Currently, if I want to compare pressure under each of the paws of a dog, I only compare the pressure underneath each of the toes. But I want to try and compare the pressures underneath the entire paw. But to do so I have to rotate them, so the toes overlap (better). Because most of the times the left and right paws are slightly rotated externally, so if you can't simply project one on to
目前，如果我想比较一只狗的每只爪子的压力，我只比较每只脚趾下面的压力。 但我想尝试比较整个爪子下面的压力。 但要做到这一点，我必须旋转它们，所以脚趾重叠（更好）。 因为大多数时候左右爪都是从外部轻微旋转的，所以如果你不能简单地将其中一个放在另一个之上。 因此，我想旋转爪子，所以它们都以同样的方式对齐。 目前，我通过使用脚趾检测来查看两个中间脚趾和后方脚趾来计算旋转角度，然后计算黄色线（脚趾绿色
In my previous question I got an excellent answer that helped me detect where a paw hit a pressure plate, but now I'm struggling to link these results to their corresponding paws: I manually annotated the paws (RF=right front, RH= right hind, LF=left front, LH=left hind). As you can see there's clearly a repeating pattern and it comes back in almost every measurement. Here's a li
在我之前的问题中，我得到了一个很好的答案，帮助我发现爪子碰到压力板的位置，但现在我正在努力将这些结果与他们相应的爪子联系起来： 我手动注释了爪子（RF =右前方，RH =右后方，LF =左前方，LH =左后方）。 正如你所看到的那样，显然有一种重复的模式，几乎每次测量都会回来。 以下是一个手动注释的6个试验的演示文稿的链接。 我最初的想法是使用启发式进行排序，如： 前爪和后爪之间的重量承受比例约为60-40％。
I've had some really awesome help on my previous questions for detecting paws and toes within a paw, but all these solutions only work for one measurement at a time. Now I have data that consists off: about 30 dogs; each has 24 measurements (divided into several subgroups); each measurement has at least 4 contacts (one for each paw) and each contact is divided into 5 parts and has
对于我以前的问题，在爪子内部检测爪子和脚趾方面，我有一些非常棒的帮助，但所有这些解决方案一次只能用于一次测量。 现在我拥有包含关闭的数据： 约30只狗; 每个测量有24个测量值（分成几个子组）; 每个测量至少有4个触点（每个爪子一个）和 每个联系人分为5个部分和 有几个参数，如接触时间，位置，总力等。 显然，将所有东西都粘在一个大对象上并不会削减它，所以我想我需要使用类而不是当前的函数。 但是，
After my previous question on finding toes within each paw, I started loading up other measurements to see how it would hold up. Unfortunately, I quickly ran into a problem with one of the preceding steps: recognizing the paws. You see, my proof of concept basically took the maximal pressure of each sensor over time and would start looking for the sum of each row, until it finds on that != 0.0
在我之前关于在每个爪子内找到脚趾的问题之后，我开始加载其他测量以了解它将如何保持。 不幸的是，我很快遇到了上述步骤之一的问题：识别爪子。 你看，我的概念验证基本上是随着时间的推移逐个传感器的最大压力，并开始寻找每一行的总和，直到找到！= 0.0。 然后它对列执行相同的操作，只要它发现多于2行的数据再次为零。 它将最小和最大行数和列值存储到某个索引。 正如您在图中所看到的，在大多数情况下，这种方式非常
In my previous question a lot of users wanted me to give some more data to toy with. So I got working on exporting all my data and processing it with Python, but then I realized: where do I leave all this data? Well I decided the best thing would be to stick them in a database, so at least I don't have to parse the raw files every time. But since I know nothing about databases this is tur
在我之前的问题中，很多用户希望我提供更多的数据供玩家使用。 所以我开始着手导出所有数据并使用Python进行处理，但后来我意识到：我在哪里可以保留所有这些数据？ 那么我决定最好的办法是将它们粘在一个数据库中，所以至少我不必每次都解析原始文件。 但由于我对数据库一无所知，这变得相当混乱。 我尝试了一些教程来创建一个sqlite数据库，添加一个表和字段，并尝试插入我的numpy.arrays，但它不能让它工作。 通常我每
