2D array of 1s and 0s. How to label every group of 1s with a unique number?
I’m stuck on this problem for a while now. 1s can be grouped vertically, horizontally and diagonally. How can you go about solving this? For example,
0 0 1 1 0
0 1 1 0 0
0 0 0 0 1
0 0 0 1 0
Should be transformed to
0 0 x x 0
0 x x 0 0
0 0 0 0 y
0 0 0 y 0
x, y can be any unique numbers.
Appreciate it.
Here is what I have so far for iterative: https://i.imgur.com/oCmYC02.png
But the result is a bit off because it only checks for immediate adjacent 1's: https://i.imgur.com/DAtTBmM.png
Anyone have any idea how to fix this?
I'd do it like this:
Scan 2D array sequentially, row by row, column by column
If 1 found, use variation of the flood fill algorithm, which moves in 8 directions instead of 4, from that starting point (see normal 4-direction algorithm at https://en.wikipedia.org/wiki/Flood_fill), since you have diagonal example with "y", each time using new filler number.
Repeat 1 and 2 until no more ones left.
Related
I want to pick out the elements which are
(2*pi*k),
where k=0,1,2,3... which means integer, and fill them (i1) into the other matrix.
But my problem is, I don't know how to make "k" be row. (By the way, dividends and divisors are float, so I need to find the approximations and see them as 2*pi*k).
My code, only can find the elements which are (2*pi*k), but can't order them like if k=1, then it will be put into k=1 row; if k=2, then the element should be put into k=2 row.
For example,
A = [2*pi 6 3 4;0.5*pi 0 2;3.1 7 4 8;2*pi 7 2 9;2.6 4*pi 6*pi 0]
I want the output to be
B = [0 2*pi 4*pi 6*pi;0 2*pi NaN NaN;NaN 2*pi NaN NaN]
This is my code:
k=0;
for m=380:650;
for n=277:600;
if abs((rem(abs(i(m,n)),(2*PI)))-(PI))>=3.11;
k=k+1;
B(m,k)=i1(m,n);
end
end
k=0;
end
It can find what I want but they seem not to be ordered the way I want.
As others, I'm a bit unsure what you want. Here's how I understood it and would code it:
check whether (2*pi*k) is contained in A, you want a numerical approach
output binary result
here's the code:
testPI=#(k) (2*pi*k); %generates 2*pi*k, where k is up to the user
A = [2*pi 6 3 4;0.5*pi 0 2 0;3.1 7 4 8;2*pi 7 2 9;2.6 4*pi 6*pi 0]; %A from example (fixed dimension error)
ismember(A,f(1:10)) %test if k=1:10 is contained in A
ans =
5×4 logical array
1 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
0 1 1 0
Adapt 1:10 to any value you'd like. Of course this only works if k is within reasonable range, otherwise this approach is suboptimal
I have an array A of 1s and 0s and want to see if the larger array of bits B contains those bits in that exact order?
Example: A= [0 1 1 0 0 0 0 1]
B= [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1]
would be true as A is contained in B
Most solutions I have found only determine if a value IS contained in another matrix, this is no good here as it is already certain that both matrices will be 1s and 0s
Thanks
One (albeit unusual) option, since you're dealing with integer values, is to convert A and B to character arrays and use the contains function:
isWithin = contains(char(B), char(A));
There are some obtuse vectorized ways to to do this, but by far the easiest, and likely just as efficient, is to use a loop with a sliding window,
A = [0 1 1 0 0 0 0 1];
B = [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1];
vec = 0:(numel(A)-1);
for idx = 1:(numel(B)-numel(A)-1)
if all(A==B(idx+vec))
fprintf('A is contained in B\n');
break; % exit the loop as soon as 1 match is found
end
end
Or if you want to know the location(s) in B (of potentially multiple matches) then,
A = [0 1 1 0 0 0 0 1];
B = [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1];
C = false(1,numel(B)-numel(A)-1);
vec = 0:(numel(A)-1);
for idx = 1:numel(C)
C(idx) = all(A==B(idx+vec));
end
if any(C)
fprintf('A is contained in B\n');
end
In this case
>> C
C =
1×6 logical array
0 0 0 1 0 0
You can use the cross-correlation between two signals for this, as a measure of local similarity.
For achieving good results, you need to shift A and B so that you don't have the value 0 any more. Then compute the correlation between the two of them with conv (keeping in mind that the convolution is the cross-correlation with one signal flipped), and normalize with the energy of A so that you get a perfect match whenever you get the value 1:
conv(B-0.5, flip(A)-0.5, 'valid')/sum((A-0.5).^2)
In the normalization term, flipping is removed as it does not change the value.
It gives:
[0 -0.5 0.25 1 0 0 -0.25 0]
4th element is 1, so starting from index equal to 4 you get a perfect match.
