Numpy where 函数多个条件 [英] Numpy where function multiple conditions

问题描述

我有一个称为 dist 的距离数组.我想选择介于两个值之间的 dist.我编写了以下代码行来做到这一点:

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))]

不过这仅针对条件选择

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

如果我使用临时变量按顺序执行命令，它工作正常.为什么上面的代码不起作用，我如何让它工作?

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?

干杯

推荐答案

您的特殊情况的最佳方法就是将您的两个标准更改为一个标准:

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.]

它只创建一个布尔数组，在我看来更容易阅读，因为它说，is dist 在 dr 或 r?(虽然我将 r 重新定义为您感兴趣区域的中心而不是开头，所以 r = r + dr/2.) 但这并不能回答你的问题.

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:
如果您只是想过滤掉不符合条件的 dists 元素，则实际上不需要 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 & 会给你一个元素的 和 (括号是必需的).

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

或者，如果您出于某种原因确实想使用 where，您可以这样做:

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

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

<小时>

原因:
它不起作用的原因是因为 np.where 返回一个索引列表，而不是一个布尔数组.您正在尝试在两个数字列表之间获取 and，这当然没有您期望的 True/False 值.如果a 和b 都是True 值，则a 和b 返回b.所以说[0,1,2]和[2,3,4]之类的东西只会给你[2,3,4].这是在行动:

---

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]),)

您期望比较的只是布尔数组，例如

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)

现在你可以在组合的布尔数组上调用 np.where :

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. ])

或者简单地使用花式索引

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

这篇关于Numpy where 函数多个条件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！