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I want to sort 2d array column-wise consequently, so if the values in one column are equal then sorting is performed by next column.
For example array
[[1, 0, 4, 2, 3]
[0, 1, 5, 7, 4]
[0, 0, 6, 1, 0]]
must be sorted as
[[0, 0, 6, 1, 0]
[0, 1, 5, 7, 4]
[1, 0, 4, 2, 3]]
So rows must not be changed, only their order. How can I do that?
This should work
import numpy as np
a = np.array([[1, 0, 4, 2, 3],[0, 1, 5, 7, 4],[0, 0, 6, 1, 0]])
np.sort(a.view('i8,i8,i8,i8,i8'), order=['f0'], axis=0).view(np.int)
I get
array([[0, 0, 6, 1, 0],
[0, 1, 5, 7, 4],
[1, 0, 4, 2, 3]])
f0 is the column which you want to sort by.
I'm trying to mark the value and indices of max values in a 3D array, getting the max in the third axis.
Now this would have been obvious in a lower dimension:
argmaxes=np.argmax(array)
maximums=array[argmaxes]
but NumPy doesn't understand the second syntax properly for higher than 1D.
Let's say my 3D array has shape (8,8,250). argmaxes=np.argmax(array,axis=-1)would return a (8,8) array with numbers between 0 to 250. Now my expected output is an (8,8) array containing the maximum number in the 3rd dimension. I can achieve this with maxes=np.max(array,axis=-1) but that's repeating the same calculation twice (because I need both values and indices for later calculations)
I can also just do a crude nested loop:
for i in range(8):
for j in range(8):
maxes[i,j]=array[i,j,argmaxes[i,j]]
But is there a nicer way to do this?
You can use advanced indexing. This is a simpler case when shape is (8,8,3):
arr = np.random.randint(99, size=(8,8,3))
x, y = np.indices(arr.shape[:-1])
arr[x, y, np.argmax(array,axis=-1)]
Sample run:
>>> x
array([[0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1, 1],
[2, 2, 2, 2, 2, 2, 2, 2],
[3, 3, 3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4, 4, 4],
[5, 5, 5, 5, 5, 5, 5, 5],
[6, 6, 6, 6, 6, 6, 6, 6],
[7, 7, 7, 7, 7, 7, 7, 7]])
>>> y
array([[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7],
[0, 1, 2, 3, 4, 5, 6, 7]])
>>> np.argmax(arr,axis=-1)
array([[2, 1, 1, 2, 0, 0, 0, 1],
[2, 2, 2, 1, 0, 0, 1, 0],
[1, 2, 0, 1, 1, 1, 2, 0],
[1, 0, 0, 0, 2, 1, 1, 0],
[2, 0, 1, 2, 2, 2, 1, 0],
[2, 2, 0, 1, 1, 0, 2, 2],
[1, 1, 0, 1, 1, 2, 1, 0],
[2, 1, 1, 1, 0, 0, 2, 1]], dtype=int64)
This is a visual example of array to help to understand it better:
I have a quick question about the numpy unique function. I want to return the unique column values for each row
import numpy as np
a = np.array([[3, 2, 3, 2, 1, 3, 1, 2, 1, 3, 1, 2, 2, 2, 3, 3],
[3, 2, 3, 2, 3, 3, 3, 3, 2, 2, 3, 1, 2, 1, 2, 1],
[3, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 3, 1, 1]]) # a.shape is (3,16)
np.unique(a)
array([1, 2, 3]) # not what I want
np.unique(a,axis=1)
array([[1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3],
[2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3],
[2, 3, 2, 3, 2, 3, 2, 1, 1, 2, 3]]) # also not what I want, and I'm not even sure what its doing
np.apply_along_axis(np.unique,1,a)
array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3]]) # this is what I want
The problem is that I also want to use other features of np.unqiue, like returning index values. Can anyone help me to get np.unique to work by itself?
You can loop over rows and collect unique values:
import numpy as np
a = np.array([[3, 2, 3, 2, 1, 3, 1, 2, 1, 3, 1, 2, 2, 2, 3, 3],
[3, 2, 3, 2, 3, 3, 3, 3, 2, 2, 3, 1, 2, 1, 2, 1],
[3, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 3, 1, 1]])
arr = np.empty((0,3), int)
for row in a:
arr = np.append(arr, np.array([np.unique(a)]), axis=0)
Output:
[[1 2 3]
[1 2 3]
[1 2 3]]
numpy will not be able to return a matrix with rows of different sizes. your example has exactly 3 distinct values per row which makes np.apply_along_axis work but if you had a value of 4 in one of the rows or only 1s and 2s on a row it would fail.
