I have an issue with numpy arrays and I can't understand what I am doing wrong. I need to create a 100x100 matrix with random int (non zero) and the last row should be the combination of all previous rows. Here is my code:
non_zero_m = np.random.randint(0,10,(99,100))
arr = non_zero_m.sum(axis=0)
singular_m = np.concatenate((non_zero_m, arr))
ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 2 dimension(s) and the array at index 1 has 1 dimension(s)
I can't understand why python shows that arrays has different dimensions
The problem is that arr is a 1-dimensional array, and you are trying to concatenate it to a matrix (2-dimensional).
Just replace the second line with:
arr = non_zero_m.sum(axis=0).reshape(1, -1)
This reshapes arr to a 2-dimensonal array, such that the first axis has dimension 1 (thus making arr effectively a row vector), and the second axis has the required dimension to keep all of arr's elements (this is the meaning of -1 in this context).
Related
I've started to learn NumPy, when I create an array and then invoke the .shape function, I understand how it works for most cases. However, the result does not make sense to me for a single-dimensional array. Can someone please explain the outcome?
array = np.array([4,5,6])
print(array.shape)
The outcome is (3,)
Output tulpe of ints in the "np.shape" function gives the lengths of the corresponding array dimension, and this tuple will be (n,m), in which n and m indicate row and columns, respectively. For a single dimension array, this Tuple will be just (n,), in which n indicates the number of array elements.
I have two arrays A1(5,3) and A2(5,2) and want to create a new array A(5,5). I've try to use A=np.concatenate((A1,A2)) but it gives me the error
all the input array dimensions for the concatenation axis must match exactly, but along dimension 1, the array at index 0 has size 3 and the array at index 1 has size 2
How can I solve this?
To use np.concatenate you should specify on which axis the concatenation is happening. In this case, you should use:
A = np.concatenate((A1, A2), axis=1)
Other way to solve this issue is to use horizontal stack:
A = np.hstack((A1, A2))
I want to reshape this array: np.array(np.arange(15)) to a 3d array that is built from a 3x3 array and a 3x2 array.
I've tried to do it with the reshape method but it didn't work.
I thought that maybe reshape can get a number of tuples maybe.
a=np.array(np.arange(15)).reshape(1,((3,2),(3,3)))
but I then I saw it cant.
How can I reshape it then? is there a nice way?
a multidimensional array can't have dimensions with different size.
but if you want a tuple you will need to split the array in 2 parts, the first that match in size with the 3x3 array and the second that match the 3x2, at this point you'll have 2 one dimensional array, then reshape them
arr1 = arr1.reshape((3,3))
arr2 = arr2.reshape((3,2))
tuple = arr1, arr2
I have a 2D numpy array with the shape (3024, 4032).
I have a 3D numpy array with the shape (3024, 4032, 3).
2D numpy array is filled with 0s and 1s.
3D numpy array is filled with values between 0 and 255.
By looking at the 2D array values, I want to change the values in 3D array. If a value in 2D array is 0, I will change the all 3 pixel values in 3D array into 0 along the last axes. If a value in 2D array is 1, I won't change it.
I have checked this question, How to filter a numpy array with another array's values, but it applies for 2 arrays which have same dimensions. In my case, dimensions are different.
How the filtering is applied in two arrays, with same size on 2 dimensions, but not size on the last dimension?
Ok, I'll answer this to highlight one pecularity regarding "missing" dimensions. Lets' assume a.shape==(5,4,3) and b.shape==(5,4)
When indexing, existing dimensions are left aligned which is why #Divakar's solution a[b == 0] = 0 works.
When broadcasting, existing dimensions are right aligned which is why #InvaderZim's a*b does not work. What you need to do is a*b[..., None] which inserts a broadcastable dimension at the right
I think this one is very simple:
If a is a 3D array (a.shape == (5, 4, 3)) filled with values, and b is a 2D array (b.shape == (5, 4)) filled with 1 and 0, then reshape b and multiply them:
a = a * b.reshape(5, 4, 1)
Numpy will automatically expand the arrays as needed.
Note: I'm using numpy
import numpy as np
Given 4 arrays of the same (but arbitrary) shape, I am trying to write a function that forms 2x2 matrices from each corresponding element of the arrays, finds the eigenvalues, and returns two arrays of the same shape as the original four, with its elements being eigenvalues (i.e. the resulting arrays would have the same shape as the input, with array1 holding all the first eigenvalues and array2 holding all the second eigenvalues).
I tried doing the following, but unsurprisingly, it gives me an error that says the array is not square.
temp = np.linalg.eig([[m1, m2],[m3, m4]])[0]
I suppose I can make an empty temp variable in the same shape,
temp = np.zeros_like(m1)
and go over each element of the original arrays and repeat the process. My problem is that I want this generalised for arrays of any arbitrary shape (need not be one dimensional). I would guess that finding the shape of the arrays and designing loops to go over each element would not be a very good way of doing it. How do I do this efficiently?
Construct a 2x2x... array:
temp = np.array([[m1, m2], [m3, m4]])
Move the first two dimensions to the end for a ...x2x2 array:
for _ in range(2):
temp = np.rollaxis(temp, 0, temp.ndim)
Call np.linalg.eigvals (which broadcasts) for a ...x2 array of eigenvalues:
eigvals = np.linalg.eigvals(temp)
And split this into an array of first eigenvalues and an array of second eigenvalues:
eigvals1, eigvals2 = eigvals[..., 0], eigvals[..., 1]