I am not able to understand how complex indexing - non contiguous indexing of a tensor works. Here is a sample code and its output
import torch
def describe(x):
print("Type: {}".format(x.type()))
print("Shape/size: {}".format(x.shape))
print("Values: \n{}".format(x))
indices = torch.LongTensor([0,2])
x = torch.arange(6).view(2,3)
describe(torch.index_select(x, dim=1, index=indices))
Returns output as
Type: torch.LongTensor Shape/size: torch.Size([2, 2]) Values:
tensor([[0, 2],
[3, 5]])
Can someone explain how did it arrive to this output tensor?
Thanks!
You are selecting the first (indices[0] is 0) and third (indices[1] is 2) tensors from x on the first axis (dim=0). Essentially, torch.index_select with dim=1 works the same as doing a direct indexing on the second axis with x[:, indices].
>>> x
tensor([[0, 1, 2],
[3, 4, 5]])
So selecting columns (since you're looking at dim=1 and not dim=0) which indices are in indices. Imagine having a simple list [0, 2] as indices:
>>> indices = [0, 2]
>>> x[:, indices[0]] # same as x[:, 0]
tensor([0, 3])
>>> x[:, indices[1]] # same as x[:, 2]
tensor([2, 5])
So passing the indices as a torch.Tensor allows you to index on all elements of indices directly, i.e. columns 0 and 2. Similar to how NumPy's indexing works.
>>> x[:, indices]
tensor([[0, 2],
[3, 5]])
Here's another example to help you see how it works. With x defined as x = torch.arange(9).view(3, 3) so we have 3 rows (a.k.a. dim=0) and 3 columns (a.k.a. dim=1).
>>> indices
tensor([0, 2]) # namely 'first' and 'third'
>>> x = torch.arange(9).view(3, 3)
tensor([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
>>> x.index_select(0, indices) # select first and third rows
tensor([[0, 1, 2],
[6, 7, 8]])
>>> x.index_select(1, indices) # select first and third columns
tensor([[0, 2],
[3, 5],
[6, 8]])
Note: torch.index_select(x, dim, indices) is equivalent to x.index_select(dim, indices)
Related
Given a 1-d tensor:
A = torch.tensor([1, 2, 3, 4])
suppose we have some "indexer tensor"
ind1 = torch.tensor([3, 0, 1])
ind2 = torch.tensor([[3, 0], [1, 2]])
as we run A[ind1] & A[ind2]
we get results tensor([4, 1, 2]) & tensor([[4, 1],[2, 3]])
which is the same shape of the indexed tensor (ind1 and ind2) and its value are mapped from tensor A.
I want to ask how can I index on higher dimension tensors?
Currently I have one solution:
For a N-d tensor A, suppose we have the indexer tensor IND,
IND is like [[i11, i12, ... i1N], [i21, i22, ... i2N], ...[iM1, i22, ... iMN], where M is the number of indexed elements.
We can divide IND into N tensors, where
IND_1 = torch.tensor([i11, i21, ... iM1])
...
IND_N = torch.tensor([i1N, i2N, ... iMN])
as we run A[IND_1, ... IND_N], we got tensor(v1, v2, ... vM)
Example:
A = tensor([[1, 2], [3, 4]], [[5, 6], [7, 8]]]) # [2 * 2 * 2]
ind1 = tensor([1, 0, 1])
ind2 = tensor([1, 1, 0])
ind3 = tensor([0, 1, 0])
A[ind1, ind2, ind3]
=> tensor([7, 4, 5])
# and the good thing is you can control the shape of result tensor by modifying the inds' shape.
ind1 = tensor([[0, 0], [1, 0]])
ind2 = tensor([[1, 1], [0, 1]])
ind3 = tensor([[0, 1], [0, 0]])
A[ind1, ind2, ind3]
=> tensor([[3, 4],[5, 3]]) # same as inds' shape
Anyone has more elegant solutions?
1- Manual approach using unraveled indices on flattened input.