To obtain what you are looking for you will need to use a normal Python list as the result. You can build it using a list comprehension:
import numpy as np
a = np.array([[1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 1],
[3, 2, 3, 2, 3, 3, 3, 3, 2, 2, 3, 1, 2, 1, 2, 1],
[3, 3, 3, 2, 3, 3, 4, 2, 2, 2, 3, 2, 2, 3, 1, 1]])
r = [ np.unique(row) for row in a ]
print(r)
# [array([1, 2]), array([1, 2, 3]), array([1, 2, 3, 4])]
r = [ np.unique(row,return_index=True)for row in a ]
print(r)
# [(array([1, 2]), array([0, 1])),
# (array([1, 2, 3]), array([11, 1, 0])),
# (array([1, 2, 3, 4]), array([14, 3, 0, 6]))]
One thing you could do is build a mask of the values that are the first of their kind on each row. This can be done using numpy.
Here's one way to do it (hopefully, numpy experts could suggest something less convoluted):
np.sum(np.cumsum(np.cumsum(a==np.unique(a)[:,None,None],axis=2),axis=2)==1,axis=0)
array([[1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0]])
Such a mask offers many processing options such as finding indices of the first occurrence on each line (using np.argwhere), erasing/assigning first or subsequent occurrences, and more.
I generated a lower triangular matrix, and I want to complete the matrix using the values in the lower triangular matrix to form a square matrix, symmetrical around the diagonal zeros.
lower_triangle = numpy.array([
[0,0,0,0],
[1,0,0,0],
[2,3,0,0],
[4,5,6,0]])
I want to generate the following complete matrix, maintaining the zero diagonal:
complete_matrix = numpy.array([
[0, 1, 2, 4],
[1, 0, 3, 5],
[2, 3, 0, 6],
[4, 5, 6, 0]])
Thanks.
You can simply add it to its transpose:
>>> m
array([[0, 0, 0, 0],
[1, 0, 0, 0],
[2, 3, 0, 0],
[4, 5, 6, 0]])
>>> m + m.T
array([[0, 1, 2, 4],
[1, 0, 3, 5],
[2, 3, 0, 6],
[4, 5, 6, 0]])
You can use the numpy.triu_indices or numpy.tril_indices:
>>> a=np.array([[0, 0, 0, 0],
... [1, 0, 0, 0],
... [2, 3, 0, 0],
... [4, 5, 6, 0]])
>>> irows,icols = np.triu_indices(len(a),1)
>>> a[irows,icols]=a[icols,irows]
>>> a
array([[0, 1, 2, 4],
[1, 0, 3, 5],
[2, 3, 0, 6],
[4, 5, 6, 0]])
This question already has answers here:
Python — How can I find the square matrix of a lower triangular numpy matrix? (with a symmetrical upper triangle)
(2 answers)
Closed 9 years ago.
I generated a lower triangular matrix, and I want to complete the matrix using the values in the lower triangular matrix to form a square matrix.
lower_triangle = numpy.array([
[0,0,0,0],
[1,0,0,0],
[2,3,0,0],
[4,5,6,0]])
I want to generate the following complete matrix, maintaining the zero diagonal:
complete_matrix = numpy.array([
[0, 6, 5, 4],
[1, 0, 3, 2],
[2, 3, 0, 1],
[4, 5, 6, 0]])
Thanks.
How about:
>>> m
array([[0, 0, 0, 0],
[1, 0, 0, 0],
[2, 3, 0, 0],
[4, 5, 6, 0]])
>>> np.rot90(m,2)
array([[0, 6, 5, 4],
[0, 0, 3, 2],
[0, 0, 0, 1],
[0, 0, 0, 0]])
>>> m + np.rot90(m, 2)
array([[0, 6, 5, 4],
[1, 0, 3, 2],
[2, 3, 0, 1],
[4, 5, 6, 0]])
See also fliplr(m)[::-1], etc.
without any addition:
>>> a=np.array([[0, 0, 0, 0],
... [1, 0, 0, 0],
... [2, 3, 0, 0],
... [4, 5, 6, 0]])
>>> irows,icols = np.triu_indices(len(a),1)
>>> a[irows,icols]=a[icols,irows]
>>> a
array([[0, 1, 2, 4],
[1, 0, 3, 5],
[2, 3, 0, 6],
[4, 5, 6, 0]])