Given K, I need to have all the possibile combinations of K x 2 numpy matrices so that in each matrix there are all 0s except for two 1s in different rows and columns.
Something like this for K = 5:
[[1,0],[0,1],[0,0],[0,0][0,0]]
[[1,0],[0,0],[0,1],[0,0][0,0]]
[[1,0],[0,0],[0,0],[0,1][0,0]]
[[1,0],[0,0],[0,0],[0,0][0,1]]
[[0,0],[1,0],[0,1],[0,0][0,0]]
[[0,0],[1,0],[0,0],[0,1][0,0]]
... and so on
So the resulting array should be a K x 2 x (K*(K-1)/2).
I want to avoid loops since it's not an efficient way when K is big enough (in my specific case K = 300)
I can't think of an elegant solution but here's a not-so-elegant pure numpy one:
import numpy as np
def combination_matrices(K):
# get combination indices
i, j = np.indices((K, K))
comb_indices = np.transpose((i < j).nonzero()) # (num_combs, 2) array where ones are
num_combs = comb_indices.shape[0] # K*(K-1)/2
# create a matrix of the desired shape, first axis enumerates combinations
matrices = np.zeros((num_combs, K, 2), dtype=int)
# broadcasting assignment of ones
comb_range, col_index = np.ogrid[:num_combs, :2]
matrices[comb_range, comb_indices, col_index] = 1
return matrices
This first uses the indices of a (K, K)-shaped array to find the index pairs for every combination (these are indices that encode the upper triangle of the array, excluding the diagonal). Then we use a bit tricky broadcasting assignment (heavy fancy indexing) to set each corresponding element of the pre-allocated output array to 1.
Note that I put the K*(K-1)/2-sized axis first, because this makes the most sense in numpy with C-contiguous memory layout. This way when you take the matrix for combination index 3, arr[3, ...] will be a contiguous chunk of memory of shape (K, 2) that's fast to work with in vectorised operations.
The output for K = 4:
[[[1 0]
[0 1]
[0 0]
[0 0]]
[[1 0]
[0 0]
[0 1]
[0 0]]
[[1 0]
[0 0]
[0 0]
[0 1]]
[[0 0]
[1 0]
[0 1]
[0 0]]
[[0 0]
[1 0]
[0 0]
[0 1]]
[[0 0]
[0 0]
[1 0]
[0 1]]]
This is an oddly specific question, but an interesting problem, I'd love to know what the context is?
You are looking for all permutations of a multiset , python's itertools doesn't currently support this. So simplest solution is to use the multiset tools of the sympy library.
The following code took about ~2.5 minutes to run on my machine, so is fairly fast for a single thread. You're looking at 179700 unique permutations for K=300.
(I took inspiration from https://stackoverflow.com/a/40289807/10739860)
from collections import Counter
from math import factorial, prod
import numpy as np
from sympy.utilities.iterables import multiset_permutations
from tqdm import tqdm
def No_multiset_permutations(multiset: list) -> int:
"""Calculates the No. possible permutations given a multiset.
See: https://en.wikipedia.org/wiki/Permutation#Permutations_of_multisets
:param multiset: List representing a multiset.
"""
value_counts = Counter(multiset).values()
denominator = prod([factorial(val) for val in value_counts])
return int(factorial(len(multiset)) / denominator)
def multiset_Kx2_permutations(K: int) -> np.ndarray:
"""This will generate all possible unique Kx2 permutations of an array
withsize K where two values are 1 and the rest are 0.
:param K: The size of the array.
"""
# Construct number multiset, e.g. K=5 gives [1, 1, 0, 0, 0, 0, 0, 0, 0, 0]
numbers = [1, 1] + [0] * (K - 1) * 2
# Use sympy's multiset_permutations to get a multiset permutation generator
generator = multiset_permutations(numbers)
# Calculate the No. possible permutations
number_of_perms = No_multiset_permutations(numbers)
# Get all permutations, bonus progress bar is included :)
unique_perms = [next(generator) for _ in tqdm(range(number_of_perms))]
# Reshape each permutation to Kx2
unique_perms = np.array(unique_perms, dtype=np.int8)
return unique_perms.reshape(-1, K, 2)
if __name__ == "__main__":
solution = multiset_Kx2_permutations(300)
Another possibility (with rearranged axes for clearer output):
from itertools import combinations
import numpy as np
k = 4
x = list(combinations(range(k), 2))
out = np.zeros((n := len(x), k, 2), dtype=int)
out[np.c_[:n], x, [0, 1]] = 1
print(out)
It gives:
[[[1 0]
[0 1]
[0 0]
[0 0]]
[[1 0]
[0 0]
[0 1]
[0 0]]
[[1 0]
[0 0]
[0 0]
[0 1]]
[[0 0]
[1 0]
[0 1]
[0 0]]
[[0 0]
[1 0]
[0 0]
[0 1]]
[[0 0]
[0 0]
[1 0]
[0 1]]]
Related
I have this array and I want to return unique array combinations. I tried meshgrid but it creates duplicates and inverse array values
>> import numpy as np
>> array = np.array([0,1,2,3])
>> combinations = np.array(np.meshgrid(array, array)).T.reshape(-1,2)
>> print(combinations)
[[0 0]
[0 1]
[0 2]
[0 3]
[1 0]
[1 1]
[1 2]
[1 3]
[2 0]
[2 1]
[2 2]
[2 3]
[3 0]
[3 1]
[3 2]
[3 3]]
What I want to exclude are the repeating arrays: [0,0] [1,1] [2,2] [3,3] and the inverse arrays when [2,3] is returned exclude [3,2] in the output.
