I'm a little confused on indexing in python. I have the following array
[[ 2 4]
[ 6 8]
[10 12]
[14 16]]
and I want to obtain array([4, 2]) as my output. I tried using
Q4= [Q3[0,1],Q3[0,0]]
and my output come out as [4, 2]. I'm missing "array ("Any pointers on indexing in Python ? Thanks!!
Just slice the 1st row reversed:
Q3[0,::-1]
array([4, 2])
While one option would be to just wrap your result in another call to numpy.array(),
np.array([Q3[0,1],Q3[0,0]])
it would be better practice and probably more performant to use integer advanced indexing. In that case, you can just use your indices from before to make one vectorized call to numpy.ndarray.__getitem__ (instead of two independent calls).
Q3[[0, 0], [1, 0]]
Edit: RomanPerekhrest's answer is definitely better in this situation, my answer would only be useful for arbitrary array indices.
Related
Given I have the number of axes, can I specify the number of axes to the type hint npt.NDArray (from import numpy.typing as npt)
i.e. if I know it is a 3D array, how can I do npt.NDArray[3, np.float64]
On Python 3.9 and 3.10 the following does the job for me:
data = [[1, 2, 3], [4, 5, 6]]
arr: np.ndarray[Tuple[Literal[2], Literal[3]], np.dtype[np.int_]] = np.array(data)
It is a bit cumbersome, but you might follow numpy issue #16544 for future development on easier specification.
In particular, for now you must declare the full shape and can't only declare the rank of the array.
In the future something like ndarray[Shape[:, :, :], dtype] should be available.
According to the documentation of numpy.ravel,
Return a contiguous flattened array.
A 1-D array, containing the elements of the input, is returned. A copy is made only if needed.
For convenience and efficiency of indexing, I would like to have a one-dimensional view of a 2-dimensional array. I am using ravel for creating the view, and so far so good.
However, it is not clear to me what is meant by "A copy is made only if needed." If some day a copy is created while my code is executed, the code will stop working.
I know that there is numpy.reshape, but its documentation says:
It is not always possible to change the shape of an array without copying the data.
In any case, I would like the data to be contiguous.
How can I reliably create at 2-dimensional array and a 1-dimensional view into it? I would like the data to be contiguous in memory (for efficiency). Are there any attributes to specify when creating the 2-dimensional array to assure that it is contiguous and ravel will not need to copy it?
Related question: What is the difference between flatten and ravel functions in numpy?
The warnings for ravel and reshape are the same. ravel is just reshape(-1), to 1d. Conversely reshape docs tells us that we can think of it as first doing a ravel.
Normal array construction produces a contiguous array, and reshape with the same order will produce a view. You can visually test that by looking at the ravel and checking if the values appear in the expected order.
In [348]: x = np.arange(6).reshape(2,3)
In [349]: x
Out[349]:
array([[0, 1, 2],
[3, 4, 5]])
In [350]: x.ravel()
Out[350]: array([0, 1, 2, 3, 4, 5])
I started with the arange, reshaped it to 2d, and back to 1d. No change in order.
But if I make a sliced view:
In [351]: x[:,:2]
Out[351]:
array([[0, 1],
[3, 4]])
In [352]: x[:,:2].ravel()
Out[352]: array([0, 1, 3, 4])
This ravel has a gap, and thus is a copy.
Transpose is also a view, which cannot be reshaped to a view:
In [353]: x.T
Out[353]:
array([[0, 3],
[1, 4],
[2, 5]])
In [354]: x.T.ravel()
Out[354]: array([0, 3, 1, 4, 2, 5])
Except, if we specify the right order, the ravel is a view.
In [355]: x.T.ravel(order='F')
Out[355]: array([0, 1, 2, 3, 4, 5])
reshape has a extensive discussion of order. And transpose actually works by returning a view with different shape and strides. For a 2d array transpose produces a order F array.
So as long as you are aware of manipulations like this, you can safely assume that the reshape/ravel is contiguous.
Note that even though [354] is a copy, assignment to the flat changes the original
In [361]: x[:,:2].flat[:] = [3,4,2,1]
In [362]: x
Out[362]:
array([[3, 4, 2],
[2, 1, 5]])
x[:,:2].ravel()[:] = [10,11,2,3] does not change x. In cases like this y = x[:,:2].flat may be more useful than the ravel equivalent.
I am using an ndarray to slice another ndarray.
Normally I use arr[ind_arr]. numpy seems to not like this and raises a FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated use arr[tuple(seq)] instead of arr[seq].
What's the difference between arr[tuple(seq)] and arr[seq]?
Other questions on StackOverflow seem to be running into this error in scipy and pandas and most people suggest the error to be in the particular version of these packages. I am running into the warning running purely in numpy.
