python - 尝试/除外未捕获错误

Question

This question already has answers here:

How do I catch a numpy warning like it's an exception (not just for testing)?

(4个答案)


2年前关闭。




我正在尝试使用try/except来解决被零除的错误，但是except似乎并没有捕获到该错误，因此我不知道为什么。

import numpy as np

A = np.array([[0.5,0,0.25,0.125,0.05],[0.5,0.5,0.5,0.5,0.5]])
Y = np.array([[0,1,0,1,1],[0,0,0,0,1]])
p = np.shape(A)[0]
q = np.shape(A)[1]
m = 2
try:
    dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A)) 
except Exception:
    dA = np.zeros((p,q))

它返回一个零除错误和一个乘法中遇到的无效值(由于零除错误)，但是它杀死了程序，而不是执行除此以外的操作。

RuntimeWarning: divide by zero encountered in true_divide
  dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A))
RuntimeWarning: invalid value encountered in multiply
  dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A))

我尝试指定错误，而不只是异常，但这也不起作用。还有什么可能是问题？

最佳答案

如果为numpy数组，则除以零不会引发Exception，而只会给出警告并给出inf值。

您可以使用numpy.isfinite来检查数组中的值是否为有限值(而不是nan或inf)。

A = np.array([[0.5,0,0.25,0.125,0.05],[0.5,0.5,0.5,0.5,0.5]])
Y = np.array([[0,1,0,1,1],[0,0,0,0,1]])
p = np.shape(A)[0]
q = np.shape(A)[1]
m = 2
try:
    dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A)) 
except Exception:
    dA = np.zeros((p,q))

if not np.isfinite(dA).all():
    dA = np.zeros((p,q))

如果确定您的元素是numpy数组，则不需要异常(exception):

dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A)) 
if not np.isfinite(dA).all():
    dA = np.zeros((p,q))

关于python - 尝试/除外未捕获错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48974703/

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Answer 1

如果为numpy数组，则除以零不会引发Exception，而只会给出警告并给出inf值。

您可以使用numpy.isfinite来检查数组中的值是否为有限值(而不是nan或inf)。

A = np.array([[0.5,0,0.25,0.125,0.05],[0.5,0.5,0.5,0.5,0.5]])
Y = np.array([[0,1,0,1,1],[0,0,0,0,1]])
p = np.shape(A)[0]
q = np.shape(A)[1]
m = 2
try:
    dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A)) 
except Exception:
    dA = np.zeros((p,q))

if not np.isfinite(dA).all():
    dA = np.zeros((p,q))

如果确定您的元素是numpy数组，则不需要异常(exception):

dA = (-1 / m) * ((Y - A) / (A - np.square(A))) * (A * (1 - A)) 
if not np.isfinite(dA).all():
    dA = np.zeros((p,q))