I have the following problem:
Let's say I have an array defined like this:
A = np.array([[1,2,3],[4,5,6],[7,8,9]])
What I would like to do is to make use of Numpy multiple indexing and set several elements to 0. To do that I'm creating a vector:
indices_to_remove = [1, 2, 0]
What I want it to mean is the following:
Remove element with index '1' from the first row
Remove element with index '2' from the second row
Remove element with index '0' from the third row
The result should be the array [[1,0,3],[4,5,0],[0,8,9]]
I've managed to get values of the elements I would like to modify by following code:
values = np.diagonal(np.take(A, indices, axis=1))
However, that doesn't allow me to modify them. How could this be solved?
You could use integer array indexing to assign those zeros -
A[np.arange(len(indices_to_remove)), indices_to_remove] = 0
Sample run -
In [445]: A
Out[445]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
In [446]: indices_to_remove
Out[446]: [1, 2, 0]
In [447]: A[np.arange(len(indices_to_remove)), indices_to_remove] = 0
In [448]: A
Out[448]:
array([[1, 0, 3],
[4, 5, 0],
[0, 8, 9]])
Related
I have a sparse coo matrix built in python using the scipy library. An example data set looks something like this:
>>> v.toarray()
array([[1, 0, 2, 4],
[0, 0, 3, 1],
[4, 5, 6, 9]])
I would like to add the 0th index and 2nd index together and the 1st index and the and 3rd index together so the shape would change from 3, 4 to 3, 2.
However looking at the docs their sum function doesn't support slicing of some sort. So the only way I have thought of a way to do something like that would be to loop the matrix as an array then use numpy to get the summed values like so:
a_col = []
b_col = []
for x in range(len(v.toarray()):
a_col.append(np.sum(v.toarray()[x, [0, 2]], axis=0))
b_col.append(np.sum(v.toarray()[x, [1, 3]], axis=0))
Then use those values for a_col and b_col to create the matrix again.
But surely there should be a way to handle it with the sum method?
You can add the values with a simple loop and 2d slicing and than take the columns you want
v = np.array([[1, 0, 2, 4],
[0, 0, 3, 1],
[4, 5, 6, 9]])
for i in range(2):
v[:, i] = v[:, i] + v[:, i+2]
print(v[:, :2])
Output
[[ 3 4]
[ 3 1]
[10 14]]
You can use csr_matrix.dot with a special matrix to achieve the same,
csr = csr_matrix(csr.dot(np.array([[1,0,1,0],[0,1,0,1]]).T))
#csr.data
#[ 3, 4, 3, 1, 10, 14]
I have the following numpy array
np.array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
I want to compute the elementwise square of second column and third column only by retaining the first column as it is, yielding the result,
np.array([[1, 4, 9],
[4, 25, 36]],
[7, 64, 81]])
What have I tried
I extracted the first column.
then I extracted the second and third columns. found square, using numpy.square function.
arr1 = arr[:, 0]
arr2 = np.square(arr[:, 1:])
and then concatenated them
np.c_[arr1, arr2]
Is there a single step solution?
Select all elements of the non-first column using [:, 1:].
arr[:, 1:] **= 2
There is a 2-d array like this:
img = [
[[1, 2, 3], [4, 5, 6], [7, 8, 9]],
[[2, 2, 2], [3, 2, 3], [6, 7, 6]],
[[9, 8, 1], [9, 8, 3], [9, 8, 5]]
]
And i just want to get the sum of certain indices which are like this:
indices = [[0, 0], [0, 1]] # which means img[0][0] and img[0][1]
# means here is represents
There was a similar ask about 1-d array in stackoverflow in this link, but it got a error when I tried to use print(img[indices]). Because I want to make it clear that the element of img are those which indicates by indices, and then get the mean sum of it.
Expected output
[5, 7, 9]
Use NumPy:
import numpy as np
img = np.array(img)
img[tuple(indices)].sum(axis = 0)
#array([5, 7, 9])
If the result would be [5, 7, 9] which is sum over the column of the list. Then easy:
img = np.asarray(img)
indices = [[0, 0], [0, 1]]
img[(indices)].sum(axis = 0)
Result:
array([5, 7, 9])
When you supply a fancy index, each element of the index tuple represents a different axis. The shape of the index arrays broadcasts to the shape of the output you get.
In your case, the rows of indices.T are the indices in each axis. You can convert them into an index tuple and append slice(None), which is the programmatic equivalent of :. You can take the mean of the resulting 2D array directly:
img[tuple(indices.T) + (slice(None),)].sum(0)
Another way is to use the splat operator:
img[(*indices.T, slice(None))].sum(0)
Given indexes for each row, how to return the corresponding elements in a 2-d matrix?
For instance, In array of np.array([[1,2,3,4],[4,5,6,7]]) I expect to see the output [[1,2],[4,5]] given indxs = np.array([[0,1],[0,1]]). Below is what I've tried:
a= np.array([[1,2,3,4],[4,5,6,7]])
indxs = np.array([[0,1],[0,1]]) #means return the elements located at 0 and 1 for each row
#I tried this, but it returns an array with shape (2, 2, 4)
a[idxs]
The reason you are getting two times your array is that when you do a[[0,1]] you are selecting the rows 0 and 1 from your array a, which are indeed your entire array.
In[]: a[[0,1]]
Out[]: array([[1, 2, 3, 4],
[4, 5, 6, 7]])
You can get the desired output using slides. That would be the easiest way.
a = np.array([[1,2,3,4],[4,5,6,7]])
a[:,0:2]
Out []: array([[1, 2],
[4, 5]])
In case you are still interested on indexing, you could also get your output doing:
In[]: [list(a[[0],[0,1]]),list(a[[1],[0,1]])]
Out[]: [[1, 2], [4, 5]]
The NumPy documentation gives you a really nice overview on how indexes work.
In [120]: indxs = np.array([[0,1],[0,1]])
In [121]: a= np.array([[1,2,3,4],[4,5,6,7]])
...: indxs = np.array([[0,1],[0,1]]) #
You need to provide an index for the first dimension, one that broadcasts with with indxs.
In [122]: a[np.arange(2)[:,None], indxs]
Out[122]:
array([[1, 2],
[4, 5]])
indxs is (2,n), so you need a (2,1) array to give a (2,n) result
Let say i have the following:
import numpy as np
data = np.array([
[1,2,3],
[1,2,3],
[1,2,3],
[4,5,6],
])
How would I go about changing values in column 3 based on values in column 2? For instance, If column 3 == 3, column 2 = 9.
[[1,9,3],
[1,9,3],
[1,9,3],
[4,5,6]]
I've looked at np.any(), but I can't figure out how to alter the array in place.
You can use Numpy's slicing and indexing to achieve this. Take all the rows where the third column is 3, and change the second column of each of those rows to 9:
>>> data[data[:, 2] == 3, 1] = 9
>>> data
array([[1, 9, 3],
[1, 9, 3],
[1, 9, 3],
[4, 5, 6]])