I'm pretty new to all programming and have been trying to get a grasp on Python (2.7.6), OpenCV, and ROS to be able to use a neural network in a robot. I am now stuck in an OpenCV (2.4.8) problem with getting the correct HSV-values, in short, it suddenly changes from red (around 5 for Hue) or blue (about 110 for Hue) to green (about 40/50) while I'm only moving the camera/presenting the
我对所有编程都很陌生，一直试图掌握Python（2.7.6），OpenCV和ROS，以便能够在机器人中使用神经网络。 我现在被困在OpenCV（2.4.8）的问题中，获得正确的HSV值，简而言之，它突然从红色（约为Hue 5）或蓝色（Hue约为110）变为绿色（约为40/50 ），而我只是移动相机/在屏幕的另一侧显示图像。 class CV: def __init__(self): # Initialize object self.cap = cv2.VideoCapture(0) # '0' should stay or w
I'm trying to find the shortest path from a vertex to another of a connected, unweighted graph. In this problem,the distance from a vertex to its adjacent vertex will be equal to 1.ie., if a graph with edges (a,b),(a,c) is considered, the distance from a to b and c will be 1 and the distance from b to c will be 2. Also, an adjacency list is maintained to store all the adjacent vertices of e
我试图找到从顶点到另一个连接的未加权图形的最短路径。 在这个问题中，如果考虑具有边（a，b），（a，c）的图，则从顶点到其相邻顶点的距离将等于1.ie。从a到b和c的距离将会为1并且从b到c的距离将为2.此外，维护邻接列表以存储每个顶点的所有相邻顶点。 那么，是否有任何算法可以找到给定问题的所有最短路径？ 你可以使用dijkstra的算法来找到距离。 这是使用networkx的一种方法 In [28]: import networkx as nx 创建
I tried the following: #!/usr/bin/env python import keras from keras.models import model_from_yaml model_file_path = 'model-301.yaml' weights_file_path = 'model-301.hdf5' # Load network with open(model_file_path) as f: yaml_string = f.read() model = model_from_yaml(yaml_string) model.load_weights(weights_file_path) model.compile(optimizer='adagrad', loss='binary_crossentropy') # Visualiz
我尝试了以下内容： #!/usr/bin/env python import keras from keras.models import model_from_yaml model_file_path = 'model-301.yaml' weights_file_path = 'model-301.hdf5' # Load network with open(model_file_path) as f: yaml_string = f.read() model = model_from_yaml(yaml_string) model.load_weights(weights_file_path) model.compile(optimizer='adagrad', loss='binary_crossentropy') # Visualize fr
In Python, how can I convert a string like this: Thu, 16 Dec 2010 12:14:05 +0000 to ISO 8601 format, while keeping the timezone? Please note that the orginal date is string, and the output should be string too, not datetime or something like that. I have no problem to use third parties libraries, though. 使用dateutil： import dateutil.parser as parser text = 'Thu, 16 Dec 2010 12:14:05 +0
在Python中，我如何转换一个像这样的字符串： Thu，16 Dec 2010 12:14:05 +0000 到ISO 8601格式，同时保持时区？ 请注意，原始日期是字符串，输出也应该是字符串，而不是datetime或类似的东西。 不过，我没有问题使用第三方库。 使用dateutil： import dateutil.parser as parser text = 'Thu, 16 Dec 2010 12:14:05 +0000' date = parser.parse(text) print(date.isoformat()) # 2010-12-16T12:14:05+00:00 Python内
This question already has an answer here: How to parse an ISO 8601-formatted date? 23 answers I prefer using the dateutil library for timezone handling and generally solid date parsing. If you were to get an ISO 8601 string like: 2010-05-08T23:41:54.000Z you'd have a fun time parsing that with strptime, especially if you didn't know up front whether or not the timezone was included.