I'm trying to remove the rows which has duplicates in sequence. I have only 2 possible values which are 0 and 1. I have nXm which n shows possible number of bits and m is not important for my question. My goal is to find an matrix which is nX(m-a). The rows a which has the property which includes duplicates in sequence. For example:
My matrix is :
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0]
I want to remove the rows has t duplicates in sequence for 0. In this question let's assume t is 3. So I want the matrix which:
B=[0 1 0 1 0 1;
0 0 1 0 0 1;
0 1 0 0 1 0]
2nd and 5th rows are removed.
I probably need to use diff.
So you want to remove rows of A that contain at least t zeros in sequence.
How about a single line?
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==-t, 2),:);
How this works:
Transform A to bipolar form (2*A-1)
Convolve each row with a sequence of t ones (conv2(...))
Keep only rows for which the convolution does not contain -t (~any(...)). The presence of -t indicates a sequence of t zeros in the corresponding row of A.
To remove rows that contain at least t ones, just change -t to t:
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==t, 2),:);
Here is a generalized approach which removes any rows which has given number of consecutive duplicates (not just zero. could be any number).
t = 3;
row_mask = ~any(all(~diff(reshape(im2col(A,[1 t],'sliding'),t,size(A,1),[]))),3);
out = A(row_mask,:)
Sample Run:
>> A
A =
0 1 0 1 0 1
0 0 1 5 5 5 %// consecutive 3 5's
0 0 1 0 0 1
0 1 0 0 1 0
1 1 1 0 0 1 %// consecutive 3 1's
>> out
out =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
How about an approach using strings? This is certainly not as fast as Luis Mendo's method where you work directly with the numerical array, but it's thinking a bit outside of the box. The basis of this approach is that I consider each row of A to be a unique string, and I can search each string for occurrences of a string of 0s by regular expressions.
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0];
t = 3;
B = sprintfc('%s', char('0' + A));
ind = cellfun('isempty', regexp(B, repmat('0', [1 t])));
B(~ind) = [];
B = double(char(B) - '0');
We get:
B =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
Explanation
Line 1: Convert each line of the matrix A into a string consisting of 0s and 1s. Each line becomes a cell in a cell array. This uses the undocumented function sprintfc to facilitate this cell array conversion.
Line 2: I use regular expressions to find any occurrences of a string of 0s that is t long. I first use repmat to create a search string that is full of 0s and is t long. After, I determine if each line in this cell array contains this sequence of characters (i.e. 000....). The function regexp helps us perform regular expressions and returns the locations of any matches for each cell in the cell array. Alternatively, you can use the function strfind for more recent versions of MATLAB to speed up the computation, but I chose regexp so that the solution is compatible with most MATLAB distributions out there.
Continuing on, the output of regexp/strfind is a cell array of elements where each cell reports the locations of where we found the particular string. If we have a match, there should be at least one location that is reported at the output, so I check to see if any matches are empty, meaning that these are the rows we don't want to remove. I want to turn this into a logical array for the purposes of removing rows from A, and so this is wrapped with a cellfun call to determine the cells that are empty. Therefore, this line returns a logical array where a 0 means that remove this row and a 1 means that we don't.
Line 3: I take the logical array from Line 2 and invert it because that's what we really want. We use this inverted array to index into the cell array and remove those strings.
Line 4: The output is still a cell array, so I convert it back into a character array, and finally back into a numerical array.
I'm trying to write code that implements median filtering on a two-dimensional array.
Here's an image to illustrate:
The program starts at the beginning of the array. The maximum array size is 100. I know that I can use an array like:
int a[100][100];
to store the input, and that I can iterate over a part of this array using two for loops like this:
for(i=0;i<size_filter;i++)
for(j=0;j<size_filter;j++)
temp[i][j]=a[i][j] // not so sure
But how can I make this code loop over the neighbors of every element in the array, calculate their median, and replace the center element with the median?
For some examples of what I'm trying to do, let's say that the input is a 5x5 matrix, so the input size is 5. And I want to run a 3x3 median filter on it, i.e. each element should be replaced by the median of the 3x3 elements surrounding it.
The program starts at the corner index (0,0). For this index, it scans the 3x3 region surrounding it (of which only four indexes actually lie within the input array), which contains the values 0, 0, 1, and 0. The median of these values is 0, so that's what the code should output for this array index.
In the picture below, the number in bold italics is the center cell, and the plain bold numbers are its neighbors within the 3x3 region surrounding it:
0 0 0 0 0
1 0 0 1 0
1 1 0 0 0
0 1 1 0 0
0 0 0 0 0
Here's another example, this time with the center index (0,1):
0 0 0 0 0
1 0 0 1 0
1 1 0 0 0
0 1 1 0 0
0 0 0 0 0
This time, the elements in the 3x3 region (excluding those outside the input array) have the values 0, 0, 0, 1, 0, and 0, and again, their median is therefore 0.