Take a look at this combination calculator, this is the output that I like but how can I create it in NumPy?
you could use combinations from itertools
import numpy as np
from itertools import combinations
array = np.array([0,1,2,3])
combs = np.array(list(combinations(arr, 2)))
I am trying to implement a quick tool to find the combinations of a set of numbers (0...k), for j lots of this array, where the sum across the row is equal to k-j, and k>=j
For instance, for k=3, and j=2, I have all the following combinations:
[[0 0]
[0 1]
[0 2]
[1 0]
[1 1]
[1 2]
[2 0]
[2 1]
[2 2]]
There are only two elements that sum up to k-j=1:
[[0 1]
[1 0]]
My current implementation is slow as I approach high numbers of k and j:
from itertools import product
from numpy import
combs = np.array(list(product(np.arange(0, k), repeat=j)))
combs = combs[np.sum(combs, axis=1)==k-j]
print(combs)
Can anyone please suggest a more efficient algorithm than I have at the moment?
For one, you don't have to consider the range up to k, but only up to k-j. Further, you could use itertools.combinations (if you are only interested in the set and not the order), as follows:
combs = np.array(list(combinations(range(k-j+1), j)))
combs = combs[np.sum(combs, axis=1)==k-j]
Like the title says, I want to compare two sitk arrays that have 1s and 0s as elements, and create a 3rd array that has 1s for where both arrays have 1 and 0s for any other cases. The arrays are the same size and are 3 dimensional, but is there a more efficient way to do this than iterating through them with nested for-loops?
import numpy as np
a = np.random.randint(low=0, high=2, size=(2,3,4), dtype=np.int)
print(a)
b = np.random.randint(low=0, high=2, size=(2,3,4), dtype=np.int)
print(b)
c = np.logical_and(a,b).astype(int)
print(c)
Is that what you're looking for?
arr_shape = (1,4,3)
a = np.random.randint(low=0,high=2, size=arr_shape)
print(a)
b = np.random.randint(low=0,high=2, size=arr_shape)
print(b)
# the new array. subtract a and b and get the absolute value.
# then invert to get the required array
d = (~abs(b - a).astype(bool)).astype(int)
print(d)
output:
[[[1 1 0]
[1 0 0]
[0 1 1]
[1 1 0]]]
[[[0 1 0]
[0 1 0]
[1 0 0]
[0 0 1]]]
array([[[0, 1, 1],
[0, 0, 1],
[0, 0, 0],
[0, 0, 0]]])
If you have SimpleITK Images, you can use the And function.
import SimpleITK as sitk
result = sitk.And(image1, image2)
This is a functional version of the AndImageFilter. You can read the docs for the class here:
https://simpleitk.org/doxygen/latest/html/classitk_1_1simple_1_1AndImageFilter.html
I have this line:
x = np.zeros((10, 2), int)
and a for loop that fills this for the first 6 elements.
Is there a way to remove the remaining 0 elements?
If you want to remove the last all zero rows:
x = x[:(np.where(x.any(axis=1))[0]).max()+1]
example:
x:
[[1 2]
[0 0]
[1 2]
[0 0]]
output:
[[1 2]
[0 0]
[1 2]]
Try this:
x = x[:6]
Note that this solution will create 6 element long view to original array. Modifying view will modify original array.
I have a numpy array of N integers ranging from 0 to M inclusive. I wish to treat them as indexes into an NxM matrix that contains a 1 in every position indicated by the array and a 0 everywhere else. For example, if given N=4, M=2 I have the following array
[1, 0, 2, 1]
I want to get this matrix
[0 1 0]
[1 0 0]
[0 0 1]
[0 1 0]
i.e. the row 0 has a one in column 1, row 1 has a 1 in column 0, etc.
How do I make this transformation in numpy?
This requires multi-dimensional array indexing.
a = np.array([1, 0, 2, 1])
z = np.zeros(12, dtype=int).reshape(4,3)
z[np.arange(a.size), a] = 1