R和Python在处理基本数组操作方面的区别 [英] Difference between R and Python in handling basic arrays operations

问题描述

我在Python中具有以下代码：

I have the following code in Python :

n = 3
m = 2

y = np.random.normal(loc = 0, scale = 1, size = (n, m))
y_diff = np.expand_dims(y, 1) - np.expand_dims(y, 0)

我想将其转换为R。据我所知，它使用$ y_i创建了一个$ nxnxm $数组- y_j $作为值。

Which I want to translate to R. As I understand it creates a $ n x n x m $ array, with $y_i - y_j$ as values.

我找到了一种将expand_dims从python转换为R的方法（请参阅：将Python np.expand_dims转换为R ）。

I have found a way to translate expand_dims from python to R (see: Translating Python np.expand_dims to R).

现在在R中具有以下代码：

And now have the following code in R:

m = 2
n = 3
x = array(rnorm(m*n),c(m,n))

以下两行似乎都起作用：

Both following lines appear to work :

expand_dims(x,1)
expand_dims(x,2)

但不是：

expand_dims(x,2) - expand_dims(x,1)

返回：

Error in expand_dims(x, 2) - expand_dims(x, 1) : non-conformable arrays

我的直觉是Python和R处理数组基本操作的方式有所不同。

My intuition is that there is a difference in how Python and R handle basic operations on their arrays.

关于如何使其在R中工作的任何想法？

Any idea on how to make it work in R ?

推荐答案

我认为 listarrays :: expand_dims（）不能真正地按照您的预期工作。我认为这就是您的问题所在。您应该可以通过比较

I think listarrays::expand_dims() is not truly working how you expect it to; I think that's where your issue is. You should be able to see this by comparing

np.expand_dims(y, 1)

与

listarrays::expand_dims(x, 2)

Python的 numpy 和R都按元素相减，所以这不是问题。我认为您最好直接在R中操作数组。出于说明的目的，我将使用更简单的nxm矩阵

Python's numpy and R both subtract element-wise, so that's not the issue. I think you're better off just manipulating the array directly in R. I will use a simpler n x m matrix for the purposes of exposition

1 2
3 4
5 6

然后在Python中我们有了

Then in Python we have

z = np.array([[1, 2], [3, 4], [5, 6]])
z

array([[1, 2],
       [3, 4],
       [5, 6]])

np.expand_dims(z, 1) - np.expand_dims(z, 0)

array([[[ 0,  0],
        [-2, -2],
        [-4, -4]],

       [[ 2,  2],
        [ 0,  0],
        [-2, -2]],

       [[ 4,  4],
        [ 2,  2],
        [ 0,  0]]])

和R

n <- 3
m <- 2
z <- matrix(1:(n*m), nrow = n, byrow = TRUE)
z
#      [,1] [,2]
# [1,]    1    2
# [2,]    3    4
# [3,]    5    6
array(rep(t(z), each = 3), dim = c(n, m, n)) - array(z, dim = c(n, m, n))
# , , 1
# 
#      [,1] [,2]
# [1,]    0    0
# [2,]   -2   -2
# [3,]   -4   -4
# 
# , , 2
# 
#      [,1] [,2]
# [1,]    2    2
# [2,]    0    0
# [3,]   -2   -2
# 
# , , 3
# 
#      [,1] [,2]
# [1,]    4    4
# [2,]    2    2
# [3,]    0    0

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