If you want to index on an arbitrary number of axes (all axes of A) then one straightforward approach is to flatten all dimensions and unravel the indices. Let's assume that A is 3D and we want to index it using a stack of ind1, ind2, and ind3:
>>> ind = torch.stack((ind1, ind2, ind3))
You can first unravel the indices using A's strides:
>>> unraveled = torch.tensor(A.stride()) # ind.flatten(1)
Then flatten A, index it with unraveled and reshape to the final form:
>>> A.flatten()[unraveled].reshape_as(ind[0])
2- Using a simple split of ind.
You can actually perform the same operation using torch.chunk:
>>> A[ind.chunk(len(ind))][0]
Or alternatively torch.split which is identical:
>>> A[ind.split(1)][0]
3- Initial answer for single-axis indexing.
Let's take a minimal multi-dimensional example with A being a 2-D tensor defined as:
>>> A = torch.tensor([[1, 2, 3, 4],
[5, 6, 7, 8]])
From your description of the problem:
the same shape of index tensor and its value are mapped from tensor A.
Then the indexer tensor would require to have the same shape as the indexed tensor A, since this one is no longer flat. Otherwise, what would the result of A (shaped (2, 4)) indexed by ind1 (shape (3,)) be?
If you are indexing on a single dimension then you can utilize torch.gather:
>>> A.gather(1, ind2)
tensor([[4, 1],
[6, 7]])
I was wondering how I would vectorize this for loop. Given a 2x2x2 array x and an array where each element is the ith, jth, and kth element of the array I want to get x[i,j,k]
Given an arrays x and y
x = np.arange(8).reshape((2, 2, 2))
y = [[0, 1, 1], [1, 1, 0]]
I want to get:
x[0, 1, 1] = 3 and x[1, 1, 0] = 6
I tried:
print(x[y])
But it prints:
array([[2, 3],
[6, 7],
[4, 5]])
So I ended up doing:
for y_ in y:
print(x[y_[0], y_[1], y_[2]])
Which works, but I can't help but think there is a better way.
Use transposed y i.e zip(*y) as the index; You need to have the indices for each dimension as an element for advanced indexing to work:
x[tuple(zip(*y))]
# array([3, 6])
How can the rows in an array be sorted without that the values in each row will changed?
Furthermore: how to get the indicies of this sort-process?
input:
a = np.array([[4,3],[0,3],[3,0],[1,3],[1,2],[2,0]])
required sorting arrray:
b = np.array([1,4,3,5,2,0])
a = a[b]
output:
a = np.array([[0,3],[1,2],[1,3][2,0],[3,0],[4,3]])
How do I get the array b ?
You need lexsort here:
b = np.lexsort((a[:, 1], a[:, 0]))
# array([1, 4, 3, 5, 2, 0], dtype=int64)
And applied to your initial array:
>>> a[b]
array([[0, 3],
[1, 2],
[1, 3],
[2, 0],
[3, 0],
[4, 3]])
As #miradulo pointed out, you may also use:
b = np.lexsort(np.fliplr(a).T)
Which is less verbose than explicitly stating the columns to sort on.
For example
x = np.repeat(np.array([[1,2],[3,4]]), 2, axis=1)
gives you
x = array([[1, 1, 2, 2],
[3, 3, 4, 4]])
but is there something which can perform
x = np.*inverse_repeat*(np.array([[1, 1, 2, 2],[3, 3, 4, 4]]), axis=1)
and gives you
x = array([[1,2],[3,4]])
Regular slicing should work. For the axis you want to inverse repeat, use ::number_of_repetitions
x = np.repeat(np.array([[1,2],[3,4]]), 4, axis=0)
x[::4, :] # axis=0
Out:
array([[1, 2],
[3, 4]])
x = np.repeat(np.array([[1,2],[3,4]]), 3, axis=1)
x[:,::3] # axis=1
Out:
array([[1, 2],
[3, 4]])
x = np.repeat(np.array([[[1],[2]],[[3],[4]]]), 5, axis=2)
x[:,:,::5] # axis=2
Out:
array([[[1],
[2]],
[[3],
[4]]])
This should work, and has the exact same signature as np.repeat:
def inverse_repeat(a, repeats, axis):
if isinstance(repeats, int):
indices = np.arange(a.shape[axis] / repeats, dtype=np.int) * repeats
else: # assume array_like of int
indices = np.cumsum(repeats) - 1
return a.take(indices, axis)
Edit: added support for per-item repeats as well, analogous to np.repeat
For the case where we know the axis and the repeat - and the repeat is a scalar (same value for all elements) we can construct a slicing index like this:
In [1117]: a=np.array([[1, 1, 2, 2],[3, 3, 4, 4]])
In [1118]: axis=1; repeats=2
In [1119]: ind=[slice(None)]*a.ndim
In [1120]: ind[axis]=slice(None,None,a.shape[axis]//repeats)
In [1121]: ind
Out[1121]: [slice(None, None, None), slice(None, None, 2)]
In [1122]: a[ind]
Out[1122]:
array([[1, 2],
[3, 4]])
#Eelco's use of take makes it easier to focus on one axis, but requires a list of indices, not a slice.
