I have an image 640x480 img, and I want to replace pixels having values not in this list or array x=[1, 2, 3, 4, 5] with a certain value 10, so that any pixel in img which doesn't have the any of the values in x will be replaced with 10. I already know how to replace only one value using img(img~=1)=10 or multiple values using this img(img~=1 & img~=2 & img~=3 & img~=4 & img~=5)=10 but I when I tried this img(img~=x)=10 it gave an error saying Matrix dimensions must agree. So if anyone could please advise.
You can achieve this very easily with a combination of permute and bsxfun. We can create a 3D column vector that consists of the elements of [1,2,3,4,5], then use bsxfun with the not equals method (#ne) on your image (assuming grayscale) so that we thus create a 3D matrix of 5 slices. Each slice would tell you whether the locations in the image do not match an element in x. The first slice would give you the locations that don't match x = 1, the second slice would give you the locations that don't match x = 2, and so on.
Once you finish this, we can use an all call operating on the third dimension to consolidate the pixel locations that are not equal to all of 1, 2, 3, 4 or 5. The last step would be to take this logical map, which that tells you the locations that are none of 1, 2, 3, 4, or 5 and we'd set those locations to 10.
One thing we need to consider is that the image type and the vector x must be the same type. We can ensure this by casting the vector to be the same class as img.
As such, do something like this:
x = permute([1 2 3 4 5], [3 1 2]);
vals = bsxfun(#ne, img, cast(x, class(img)));
ind = all(vals, 3);
img(ind) = 10;
The advantage of the above method is that the list you want to use to check for the elements can be whatever you want. It prevents having messy logical indexing syntax, like img(img ~= 1 & img ~= 2 & ....). All you have to do is change the input list at the beginning line of the code, and bsxfun, permute and any should do the work for you.
Here's an example 5 x 5 image:
>> rng(123123);
>> img = randi(7, 5, 5)
img =
3 4 3 6 5
7 2 6 5 1
3 1 6 1 7
6 4 4 3 3
6 2 4 1 3
By using the code above, the output we get is:
img =
3 4 3 10 5
10 2 10 5 1
3 1 10 1 10
10 4 4 3 3
10 2 4 1 3
You can most certainly see that those elements that are neither 1, 2, 3, 4 or 5 get set to 10.
Aside
If you don't like the permute and bsxfun approach, one way would be to have a for loop and with an initially all true array, keep logical ANDing the final result with a logical map that consists of those locations which are not equal to each value in x. In the end, we will have a logical map where true are those locations that are neither equal to 1, 2, 3, 4 or 5.
Therefore, do something like this:
ind = true(size(img));
for idx = 1 : 5
ind = ind & img ~= idx;
end
img(ind) = 10;
If you do this instead, you'll see that we get the same answer.
Approach #1
You can use ismember,
which according to its official documentation for a case of ismember(A,B) would output a logical array of the same size as A and with 1's where
any element from B is present in A, 0's otherwise. Since, you are looking to detect "not in the list or array", you need to invert it afterwards, i.e. ~ismember().
In your case, you have img as A and x as B, so ~ismember(img,x) would give you those places where img~=any element in x
You can then map into img to set all those in it to 10 with this final solution -
img(~ismember(img,x)) = 10
Approach #2
Similar to rayryeng's solution, you can use bsxfun, but keep it in 2D which could be more efficient as it would also avoid permute. The implementation would look something like this -
img(reshape(all(bsxfun(#ne,img(:),x(:).'),2),size(img))) = 10
Related
I'm trying to extract a matrix with two columns. The first column is the data that I want to group into a vector, while the second column is information about the group.
