python numpy:在不使用循环的情况下针对不同的条件进行迭代

问题描述

我正在尝试遍历numpy数组并生成一个输出，该输出的条件类似于以下所述:

I am trying to iterate over numpy arrays and generate an output, which has conditions similar to that described below:

min1 = 3
max1 = 1
a1 = np.array([1, 2, 5, 3, 4])
a2 = np.array([5, 2, 6, 2, 1])
output = np.zeros(5)
for i in range(0, 5):
  if((a1[i] - a2[i]) > min1):
    output[i] = 3 * (a1[i] - a2[i])

  if((a1[i] - a2[i]) < max1):
    output = 5 * (a1[i] - a2[i])

我需要优化上面的代码，以便我将numpy功能发挥到最好，并避免使用循环.我该怎么办?

I need to optimize the above code, so that I utilize the numpy functionalities as the best and also avoid using a loop. How should I do it?

推荐答案

虽然像select和where这样的函数可以压缩代码，但我认为了解如何使用基本布尔掩码来实现此目的是一个好主意.它适用于很多情况，并且几乎总是一样快.

While functions like select and where can condense the code, I think it's a good idea to know how to do this with basic boolean masking. It's applicable in many cases, and nearly always as fast.

计算多次使用的差异:

In [432]: diff = a1-a2
In [433]: diff
Out[433]: array([-4,  0, -1,  1,  3])
In [435]: output = np.zeros_like(a1)

找到满足第一个条件的情况，并设置output的相应元素:

find those cases where it meets the first condition, and set the corresponding elements of output:

In [436]: mask1 = diff>min1
In [437]: output[mask1] = 3*diff[mask1]

重复第二个条件

In [438]: mask2 = diff<max1
In [439]: output[mask2] = 5*diff[mask2]

，如果还有更多条件，请再次输入.

and again if there are more conditions.

In [440]: output
Out[440]: array([-20,   0,  -5,   0,   0])

在此示例中，只有-4和-1满足条件2，而没有条件1.

In this example, only the -4 and -1 met condition 2, and none condition 1.

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