I have the following Matrix made of 0s and 1s which I want to identify its spots(elements with the value 1 and connected to eachothers).
M = np.array([[1,1,1,0,0,0,0,0,0,0,0],
[1,1,1,0,0,0,0,0,0,1,1],
[1,1,1,0,0,0,0,0,0,1,1],
[1,1,1,0,0,1,1,1,0,0,0],
[0,0,0,0,0,1,1,1,0,0,0],
[1,1,1,0,1,1,1,1,0,0,0],
[1,1,1,0,0,1,1,1,0,0,0],
[1,1,1,0,0,1,1,1,0,0,0]])
In the matrix there are four spots.
an example of my output should seem the following
spot_0 = array[(0,0),(0,1), (0,2), (1,0),(1,1), (1,2), (2,0),(2,1), (2,2), (3,0),(3,1), (3,2)]
Nbr_0 = 12
Top_Left = (0, 0)
and that is the same process for the other 3 spots
Does anyone know how can I identify each spot with the number of its elements and top_left element, using numpy functions ?
Thanks
You can use a connected component labeling to find the spots. Then, you can use np.max so to find the number of component and np.argwhere so to find the locations of each component. Here is an example:
# OpenCV provides a similar function
from skimage.measure import label
components = label(M)
# array([[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [1, 1, 1, 0, 0, 3, 3, 3, 0, 0, 0],
# [0, 0, 0, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 3, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0]])
for i in range(1, np.max(components)+1):
spot_i = np.argwhere(components == i)
Nbr_i = len(spot_i)
Top_Left_i = spot_i[0]
Note that Top_Left only make sense for a rectangular area. If they are not rectangular this point needs to be carefully defined.
Note also that this method is only efficient with few component. If there are many component, then it is better to replace the current loop by an iteration over the components array (in this case the output structure is stored in a list l and l[components[i,j]] is updated with the information found for all item location (i,j) of components). This last algorithm will be slow unless Numba/Cython are used to speed the process up.
You could use skimage.measure.label or other tools (for instance, OpenCV or igraph) to create labels for connected components:
#from #Jérôme's answer
from skimage.measure import label
components = label(M)
# array([[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [1, 1, 1, 0, 0, 3, 3, 3, 0, 0, 0],
# [0, 0, 0, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 3, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0]])
In the later part you could create a one-dimensional view of image, sort values of pixels and find dividing points of sorted label values:
components_ravel = components.ravel()
c = np.arange(1, np.max(components_ravel) + 1)
argidx = np.argsort(components_ravel)
div_points = np.searchsorted(components_ravel, c, sorter=argidx)
# Sorted label values are:
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
# 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
# 2, 2, 2, 2
# 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
# 4, 4, 4, 4, 4, 4, 4, 4, 4
# So you find indices that divides these groups:
# [47, 59, 63, 79]
After that you could split array of indices that sorts your one-dimensional view of image at these points and convert them into two-dimensional ones:
spots = []
for n in np.split(argidx, div_points)[1:]: #in case there are no zeros, cancel `[1:]`
x, y = np.unravel_index(n, components.shape)
spots.append(np.transpose([x, y]))
It creates a list of spot coordinates of each group:
[array([[1, 0], [1, 2], [0, 2], [0, 1], [1, 1], [0, 0], [2, 2], [2, 1], [2, 0], [3, 2], [3, 1], [3, 0]]),
