To answer this question, we have to look at how indexing a multidimensional array works in Numpy. Let's first say you have the array x from your question. The buffer assigned to x will contain 16 ascending integers from 0 to 15. If you access one element, say x[i,j], NumPy has to figure out the memory location of this element relative to the beginning of the buffer. This is done by calculating in effect i*x.shape[1]+j (and multiplying with the size of an int to get an actual memory offset).
If you extract a subarray by basic slicing like y = x[0:2,0:2], the resulting object will share the underlying buffer with x. But what happens if you acces y[i,j]? NumPy can't use i*y.shape[1]+j to calculate the offset into the array, because the data belonging to y is not consecutive in memory.
NumPy solves this problem by introducing strides. When calculating the memory offset for accessing x[i,j], what is actually calculated is i*x.strides[0]+j*x.strides[1] (and this already includes the factor for the size of an int):
x.strides
(16, 4)
When y is extracted like above, NumPy does not create a new buffer, but it does create a new array object referencing the same buffer (otherwise y would just be equal to x.) The new array object will have a different shape then x and maybe a different starting offset into the buffer, but will share the strides with x (in this case at least):
y.shape
(2,2)
y.strides
(16, 4)
This way, computing the memory offset for y[i,j] will yield the correct result.
But what should NumPy do for something like z=x[[1,3]]? The strides mechanism won't allow correct indexing if the original buffer is used for z. NumPy theoretically could add some more sophisticated mechanism than the strides, but this would make element access relatively expensive, somehow defying the whole idea of an array. In addition, a view wouldn't be a really lightweight object anymore.
This is covered in depth in the NumPy documentation on indexing.
Oh, and nearly forgot about your actual question: Here is how to make the indexing with multiple lists work as expected:
x[[[1],[3]],[1,3]]
This is because the index arrays are broadcasted to a common shape. Of course, for this particular example, you can also make do with basic slicing:
x[1::2, 1::2]
To answer this question, we have to look at how indexing a multidimensional array works in Numpy. Let's first say you have the array x from your question. The buffer assigned to x will contain 16 ascending integers from 0 to 15. If you access one element, say x[i,j], NumPy has to figure out the memory location of this element relative to the beginning of the buffer. This is done by calculating in effect i*x.shape[1]+j (and multiplying with the size of an int to get an actual memory offset).
If you extract a subarray by basic slicing like y = x[0:2,0:2], the resulting object will share the underlying buffer with x. But what happens if you acces y[i,j]? NumPy can't use i*y.shape[1]+j to calculate the offset into the array, because the data belonging to y is not consecutive in memory.
NumPy solves this problem by introducing strides. When calculating the memory offset for accessing x[i,j], what is actually calculated is i*x.strides[0]+j*x.strides[1] (and this already includes the factor for the size of an int):
x.strides
(16, 4)
When y is extracted like above, NumPy does not create a new buffer, but it does create a new array object referencing the same buffer (otherwise y would just be equal to x.) The new array object will have a different shape then x and maybe a different starting offset into the buffer, but will share the strides with x (in this case at least):
y.shape
(2,2)
y.strides
(16, 4)
This way, computing the memory offset for y[i,j] will yield the correct result.
But what should NumPy do for something like z=x[[1,3]]? The strides mechanism won't allow correct indexing if the original buffer is used for z. NumPy theoretically could add some more sophisticated mechanism than the strides, but this would make element access relatively expensive, somehow defying the whole idea of an array. In addition, a view wouldn't be a really lightweight object anymore.
This is covered in depth in the NumPy documentation on indexing.
Oh, and nearly forgot about your actual question: Here is how to make the indexing with multiple lists work as expected:
x[[[1],[3]],[1,3]]
This is because the index arrays are broadcasted to a common shape. Of course, for this particular example, you can also make do with basic slicing:
x[1::2, 1::2]
As Sven mentioned, x[[[0],[2]],[1,3]] will give back the 0 and 2 rows that match with the 1 and 3 columns while x[[0,2],[1,3]] will return the values x[0,1] and x[2,3] in an array.
There is a helpful function for doing the first example I gave, numpy.ix_. You can do the same thing as my first example with x[numpy.ix_([0,2],[1,3])]. This can save you from having to enter in all of those extra brackets.
Slicing 2D Numpy arrays
NumPy 2D array slicing. Is it possible?
I can’t say I’d recommend this, but technically it is only one index…
r, c = data.shape data[[*[0]*c,*[*range(1,r-1)]*2,*[r-1]*c],[*range(c),*np.repeat([0,c-1],r-2),*range(c)]] = 0
A slightly nicer approach would be
data[np.pad(np.zeros(np.array(data.shape)-2,dtype=bool),1,constant_values=1)] = 0
An alternative approach would be to just replace the variable with the padded internals
data = np.pad(data[-1:1,-1:1], 1, constant_values=0)More on reddit.com
Problems with slicing 2D array
Numpy array slicing view vs copy confusion
So,here I come again 🤣
I don't get slicing in 2D..
In my lesson,I was taught that using this
d[1:2,1]
means the 2nd element from the last two rows,and 2nd element from 1st column should be sliced..but when I use it I get only one element.Did I do something wrong?Can some of you awesome people hook me up with an explanation?
Here's some code for your palates:
a=[[1,2,3],[4,5,6],[7,8,9]] import numpy as np d=np.array(a) d[1:2,1]