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.
Related
I'm confused by the dimension of a tensor created with tf.zeros(n). For instance, if I write: tf.zeros(6).eval.shape, this will return me (6, ). What dimension is this? is this a matrix of 6 rows and arbitrary # of columns? Or is this a matrix of 6 columns with arbitrary # of rows?
weights = tf.random_uniform([3, 6], minval=-1, maxval=1, seed=1)- this is 3X6 matrix
b=tf.zeros(6).eval- I'm not sure what dimension this is.
Why I am able to add the two like weights+b? If I understand correctly, in order for the two to be added, b needs to be 3X1 dimension.
why i am able to add the two like weights+b?
Operator + is the same as using tf.add() (<obj>.__add__() calls the tf.add() or tf.math.add()) and if you read the documentation it says:
NOTE: math.add supports broadcasting. AddN does not. More about broadcasting here
Now I'm quoting from numpy broadcasting rules (which are the same for tensorflow):
When operating on two arrays, NumPy compares their shapes element-wise. It starts with the trailing dimensions, and works its way forward. Two dimensions are compatible when
they are equal, or
one of them is 1
So you're able to add two tensors with different shapes because they have the same trailing dimensions. If you change the dimension of your weights tensor to, let's say [3, 5], you will get InvalidArgumentError exception because trailing dimensions differ.
(6,) is python syntax for a tuple with 6 as a single element. Hence the shape here is a uni-dimensional vector of length 6.
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].
Let T be a tensor of shape [n,f], which represents a batch. Now I want to slice T into m tensors along axis=0. The value of m depends on the current batch. I have another tensor I of shape [m,2] which stores pairs of indices which indicate where the slices should occur.
I am not really sure how to "iterate" over the indices to apply tf.slice. Any ideas?
Can this somehow be achieved using tf.scan?
I suppose you are looking for the split function.
Suppose I have a 20-by-10 matrix m
and a 20-by-1 vector v, where each element is an integer between 1 to 10.
Is there smart indexing command something like m[:,v]
that would give a vector, where each element i is element of m at the index [i,v[i]]?
No, it seems that you cannot do it. Documentation (http://docs.julialang.org/en/stable/manual/arrays/) says:
If all the indices are scalars, then the result X is a single element from the array A. Otherwise, X is an array with the same number of dimensions as the sum of the dimensionalities of all the indices.
So, to get 1d result from indexing operation you need to have one of the indices to have dimensionality 0, i.e. to be just a scalar -- and you won't get what you want then.
Use comprehension, as proposed in the comment to your question.
To be explicit about the comprehension approach:
[m[i,v[i]] for i = 1:length(v)]
This is concise and clear enough that having a special syntax seems unnecessary.
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.