I get the error: TypeError: __init__() takes exactly 2 arguments (3 given) When trying to instantiate an object from the class Top: super(Middle1, self).__init__(name, "middle") class Base(object): def __init__(self, name, type): pass class Middle1(Base): def __init__(self, name): super(Middle1, self).__init__(name, "middle1") class Middle2(Base): def
我得到的错误：TypeError：__init __（）只需要2个参数（给出3） 当试图从类Top实例化一个对象时： super（Middle1，self）.__ init __（name，“middle”） class Base(object): def __init__(self, name, type): pass class Middle1(Base): def __init__(self, name): super(Middle1, self).__init__(name, "middle1") class Middle2(Base): def __init__(self, name): super(Middle2
This question already has an answer here: How to invoke the super constructor? 5 answers The way you are doing it is indeed the recommended one (for Python 2.x). The issue of whether the class is passed explicitly to super is a matter of style rather than functionality. Passing the class to super fits in with Python's philosophy of "explicit is better than implicit". Pytho
这个问题在这里已经有了答案： 如何调用超级构造函数？ 5个答案 你这样做的方式确实是推荐的（对于Python 2.x）。 这个类是否被明确地传递给super是一个风格问题而不是功能问题。 传递类super符合Python的“显式优于隐式”的哲学。 Python 3包含一个改进的super（），它允许像这样使用： super().__init__(args) 你可以简单地写： class A(object): def __init__(self): print "Constructor A was called"
I'm running Python 2.5, so this question may not apply to Python 3. When you make a diamond class hierarchy using multiple inheritance and create an object of the derived-most class, Python does the Right Thing (TM). It calls the constructor for the derived-most class, then its parent classes as listed from left to right, then the grandparent. I'm familiar with Python's MRO; that&#
我正在运行Python 2.5，所以这个问题可能不适用于Python 3.当您使用多继承创建钻石类层次结构并创建派生类的对象时，Python会执行Right Thing（TM）。 它调用派生类的构造函数，然后从左到右列出父类，然后是祖父类。 我熟悉Python的MRO; 那不是我的问题。 我很好奇从超级对象返回的对象实际上是如何与父类中的超级调用进行通信的。 考虑这个示例代码： #!/usr/bin/python class A(object): def __init__(self): print
I'm going through the free Python class Google has available (Link if interested) and I'm stuck on one of the exercises. Goal: Consider dividing a string into two halves. If the length is even, the front and back halves are the same length. If the length is odd, we'll say that the extra char goes in the front half. eg 'abcde', the front half is 'abc', the back hal
我正在浏览Google提供的免费Python类（如果感兴趣，请链接），并且我被困在其中一个练习中。 目标：考虑将字符串分成两半。 如果长度均匀，则前后两半长度相同。 如果长度很奇怪，我们会说额外的字符在前半部分。 例如'abcde'，前半部分是'abc'，后半部分是'de'。 给定2个字符串a和b，返回形式为a-front + b-front + a-back + b-back的字符串 这是我到目前为止： def front_back(a, b): if
Sometimes I break long conditions in if s onto several lines. The most obvious way to do this is: if (cond1 == 'val1' and cond2 == 'val2' and cond3 == 'val3' and cond4 == 'val4'): do_something Isn't very very appealing visually, because the action blends with the conditions. However, it is the natural way using correct Python indentation of 4 spaces. For the moment I'm
有时我打破长的条件下if小号多个行。 最明显的做法是： if (cond1 == 'val1' and cond2 == 'val2' and cond3 == 'val3' and cond4 == 'val4'): do_something 视觉上不太吸引人，因为动作与条件融合在一起。 但是，这是使用4个空格的正确Python缩进的自然方式。 目前我正在使用： if ( cond1 == 'val1' and cond2 == 'val2' and cond3 == 'val3' and cond4 == 'val4'): do_something