But repeat does allow for differing repeat counts.
In [1127]: np.repeat(a1,[2,3],axis=1)
Out[1127]:
array([[1, 1, 2, 2, 2],
[3, 3, 4, 4, 4]])
Knowing axis=1 and repeats=[2,3] we should be able construct the right take indexing (probably with cumsum). Slicing won't work.
But if we only know the axis, and the repeats are unknown then we probably need some sort of unique or set operation as in #redratear's answer.
In [1128]: a2=np.repeat(a1,[2,3],axis=1)
In [1129]: y=[list(set(c)) for c in a2]
In [1130]: y
Out[1130]: [[1, 2], [3, 4]]
A take solution with list repeats. This should select the last of each repeated block:
In [1132]: np.take(a2,np.cumsum([2,3])-1,axis=1)
Out[1132]:
array([[1, 2],
[3, 4]])
A deleted answer uses unique; here's my row by row use of unique
In [1136]: np.array([np.unique(row) for row in a2])
Out[1136]:
array([[1, 2],
[3, 4]])
unique is better than set for this use since it maintains element order. There's another problem with unique (or set) - what if the original had repeated values, e.g. [[1,2,1,3],[3,3,4,1]].
Here is a case where it would be difficult to deduce the repeat pattern from the result. I'd have to look at all the rows first.
In [1169]: a=np.array([[2,1,1,3],[3,3,2,1]])
In [1170]: a1=np.repeat(a,[2,1,3,4], axis=1)
In [1171]: a1
Out[1171]:
array([[2, 2, 1, 1, 1, 1, 3, 3, 3, 3],
[3, 3, 3, 2, 2, 2, 1, 1, 1, 1]])
But cumsum on a known repeat solves it nicely:
In [1172]: ind=np.cumsum([2,1,3,4])-1
In [1173]: ind
Out[1173]: array([1, 2, 5, 9], dtype=int32)
In [1174]: np.take(a1,ind,axis=1)
Out[1174]:
array([[2, 1, 1, 3],
[3, 3, 2, 1]])
>>> import numpy as np
>>> x = np.repeat(np.array([[1,2],[3,4]]), 2, axis=1)
>>> y=[list(set(c)) for c in x] #This part remove duplicates for each array in tuple. So this will not work for x = np.repeat(np.array([[1,1],[3,3]]), 2, axis=1)=[[1,1,1,1],[3,3,3,3]. Result will be [[1],[3]]
>>> print y
[[1, 2], [3, 4]]
You dont need know to axis and repeat amount...
I would like to make an array out of specific components of a tensor. I have found the wonderful command np.argwhere(). This returns the indices of the tensor meeting a specific criteria, however it does not name them as components of the tensor, i.e. they come back as [0,0,1,1] versus x[0,0,1,1] for a tensor x.
Is there a built in or slick way to grab the components of a tensor that meet a certain criteria where the components are written with their indices and the name of the tensor attached?
You can use where instead of argwhere...
>>> x = np.arange(6).reshape(2,3)
>>> x
array([[0, 1, 2],
[3, 4, 5]])
>>> np.argwhere(x > 1)
array([[0, 2],
[1, 0],
[1, 1],
[1, 2]])
>>> np.where(x > 1)
(array([0, 1, 1, 1]), array([2, 0, 1, 2]))
>>> x[np.where(x > 1)]
array([2, 3, 4, 5])