Here's yet another example, this time from the middle of the input, at center index (3,2):
0 0 0 0 0
1 0 0 1 0
1 1 0 0 0
0 1 1 0 0
0 0 0 0 0
This time, the elements within the 3x3 region have the values 1, 0, 0, 1, 1, 0, 0, 1, and 1, and their median in therefore 1.
Final example:
<size of array><size filter> <data>
8
3
0 0 0 0 0 0 0 0
0 5 0 0 6 0 0 0
0 0 0 0 0 7 0 0
0 0 0 0 5 0 0 0
0 0 0 5 6 0 0 0
0 0 8 5 5 0 0 0
0 0 0 7 0 0 9 0
0 0 0 0 0 0 0 0
Output:
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 5 5 0 0 0
0 0 0 5 5 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
It looks like you're trying to implement a two-dimensional median filter. The straightforward way to implement such a filter is to have four nested loops: two outer loops over the x and y coordinates of the whole image, and two inner loops over the neighborhood of the center pixel.
It's perhaps easier to describe this in code than in text, so here's some Python-esque pseudocode to illustrate:
# assumptions:
# * image is a height x width array containing source pixel values
# * filtered is a height x width array to store result pixel values in
# * size is an odd number giving the diameter of the filter region
radius = (size - 1) / 2 # size = 3 -> radius = 1
for y from 0 to height-1:
top = max(y - radius, 0)
bottom = min(y + radius, height-1)
for x from 0 to width-1:
left = max(x - radius, 0)
right = min(x + radius, width-1)
values = new list
for v from top to bottom:
for u from left to right:
add image[v][u] to values
filtered[y][x] = median(values)
Translating this code into C is left as an exercise.
It's also possible to optimize this code by noting that the neighborhoods of adjacent array cells overlap significantly, so that the values of those neighboring cells can be reused across successive iterations of the outer loops. Since the performance of this algorithm on modern CPUs is essentially limited by RAM access latency, such reuse can provide a significant speedup, especially for large filter sizes.
this:
for(i=0;i<size_filter;i++)
for(j=0;j<size_filter;j++)
temp[i][j]=a[i][j];
is a good starting point.
You just iterating over every pixel of your input array, determine the median of the neighborhood and write it to an output array.
So instead of temp[i][j]=a[i][j]; you need some WhatEverType calcMedianAt(const WhatEverType a[100][100], int r, int c, int size); function.
So you can call temp[i][j]=calcMedianAt(a, i,j, 3);
the function itself has to extract the value to a list (do proper border handling) and find the median in that list (for example by calling some median function WhatEverType calcMedian(const WhatEverType* data, int len); and return it.
I have a large matrix of 1's and 0's, and am looking for a way to "fill" up areas that are locally dense with 1's.
I first did this task for an array, and counted the number of 1's within a certain radius of the element in questions. If the radius was 5, for example, and my threshold was 4, then a point that had 4 elements marked "1" within 5 elements to the left or right would be changed to a 1.
Basically I would like to generalized this to a two - dimensional array and have a resulting matrix that has "smooth" and "connected" regions of 1's and no "patchy" spots.
As an example, the matrix
1 0 0 1 0 0 0
0 0 1 0 1 0 0
0 1 0 1 0 0 0
0 0 1 1 1 0 0
would ideally be changed to
1 0 0 1 1 0 0
0 0 1 1 1 0 0
0 1 1 1 1 0 0
0 0 1 1 1 0 0
or something similar
For binary images, the morphologial operations that are implemented in MATLAB are perfect for manipulating the shape and size of connected regions. Specifically, the process of image closing is designed to fill holes in connected regions. In MATLAB, the function is imclose, which takes the image and a structuring element, similar to a filter kernel, for how neighboring pixels effect the filling of holes and gaps. A simple invocation of imclose is,
IM2 = imclose(IM,strel(ones(3)));
Larger gaps can be filled by increasing the area of the influence of of neighboring pixes, via larger structuring elements. For example, we an use a disk of radius 10 pixels:
IM2 = imclose(IM,strel('disk',10));
While, imclose supports grayscale and binary (0 and 1) images, the function bwmorph is designed for operation on binary images only but provides a generic interface to all of the morphological operations and various neat combinations of operations (e.g. 'bothat', 'tophat', etc.). The syntax for closing is simplified with bwmorph:
BW2 = bwmorph(BW,'close');
Here the structuring element is the standard ones(3).
A simple filter such as the following might do the trick:
h = [ 0 1 0
1 0 1
0 1 0];
img2=(imfilter(img,h)>2) | img;
For instance:
img =
1 0 0 1 0 0 0
0 0 1 0 1 0 0
0 1 0 1 0 0 0
0 0 1 1 1 0 0
img2 =
1 0 0 1 0 0 0
0 0 1 1 1 0 0
0 1 1 1 1 0 0
0 0 1 1 1 0 0
You can try different filters to modify the output img2.
This uses the image processing toolbox. If you don't have that, you may want to look up equivalent routines from the matlab exchange.