array([[2, 10], [1, 9], [2, 9], [1, 10]]),
array([[6, 5], [7, 5], [7, 6], [7, 7], [6, 7], [6, 6], [3, 5], [4, 6], [3, 6], [4, 5], [3, 7], [5, 7], [5, 6], [4, 7], [5, 5], [5, 4]]),
array([[5, 0], [5, 1], [5, 2], [6, 2], [7, 0], [6, 0], [6, 1], [7, 1], [7, 2]])]
Note that an order of pixels of each group is mixed. This is because np.argsort uses a sort which is not stable. You could fix it like so:
argidx = np.argsort(components_ravel, kind='stable')
In this case you'll get:
[array([[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 0], [3, 1], [3, 2]]),
array([[1, 9], [1, 10], [2, 9], [2, 10]]),
array([[3, 5], [3, 6], [3, 7], [4, 5], [4, 6], [4, 7], [5, 4], [5, 5], [5, 6], [5, 7], [6, 5], [6, 6], [6, 7], [7, 5], [7, 6], [7, 7]]),
array([[5, 0], [5, 1], [5, 2], [6, 0], [6, 1], [6, 2], [7, 0], [7, 1], [7, 2]])]
Related
I have the following arrays:
A = np.array([
[[[0, 1, 2, 3],
[3, 0, 1, 2],
[2, 3, 0, 1],
[1, 3, 2, 1],
[1, 2, 3, 0]]],
[[[9, 8, 7, 6],
[5, 4, 3, 2],
[0, 9, 8, 3],
[1, 9, 2, 3],
[1, 0, -1, 2]]],
[[[0, 7, 1, 2],
[1, 2, 1, 0],
[0, 2, 0, 7],
[-1, 3, 0, 1],
[1, 0, 1, 0]]]
])
A.shape
(3,1,5,4)
B = np.array([
[[[1, 0],
[-1, 2],
[9, 1],
[8, 2],
[7, 0]]],
[[[9, 6],
[5, 2],
[0, 3],
[1, 9],
[1, 0]]],
[[[0, 7],
[1, 0],
[0, 7],
[-1, 1],
[0, 0]]]
])
B.shape
(3,1,5,2)
Then I want to expand array A with B in the last dimension of A. Such that, the result X is:
X = np.array([
[[[0, 1, 2, 3, 1, 0],
[3, 0, 1, 2,-1, 2],
[2, 3, 0, 1, 9, 1],
[1, 3, 2, 1, 8, 2],
[1, 2, 3, 0, 7, 0]]],
[[[9, 8, 7, 6, 9, 6],
[5, 4, 3, 2, 5, 2],
[0, 9, 8, 3, 0, 3],
[1, 9, 2, 3, 1, 9],
[1, 0,-1, 2, 1, 0]]],
[[[0, 7, 1, 2, 0, 7],
[1, 2, 1, 0, 1, 0],
[0, 2, 0, 7, 0, 7],
[-1,3, 0, 1,-1, 1],
[1, 0, 1, 0, 0, 0]]]
])
X.shape
(3,1,5,6)
``
You have to concatenate the 2 arrays together along the axis you need:
C = np.concatenate((A, B), axis=3)
This is my first Question here, let me know if I could've done anything better.
I'm trying to do a element-wise operation between two arrays, but the broadcasting wont work like I want it to.
I have an array of shape (N,4).
square_list = np.array([[1,2,255,255], [255,255,4,4], [255,255,8,8], [255,255,16,16], [255,255,8,4], [255,1,8,8], [1,255,8,8]], dtype='B')
I also have an array of shape (4,).
square = np.array([1, 8, 8, 1], dtype='B')
What I am able to do is compare my square against each element in the square_list and it is being broadcast into shape (N,4) as expected.
Now I want to compare my square in each possible rotation against the square_list. I've written a function which returns an array of shape (4,4), which contains each possible rotation.
square.rotations
array([[1, 8, 8, 1],
[1, 1, 8, 8],
[8, 1, 1, 8],
[8, 8, 1, 1]], dtype=uint8)
I know how to do this using a loop. I'd prefer however to use an element-wise operator that returns my desired shape.
What I get:
rotations & square_list
ValueError: operands could not be broadcast together with shapes (4,4) (6,4)
What I'd like to get:
rotations & square_list
array([[[1, 0, 8, 1],
[1, 8, 0, 0],
[1, 8, 8, 0],
[1, 8, 0, 0],
[1, 8, 8, 0],
[1, 0, 8, 0]],
[[1, 0, 8, 8],
[1, 1, 0, 0],
[1, 1, 8, 8],
[1, 1, 0, 0],
[1, 1, 8, 0],
[1, 1, 8, 8]],
[[0, 0, 1, 8],
[8, 1, 0, 0],
[8, 1, 0, 8],
[8, 1, 0, 0],
[8, 1, 0, 0],
[8, 1, 0, 8],
[0, 1, 0, 8]],
[[0, 0, 1, 1],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 0, 0, 0],
[0, 8, 0, 0]]], dtype=uint8)
This is just to visualize what I want, I don't particularly care about the order of the axis'. A shape of either (4, N, 4) or (N, 4, 4) would be great.
I have the feeling that this can be achieved easily by just reshaping one of the input arrays but I couldn't figure it out.
Thanks in advance!
Add an extra dimension to rotations:
square_list & rotations[:,None]
output:
array([[[1, 0, 8, 1],
[1, 8, 0, 0],
[1, 8, 8, 0],
[1, 8, 0, 0],
[1, 8, 8, 0],
[1, 0, 8, 0],
[1, 8, 8, 0]],
[[1, 0, 8, 8],
[1, 1, 0, 0],
[1, 1, 8, 8],
[1, 1, 0, 0],
[1, 1, 8, 0],
[1, 1, 8, 8],
[1, 1, 8, 8]],
[[0, 0, 1, 8],
[8, 1, 0, 0],
[8, 1, 0, 8],
[8, 1, 0, 0],
[8, 1, 0, 0],
[8, 1, 0, 8],
[0, 1, 0, 8]],
[[0, 0, 1, 1],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 8, 0, 0],
[8, 0, 0, 0],
[0, 8, 0, 0]]], dtype=uint8)
num1 = [1,2,3,4,5]
num2 = [1,2,3,4,5]
arr1 = [[0]*(len(num2)+1)]*(len(num1)+1)
arr2 = [[0 for _ in range(len(num2)+1)] for _ in range(len(num1)+1)]
I get a different answer when I define arr1 and arr2.
Aren't arr1 and arr2 create the same 2D array?
They are not the same. arr1 is a list with (len(nums1)+1) references to the same list [0]*(len(nums2)+1). So when you modify an element in one of them, all references will see this change as well.
For example,
>>> arr1[0][0] += 1
>>> print(arr1)
[[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0]]
arr2 doesn't suffer from this problem because it has len(nums1)+1 distinct lists:
>>> arr2[0][0] += 1
>>> print(arr2)
[[1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0]]
A better way to see the difference is to use a random number to fill the entries.
from random import randrange
num1 = [1,2,3,4,5]
num2 = [1,2,3,4,5]
arr1 = [[randrange(10)]*(len(nums2)+1)]*(len(nums1)+1)
arr2 = [[randrange(10) for _ in range(len(nums2)+1)] for _ in range(len(nums1)+1)]
print(arr1)
print(arr2)
The output is:
[[5, 5, 5, 5, 5, 5], [5, 5, 5, 5, 5, 5], [5, 5, 5, 5, 5, 5], [5, 5, 5, 5, 5, 5], [5, 5, 5, 5, 5, 5], [5, 5, 5, 5, 5, 5]]
[[7, 4, 2, 4, 0, 3], [7, 5, 1, 0, 1, 7], [4, 4, 1, 0, 2, 1], [2, 3, 6, 2, 6, 7], [6, 6, 6, 0, 3, 3], [0, 4, 5, 0, 6, 6]]
You can see that for the arr1, it populates every entry with the same number; while for arr2, the entries are all truly random. This is because arr1 is constructed by expanding a list of just one number, which is [5] here.
Is there a numpy solution that would allow you to initialize an array based on the following conditions?
Number of elements in axis 1. (In the example below you have 4 places in each element of the array)
Sum of values. (All elements sum to 8)
Step size. (Using increments of 2)
Essentially this shows all the combinations of 4 values you can add to achieve the wanted sum (8) at a step size of 2.
My experiments fail when I set the axis 1 dimension to over 6 and the sum to over 100.
There has to be a better way to do this than what I've been trying.
array([[0, 0, 0, 8],
[0, 0, 2, 6],
[0, 0, 4, 4],
[0, 0, 6, 2],
[0, 0, 8, 0],
[0, 2, 0, 6],
[0, 2, 2, 4],
[0, 2, 4, 2],
[0, 2, 6, 0],
[0, 4, 0, 4],
[0, 4, 2, 2],
[0, 4, 4, 0],
[0, 6, 0, 2],
[0, 6, 2, 0],
[0, 8, 0, 0],
[2, 0, 0, 6],
[2, 0, 2, 4],
[2, 0, 4, 2],
[2, 0, 6, 0],
[2, 2, 0, 4],
[2, 2, 2, 2],
[2, 2, 4, 0],
[2, 4, 0, 2],
[2, 4, 2, 0],
[2, 6, 0, 0],
[4, 0, 0, 4],
[4, 0, 2, 2],
[4, 0, 4, 0],
[4, 2, 0, 2],
[4, 2, 2, 0],
[4, 4, 0, 0],
[6, 0, 0, 2],
[6, 0, 2, 0],
[6, 2, 0, 0],
[8, 0, 0, 0]], dtype=int64)
Here is a small code that will enable you to loop over the desired combinations. It takes 3 parameter:
itsize: Number of elements.
itsum: Sum of values.
itstep: Step size.
It may be necessary to optimize it if the computations you do in the FOR loop are light. I loop over more combinations than necessary (all the i,j,k,l that take values in 0,itstep,2*itstep,...,itsum) and keep only those verifying the condition that all sum up to itsum. The big size array is not computed and the rows are computed on-the-fly when iterating so you will not have the memory troubles:
class Combinations:
def __init__(self, itsize, itsum, itstep):
assert(itsum % itstep==0) # Sum is a multiple of step
assert(itsum >= itstep) # Sum bigger or equal than step
assert(itsize > 0) # Number of elements >0
self.itsize = itsize # Number of elements
self.itsum = itsum # Sum parameter
self.itstep = itstep # Step parameter
self.cvalue = None # Value of the iterator
def __iter__(self):
self.itvalue = None
return self
def __next__(self):
if self.itvalue is None: # Initialization of the iterator
self.itvalue = [0]*(self.itsize)
elif self.itvalue[0] == self.itsum: # We reached all combinations the iterator is restarted
self.itvalue = None
return None
while True: # Find the next iterator value
for i in range(self.itsize-1,-1,-1):
if self.itvalue[i]<self.itsum:
self.itvalue[i] += self.itstep
break
else:
self.itvalue[i] = 0
if sum(self.itvalue) == self.itsum:
break
return self.itvalue # Return iterator value
myiter = iter(Combinations(4,8,2))
for val in myiter:
if val is None:
break
print(val)
Output:
% python3 script.py
[0, 0, 0, 8]
[0, 0, 2, 6]
[0, 0, 4, 4]
[0, 0, 6, 2]
[0, 0, 8, 0]
[0, 2, 0, 6]
[0, 2, 2, 4]
[0, 2, 4, 2]
[0, 2, 6, 0]
[0, 4, 0, 4]
[0, 4, 2, 2]
[0, 4, 4, 0]
[0, 6, 0, 2]
[0, 6, 2, 0]
[0, 8, 0, 0]
[2, 0, 0, 6]
[2, 0, 2, 4]
[2, 0, 4, 2]
[2, 0, 6, 0]
[2, 2, 0, 4]
[2, 2, 2, 2]
[2, 2, 4, 0]
[2, 4, 0, 2]
[2, 4, 2, 0]
[2, 6, 0, 0]
[4, 0, 0, 4]
[4, 0, 2, 2]
[4, 0, 4, 0]
[4, 2, 0, 2]
[4, 2, 2, 0]
[4, 4, 0, 0]
[6, 0, 0, 2]
[6, 0, 2, 0]
[6, 2, 0, 0]
[8, 0, 0, 0]
I tried this out and also found that it slowed down significantly at that size. I think part of the problem is that the output array gets pretty large at that point. I'm not 100% sure my code is right, but the plot shows how the array size grows with condition 2 (sum of values in each row). I didn't do 100, but it looks like it would be about 4,000,000 rows
plot
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]])