这个问题在这里已经有了答案： 如何解析ISO 8601格式的日期？ 23个答案 我更喜欢使用dateutil库进行时区处理，并且通常使用固定日期解析。 如果你想得到一个ISO 8601字符串，例如：2010-05-08T23：41：54.000Z你可以用strptime解析这个有趣的时间，特别是如果你不知道时区是否包含在内。 pyiso8601有几个问题（检查他们的跟踪器），我在使用过程中碰到过，并且在几年内还没有更新。 相比之下，dateutil一直活跃并为我工
This is my code: import datetime today = datetime.date.today() print today This prints: 2008-11-22 which is exactly what I want BUT....I have a list I'm appending this to and then suddenly everything goes "wonky". Here is the code: import datetime mylist = [] today = datetime.date.today() mylist.append(today) print mylist This prints the following: [datetime.date(2008, 11, 22)
这是我的代码： import datetime today = datetime.date.today() print today 这打印出来：2008-11-22这正是我想要的，但是....我有一个列表，我将它附加到然后突然一切都会变得“不可靠”。 代码如下： import datetime mylist = [] today = datetime.date.today() mylist.append(today) print mylist 这将打印以下内容： [datetime.date(2008, 11, 22)] 我怎样才能得到像“2008-11-22”这样简单的日期？ WHY：日期是对象
I stripped some tags that I thought were unnecessary from an XML file. Now when I try to parse it, my SAX parser throws an error and says my file is not well-formed. However, I know every start tag has an end tag. The file's opening tag has a link to an XML schema. Could this be causing the trouble? If so, then how do I fix it? Edit: I think I've found the problem. My character da
我从XML文件中剥离了一些我认为不必要的标签。 现在当我尝试解析它时，我的SAX解析器会抛出一个错误，并说我的文件格式不正确。 但是，我知道每个开始标签都有一个结束标签。 该文件的开始标记具有指向XML模式的链接。 这是否会造成麻烦？ 如果是这样，那我该如何解决？ 编辑：我想我已经找到了问题。 我的角色数据包含“＆lt”和“＆gt”字符，大概来自html标签。 经过分析后，它们被转换为“<”和“>”字符，这似乎打扰
I'm building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don't, however, have the absolute path to the directory where the templates are stored. I do have a relative path from the script but when I call the script it treats that as a path relative to the current working directory. Is there a way to specify that
我正在构建一个简单的助手脚本，用于将我们的代码库中的几个模板文件复制到当前目录。 但是，我不具有存储模板的目录的绝对路径。 我确实有一个来自脚本的相对路径，但是当我调用脚本时，它会将它视为相对于当前工作目录的路径。 有没有一种方法可以指定这个相对的URL来自脚本的位置呢？ 在具有该脚本的文件中，您希望执行如下操作： import os dirname = os.path.dirname(__file__) filename = os.path.join(dirname, 'rel
I am using Kubuntu 16.04 with kde 5.6.4. I have installed anaconda python 3.5 which includes ipython qtconsole . To launch ipython qtconsole, I have to type ipython qtconsole in terminal. Is there anyway I can create a launcher for it? I know there a package, but it doesn't link to the anaconda python 3.5 and I don't want another separate python 3.5. abhishek ~ $ apt-cache search i
我使用Kubuntu 16.04和kde 5.6.4。 我已经安装了包含ipython qtconsole的 anaconda python 3.5。 要启动ipython qtconsole，我必须在终端中键入ipython qtconsole 。 无论如何，我可以为它创建一个启动器？ 我知道那里有一个软件包，但它并没有链接到anaconda python 3.5，我不希望另一个单独的python 3.5。 abhishek ~ $ apt-cache search ipython3-qtconsole ipython3 - enhanced interactive Python 3 shell ipython3-q
Update: The list are filled with strings I edited the list to show this I have 3 different list such as Section = [('1', '1.1', '1.2'), ('1', '2', '2.2', '3'), ('1', '1.2', '3.2', '3.5')] Page = [('1', '1', '3'), ('1', '2', '2', '2'), ('1', '2', '3', '5')] Titles = [('General', 'Info', 'Titles'), ('More', 'Info', 'Section', 'Here'), ('Another', 'List', 'Of', 'Strings')] I want to combine the
更新：列表中充满了我编辑列表的字符串以显示此内容 我有3个不同的列表，如 Section = [('1', '1.1', '1.2'), ('1', '2', '2.2', '3'), ('1', '1.2', '3.2', '3.5')] Page = [('1', '1', '3'), ('1', '2', '2', '2'), ('1', '2', '3', '5')] Titles = [('General', 'Info', 'Titles'), ('More', 'Info', 'Section', 'Here'), ('Another', 'List', 'Of', 'Strings')] 我想把它们结合起来，比如 Combined_List = [('1', '1.1'