Example posts:
FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated use `arr[tuple(seq)]` instead of `arr[seq]`
FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated use `arr[tuple(seq)]`
FutureWarning with distplot in seaborn
MWE reproducing warning:
import numpy as np
# generate a random 2d array
A = np.random.randint(20, size=(7,7))
print(A, '\n')
# define indices
ind_i = np.array([1, 2, 3]) # along i
ind_j = np.array([5, 6]) # along j
# generate index array using meshgrid
ind_ij = np.meshgrid(ind_i, ind_j, indexing='ij')
B = A[ind_ij]
print(B, '\n')
C = A[tuple(ind_ij)]
print(C, '\n')
# note: both produce the same result
meshgrid returns a list of arrays:
In [50]: np.meshgrid([1,2,3],[4,5],indexing='ij')
Out[50]:
[array([[1, 1],
[2, 2],
[3, 3]]), array([[4, 5],
[4, 5],
[4, 5]])]
In [51]: np.meshgrid([1,2,3],[4,5],indexing='ij',sparse=True)
Out[51]:
[array([[1],
[2],
[3]]), array([[4, 5]])]
ix_ does the same thing, but returns a tuple:
In [52]: np.ix_([1,2,3],[4,5])
Out[52]:
(array([[1],
[2],
[3]]), array([[4, 5]]))
np.ogrid also produces the list.
In [55]: arr = np.arange(24).reshape(4,6)
indexing with the ix tuple:
In [56]: arr[_52]
Out[56]:
array([[10, 11],
[16, 17],
[22, 23]])
indexing with the meshgrid list:
In [57]: arr[_51]
/usr/local/bin/ipython3:1: FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use `arr[tuple(seq)]` instead of `arr[seq]`. In the future this will be interpreted as an array index, `arr[np.array(seq)]`, which will result either in an error or a different result.
#!/usr/bin/python3
Out[57]:
array([[10, 11],
[16, 17],
[22, 23]])
Often the meshgrid result is used with unpacking:
In [62]: I,J = np.meshgrid([1,2,3],[4,5],indexing='ij',sparse=True)
In [63]: arr[I,J]
Out[63]:
array([[10, 11],
[16, 17],
[22, 23]])
Here [I,J] is the same as [(I,J)], making a tuple of the 2 subarrays.
Basically they are trying to remove a loophole that existed for historical reasons. I don't know if they can change the handling of meshgrid results without causing further compatibility issues.
For Future Readers having this FutureWarning: ... and want to correctly resolve them.
Read the answer of #hpaulj here. And notice the point is when he(she) said:
[...] returns a list of [...]
[...] returns a tuple of [...]
If you don't know why 1. is the point please read the answer: https://stackoverflow.com/a/71487259/5290519, which is also written by him(her). This answer provides:
Why, in the first place, the usage of list as index will cause a warning.
[4] is a problem because in the past, certain lists were interpreted as though they were tuples. This is a legacy case that developers are trying to cleanup, hence the FutureWarning.
A series of concise examples on the (correct) usage of list,tuple,np.array as array index.
If you still don't get the point, try my own answer after I figuring all these complicated concepts: https://stackoverflow.com/a/71493474/5290519
Given an index array index and, say, a matrix A I want a matrix B with the corresponding permutation of the columns of A.
In Numpy I would do the following,
>>> A = np.arange(6).reshape(2,3); A
array([[0, 1, 2],
[3, 4, 5]])
>>> index = [2,0,1]
>>> A[:,index]
array([[2, 0, 1],
[5, 3, 4]])
Is there a natural or efficient way to do this in MXNet? The functions pick() and take() don't seem to work in this way. I managed to come up with the following but it's not elegant.
>>> mx.nd.take(A.T, mx.nd.array([[2],[0],[1]])).T.reshape((2,3))
[[ 2. 0. 1.]
[ 5. 3. 4.]]
<NDArray 2x3 #cpu(0)>
Finally, to throw a wrench into the works, is there a way to do this in-place?
Update Here is a slightly more elegant, but presumably not as efficient (due to the transposition), version of above:
>>> mx.nd.take(A.T, mx.nd.array([2,0,1])).T
[[ 2. 0. 1.]
[ 5. 3. 4.]]
<NDArray 2x3 #cpu(0)>
What you need is the so-called advanced indexing in MXNet. There is a PR submitted for getting elements through advanced indexing from MXNet NDArray and will add the functionality of setting elements to NDArray as well. It is expected to come out in the release 1.0.
https://github.com/apache/incubator-mxnet/pull/8246
Can you intuitively explain or give more examples about tf.gather_nd for indexing and slicing into high-dimensional tensors in Tensorflow?
I read the API, but it is kept quite concise that I find myself hard to follow the function's concept.
Ok, so think about it like this:
You are providing a list of index values to index the provided tensor to get those slices. The first dimension of the indices you provide is for each index you will perform. Let's pretend that tensor is just a list of lists.
[[0]] means you want to get one specific slice(list) at index 0 in the provided tensor. Just like this:
[tensor[0]]
[[0], [1]] means you want get two specific slices at indices 0 and 1 like this:
[tensor[0], tensor[1]]
Now what if tensor is more than one dimensions? We do the same thing:
[[0, 0]] means you want to get one slice at index [0,0] of the 0-th list. Like this:
[tensor[0][0]]
[[0, 1], [2, 3]] means you want return two slices at the indices and dimensions provided. Like this:
[tensor[0][1], tensor[2][3]]
I hope that makes sense. I tried using Python indexing to help explain how it would look in Python to do this to a list of lists.
You provide a tensor and indices representing locations in that tensor. It returns the elements of the tensor corresponding to the indices you provide.
EDIT: An example
import tensorflow as tf
sess = tf.Session()
x = [[1,2,3],[4,5,6]]
y = tf.gather_nd(x, [[1,1],[1,2]])
print(sess.run(y))
[5, 6]