当条件为2时，有效地更新numpy数组

Suppose that you have a 1D numpy array containing elements that are either 3 or 4

假设您有一个包含3或4个元素的numpy数组

for example:

例如：

I would like to alter this array so that if an element is 3 then it should become some number X and if the element is 4 it should become some number Y. It is guaranteed that X is different than Y. For the above array we would get:

我想改变这个数组，这样如果一个元素是3，那么它应该变成一个数字X，如果元素是4，它应该变成一个数字Y.保证X不同于Y.对于上面的数组我们会得到：

X
Y
Y
Y
X
X
X

I was thinking about doing something like this:

我在考虑做这样的事情：

arr[arr==3]=X
arr[arr==4]=Y

however, what if X is 4? Then in the end the entire array will contain just Ys.

但是，如果X是4怎么办？然后最后整个数组将只包含Ys。

I am trying to avoid using a for loop for performance reasons, but if it's the only way to go, I can afford following that route.

我试图避免出于性能原因使用for循环，但如果这是唯一的方法，我可以负担得起这条路线。

2 个解决方案

#1

Since the input 1D array contains only 3 or 4, you can use np.where, like so -

由于输入1D数组只包含3或4，你可以使用np.where，就像这样 -

np.where(A==3,X,Y)

Sample run #1 -

样品运行＃1 -

In [55]: A
Out[55]: array([4, 3, 4, 3, 3, 3, 4, 4, 4, 4, 3, 3, 4, 3, 4])

In [56]: X = 30; Y = 40

In [57]: np.where(A==3,X,Y)
Out[57]: array([40, 30, 40, 30, 30, 30, 40, 40, 40, 40, 30, 30, 40, 30, 40])

Sample run #2 (if X is 4) -

样品运行＃2（如果X为4） -

In [60]: A
Out[60]: array([4, 3, 4, 3, 3, 3, 4, 4, 4, 4, 3, 3, 4, 3, 4])

In [61]: X = 4; Y = 40

In [62]: np.where(A==3,X,Y)
Out[62]: array([40,  4, 40,  4,  4,  4, 40, 40, 40, 40,  4,  4, 40,  4, 40])

#2

This is a typical use for the map function. A way of implementing it would be:

这是地图功能的典型用法。实现它的方法是：

result = map(lambda k: X if k==3 else Y, myarray)

result = map（lambda k：X if k == 3 else Y，myarray）

You can replace k == 3 with any function that returns a boolean like so:

你可以用任何返回boolean的函数替换k == 3，如下所示：

result = map(lambda k: X if myfun(k) else Y, myarray)

result = map（lambda k：X if myfun（k）else Y，myarray）

You may want to consider this post, however you will not get much speed up.

你可能想要考虑这篇文章，但是你不会加快速度。

#1