class Bar():
    def __init__(self):
        self.var = 4
    def addone(self):
        self.var +=1
    def showvar(self):
        print(self.var)

foo = Bar()
foo.showvar()   # 4

# you can access variable directly
print(foo.var)  # 4

# change variable using other functions
foo.addone()
foo.showvar()   # 5

Answer 2

最简单的解决方案是：

def f():
    x=np.array(([1.2],[3,4],[5,6]))
    return x
x = f()
x_shape = x.shape

或

def f():
    x=np.array(([1.2],[3,4],[5,6]))
    x_shape = x.shape
    return x, x_shape

x, x_shape = f()

或者，如果您不想修改该功能，并且将两者都放在同一行上，则可以执行以下操作：

def f():
    x=np.array(([1.2],[3,4],[5,6]))
    return x

x, x_shape = f(), f().shape

在Python中访问外部函数和类

2 个答案: