I am working with multidimensional arrays with dynamical axes. Now I want to select elements of the array along a specific axis.
For example, if I have a 3-dimensional array, I want to pick the elements like this
b = a[:, :, 1]
Now my problem is that after one iteration of code the same array becomes 4 dimensional. And again I want to pick the elements like this
b = a[:,:,1,:]
Thus I am looking for a general solution to pick all elements from the 3rd axis of the array. This is very simple if I had to choose a[1] and I could get a[1,:,:,:], but I am not aware how to chose for other axes.
Edit:
Also, I would be interested in a solution where the interested axis also changes for example with the same code and next iteration I would like to get
b = a[:,:,:,1]
Related
I have a Numpy Tensor,
X = np.arange(64).reshape((4,4,4))
I wish to grab the 2,3,4 entries of the first dimension of this tensor, which you can do with,
Y = X[[1,2,3],:,:]
Is this a simpler way of writing this instead of explicitly writing out the indices [1,2,3]? I tried something like [1,:], which gave me an error.
Context: for my real application, the shape of the tensor is something like (30000,100,100). I would like to grab the last (10000, 100,100) to (30000,100,100) of this tensor.
The simplest way in your case is to use X[1:4]. This is the same as X[[1,2,3]], but notice that with X[1:4] you only need one pair of brackets because 1:4 already represent a range of values.
For an N dimensional array in NumPy if you specify indexes for less than N dimensions you get all elements of the remaining dimensions. That is, for N equal to 3, X[1:4] is the same as X[1:4, :, :] or X[1:4, :]. Only if you want to index some dimension while getting all elements in a dimension that comes before it is that you actually need to pass :. Such as X[:, 2:4], for instance.
If you wish to select from some row to the end of array, simply use python slicing notation as below:
X[10000:,:,:]
This will select all rows from 10000 to the end of array and all columns and depths for them.
I want to have a 2D Matrix and then fill the elements of this matrix with different values. I know that I need to create a matrix first with the following definition:
Matrix = np.zeros(10,10)
Now my question is how I can fill of the elements of this matrix by a value lets say the element of [4][7] with value of 5. Thanks
Be careful, because the right sintax for a 10x10 matrix filled by zeros is Matrix = np.zeros((10,10)). Then you can simply write in a different line Matrix[4][7] = 5. I advice you to read a tutorial or a introductory book on Python.
I am trying to implement an algorithm that iteratively removes some rows and columns of a matrix and continues processing the remaining submatrix. However, I would like to know the index of a value in the original matrix rather than the remaining submatrix.
For example, assume that a matrix x is built using
x = np.arange(9).reshape(3, 3)
Now, I would like to find the index of the element that is equal to 8 in the submatrix defined below:
np.where(x[1:, 1:] == 8)
By default, numpy returns (array[1], array[1]) because it is finding the element in the sliced submatrix. What I like to be returned instead is (array[2], array[2]), which is the index of 8 in the original matrix.
What is an efficient solution to this problem?
P.S.
The submatrix may be built arbitrarily. For example, I may need to keep rows, 0 and 1, but columns 0 and 2.
Each submatrix may be sliced in next iterations to make a smaller submatrix. I still would like to have access to the index in the original matrix. In other words, I am looking for a solution that works on submatrices of submatrices as well.
I recently learned about indexing with arrays where submatrices of a matrix can be selected using another numpy array. I think what I can do to solve the problem is to map indices of the submatrix to elements of the indexing array.
For example, in the example above, the submatrix can be defined like this:
row_idx = np.array([1, 2])
col_idx = np.array([1, 2])
np.where(x[row_idx[:, None], col_idx] == 8)
This will still return the same (array[1], array[1]) output, but I can use these indices to lookup the elements of row_idx and col_idx in order to find the corresponding indices in the original matrix, i.e. row_idx[1] and col_idx[1].
I'm making the transition from MATLAB to Numpy and feeling some growing pains.
I have a 3D array, lets say it's 3x3x3 and I want the scalar sum of each plane.
In matlab, I would use:
sum_vec = sum(3dArray,3);
TIA
wbg
EDIT: I was wrong about my matlab code. Matlab only vectorizes in one dim, so a loop wold be required. So numpy turns out to be more elegant...cool.
MATLAB
for i = 1:3
sum_vec(i) = sum(sum(3dArray(:,:,i));
end
You can do
sum_vec = np.array([plane.sum() for plane in cube])
or simply
sum_vec = cube.sum(-1).sum(-1)
where cube is your 3d array. You can specify 0 or 1 instead of -1 (or 2) depending on the orientation of the planes. The latter version is also better because it doesn't use a Python loop, which usually helps to improve performance when using numpy.
You should use the axis keyword in np.sum. Like in many other numpy functions, axis lets you perform the operation along a specific axis. For example, if you want to sum along the last dimension of the array, you would do:
import numpy as np
sum_vec = np.sum(3dArray, axis=-1)
And you'll get a resulting 2D array which corresponds to the sum along the last dimension to all the array slices 3dArray[i, k, :].
UPDATE
I didn't understand exactly what you wanted. You want to sum over two dimensions (a plane). In this case you can do two sums. For example, summing over the first two dimensions:
sum_vec = np.sum(np.sum(3dArray, axis=0), axis=0)
Instead of applying the same sum function twice, you may perform the sum on the reshaped array:
a = np.random.rand(10, 10, 10) # 3D array
b = a.view()
b.shape = (a.shape[0], -1)
c = np.sum(b, axis=1)
The above should be faster because you only sum once.
sumvec= np.sum(3DArray, axis=2)
or this works as well
sumvec=3DArray.sum(2)
Remember Python starts with 0 so axis=2 represent the 3rd dimension.
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.sum.html
If you're trying to sum over a plane (and avoid loops, which is always a good idea) you can use np.sum and pass two axes as a tuple for your argument.
For example, if you have an (nx3x3) array then using
np.sum(a, (1,2))
Will give an (nx1x1), summing over a plane, not a single axis.
Many array methods return a single index despite the fact that the array is multidimensional. For example:
a = rand(2,3)
z = a.argmax()
For two dimensions, it is easy to find the matrix indices of the maximum element:
a[z/3, z%3]
But for more dimensions, it can become annoying. Does Numpy/Scipy have a simple way of returning the indices in multiple dimensions given an index in one (collapsed) dimension? Thanks.
Got it!
a = X.argmax()
(i,j) = unravel_index(a, X.shape)
I don't know of an built-in function that does what you want, but where this
has come up for me, I realized that what I really wanted to do was this:
given 2 arrays a,b with the same shape, find the element of b which is in
the same position (same [i,j,k...] position) as the maximum element of a
For this, the quick numpy-ish solution is:
j = a.flatten().argmax()
corresponding_b_element = b.flatten()[j]
Vince Marchetti