脾气暴躁的地方有多个条件 [英] Numpy where function multiple conditions

问题描述

我有一个称为dists的距离数组.我想选择两个值之间的距离.我编写了以下代码来做到这一点:

I have an array of distances called dists. I want to select dists which are between two values. I wrote the following line of code to do that:

 dists[(np.where(dists >= r)) and (np.where(dists <= r + dr))]

但这仅针对条件选择

 (np.where(dists <= r + dr))

如果我使用临时变量按顺序执行命令，则可以正常工作.上面的代码为什么不起作用，如何使它起作用?

If I do the commands sequentially by using a temporary variable it works fine. Why does the above code not work, and how do I get it to work?

欢呼

推荐答案

在您的特定情况下的最佳方法是将两个条件更改为一个条件:

The best way in your particular case would just be to change your two criteria to one criterion:

dists[abs(dists - r - dr/2.) <= dr/2.]

它只会创建一个布尔数组，在我看来更容易阅读，因为它说是dist在dr还是r内?(尽管我会重新定义r成为您感兴趣区域的中心，而不是起点，所以r = r + dr/2.)但这不能回答您的问题.

It only creates one boolean array, and in my opinion is easier to read because it says, is dist within a dr or r? (Though I'd redefine r to be the center of your region of interest instead of the beginning, so r = r + dr/2.) But that doesn't answer your question.

问题的答案:
如果您只是想过滤出不符合条件的dists元素，则实际上并不需要where:

The answer to your question:
You don't actually need where if you're just trying to filter out the elements of dists that don't fit your criteria:

dists[(dists >= r) & (dists <= r+dr)]

因为&将为您提供元素级的and(必须带括号).

Because the & will give you an elementwise and (the parentheses are necessary).

或者，如果您出于某些原因确实想使用where，则可以执行以下操作:

Or, if you do want to use where for some reason, you can do:

 dists[(np.where((dists >= r) & (dists <= r + dr)))]

---

原因:
它不起作用的原因是因为np.where返回的是索引列表，而不是布尔数组.您试图在两个数字列表之间获得and，而这些数字当然没有您期望的True/False值.如果a和b均为True值，则a and b返回b.所以说类似[0,1,2] and [2,3,4]这样的话只会给你[2,3,4].它在起作用:

---

Why:
The reason it doesn't work is because np.where returns a list of indices, not a boolean array. You're trying to get and between two lists of numbers, which of course doesn't have the True/False values that you expect. If a and b are both True values, then a and b returns b. So saying something like [0,1,2] and [2,3,4] will just give you [2,3,4]. Here it is in action:

In [230]: dists = np.arange(0,10,.5)
In [231]: r = 5
In [232]: dr = 1

In [233]: np.where(dists >= r)
Out[233]: (array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),)

In [234]: np.where(dists <= r+dr)
Out[234]: (array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12]),)

In [235]: np.where(dists >= r) and np.where(dists <= r+dr)
Out[235]: (array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12]),)

例如，您希望比较的只是布尔数组

What you were expecting to compare was simply the boolean array, for example

In [236]: dists >= r
Out[236]: 
array([False, False, False, False, False, False, False, False, False,
       False,  True,  True,  True,  True,  True,  True,  True,  True,
        True,  True], dtype=bool)

In [237]: dists <= r + dr
Out[237]: 
array([ True,  True,  True,  True,  True,  True,  True,  True,  True,
        True,  True,  True,  True, False, False, False, False, False,
       False, False], dtype=bool)

In [238]: (dists >= r) & (dists <= r + dr)
Out[238]: 
array([False, False, False, False, False, False, False, False, False,
       False,  True,  True,  True, False, False, False, False, False,
       False, False], dtype=bool)

现在您可以在组合的布尔数组上调用np.where了:

Now you can call np.where on the combined boolean array:

In [239]: np.where((dists >= r) & (dists <= r + dr))
Out[239]: (array([10, 11, 12]),)

In [240]: dists[np.where((dists >= r) & (dists <= r + dr))]
Out[240]: array([ 5. ,  5.5,  6. ])

或者使用 fancy indexing

In [241]: dists[(dists >= r) & (dists <= r + dr)]
Out[241]: array([ 5. ,  5.5,  6. ])

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