我正在调试一个过去有效的简单程序.我已经挑出了发生错误的指令,但我无法弄清楚是什么触发它.我已经阅读了与WinError 10061相关的所有问题,但我没有看到明确的答案
urllib.request.urlopen('http://www.wikipedia.org/')
Traceback (most recent call last):
File "C:\Python33\lib\urllib\request.py", line 1248, in do_open h.request(req.get_method(), req.selector, req.data, headers)
File "C:\Python33\lib\http\client.py", line 1061, in request self._send_request(method, url, body, headers)
File "C:\Python33\lib\http\client.py", line 1099, in _send_request self.endheaders(body)
File "C:\Python33\lib\http\client.py", line 1057, in endheaders self._send_output(message_body)
File "C:\Python33\lib\http\client.py", line 902, in _send_output self.send(msg)
File "C:\Python33\lib\http\client.py", line 840, in send self.connect()
File "C:\Python33\lib\http\client.py", line 818, in connect self.timeout, self.source_address)
File "C:\Python33\lib\socket.py", line 435, in create_connection raise err
File "C:\Python33\lib\socket.py", line 426, in create_connection sock.connect(sa)
ConnectionRefusedError: …我正在运行一个从网站下载文件的程序。我已经介绍了一个异常处理 urllib.error.HTTPError，但现在我不时遇到其他我不确定如何捕获的错误：http.client.IncompleteRead。我是否只需将以下内容添加到底部的代码中？
except http.client.IncompleteRead:
我必须添加多少个例外才能确保程序不会停止？并且我是否必须将它们全部添加到同一个Except 语句或多个Except 语句中。
try:
# Open a file object for the webpage
f = urllib.request.urlopen(imageURL)
# Open the local file where you will store the image
imageF = open('{0}{1}{2}{3}'.format(dirImages, imageName, imageNumber, extension), 'wb')
# Write the image to the local file
imageF.write(f.read())
# Clean up
imageF.close()
f.close()
except urllib.error.HTTPError: # The 'except' block executes if an HTTPError is thrown by the try block, then the program continues as usual.
print ("Image fetch failed.")
我有一个大小为(m,1)的向量v,其元素是从1:n中选取的整数.我想创建一个大小为(m,n)的矩阵M,如果v(i)= j则其元素M(i,j)为1,否则为0.我不想使用循环,并希望仅将其实现为简单的向量矩阵操作.
所以我首先考虑创建一个包含重复元素的矩阵
M = v * ones(1,n) % this is a (m,n) matrix of repeated v
例如,v = [1,1,3,2]'m = 4且n = 3
M =
1 1 1
1 1 1
3 3 3
2 2 2
然后我需要创建一个大小为(1,n)的比较向量c
c = 1:n
1 2 3
然后我需要进行一系列逻辑比较
M(1,:)==c % this results in [1,0,0]
.
M(4,:)==c % this results in [0,1,0]
但是,我认为应该可以在紧凑矩阵表示法中执行遍历每一行的最后步骤,但是我很难理解并且对索引不够了解.最终结果应该是
M =
1 0 0
1 0 0
0 0 1
0 1 0