A =
1 1
2 1
7 2
9 2
7 3
10 3
13 3
1 4
5 4
17 4
1 5
6 5
the result that i seek are
A1 =
1
2
A2 =
7
9
A3 =
7
10
13
A4=
1
5
17
A5 =
1
6
as an illustration, I used the eval function but it didn't give the results I wanted
Assuming that you don't actually need individually named separated variables, the following will put the values into separate cells of a cell array, each of which can be an arbitrary size and which can be then retrieved using cell index syntax. It makes used of logical indexing so that each iteration of the for loop assigns to that cell in B just the values from the first column of A that have the correct number in the second column of A.
num_cells = max (A(:,2));
B = cell (num_cells,1);
for idx = 1:max(A(:,2))
B(idx) = A((A(:,2)==idx),1);
end
B =
{
[1,1] =
1
2
[2,1] =
7
9
[3,1] =
7
10
13
[4,1] =
1
5
17
[5,1] =
1
6
}
Cell arrays are accessed a bit differently than normal numeric arrays. Array indexing (with ()) will return another cell, e.g.:
>> B(1)
ans =
{
[1,1] =
1
2
}
To get the contents of the cell so that you can work with them like any other variable, index them using {}.
>> B{1}
ans =
1
2
How it works:
Use max(A(:,2)) to find out how many array elements are going to be needed. A(:,2) uses subscript notation to indicate every value of A in column 2.
Create an empty cell array B with the right number of cells to contain the separated parts of A. This isn't strictly necessary, but with large amounts of data, things can slow down a lot if you keep adding on to the end of an array. Pre-allocating is usually better.
For each iteration of the for loop, it determines which elements in the 2nd column of A have the value matching the value of idx. This returns a logical array. For example, for the third time through the for loop, idx = 3, and:
>> A_index3 = A(:,2)==3
A_index3 =
0
0
0
0
1
1
1
0
0
0
0
0
That is a logical array of trues/falses indicating which elements equal 3. You are allowed to mix both logical and subscripts when indexing. So using this we can retrieve just those values from the first column:
A(A_index3, 1)
ans =
7
10
13
we get the same result if we do it in a single line without the A_index3 intermediate placeholder:
>> A(A(:,2)==3, 1)
ans =
7
10
13
Putting it in a for loop where 3 is replaced by the loop variable idx, and we assign the answer to the idx location in B, we get all of the values separated into different cells.
I have a problem with coming up with an algorithm for the "graph" :(
Maybe one of you would be so kind and direct me somehow <3
The task is as follows:
We have a board of at least 3x3 (it doesn't have to be a square, it can be 4x5 for example). The user specifies a sequence of moves (as in Android lock pattern). The task is to check how many points he has given are adjacent to each other horizontally or vertically.
Here is an example:
Matrix:
1 2 3 4
5 6 7 8
9 10 11 12
The user entered the code: 10,6,7,3
The algorithm should return the number 3 because:
10 is a neighbor of 6
6 is a neighbor of 7
7 is a neighbor of 3
Eventually return 3
Second example:
Matrix:
1 2 3
4 5 6
7 8 9
The user entered the code: 7,8,6,3
The algorithm should return 2 because:
7 is a neighbor of 8
8 is not a neighbor of 6
6 is a neighbor of 3
Eventually return 2
Ofc number of operations equal length of array - 1
Sorry for "ile" and "tutaj", i'm polish
If all the codes are unique, use them as keys to a dictionary (with (row/col) pairs as values). Loop thru the 2nd item in user input to the end, check if math.Abs(cur.row-prev.row)+math.Abs(cur.col-prev.col)==1. This is not space efficient but deal with user input in linear complexity.
The idea is you have 4 conditions, one for each direction. Given any matrix of the shape n,m which is made of a sequence of integers AND given any element:
The element left or right will always be + or - 1 to the given element.
The element up or down will always be + or - m to the given element.
So, if abs(x-y) is 1 or m, then x and y are neighbors.
I demonstrate this in python.
def get_neighbors(seq,matrix):
#Conditions
check = lambda x,y,m: np.abs(x-y)==1 or np.abs(x-y)==m
#Pairs of sequences appended with m
params = zip(seq, seq[1:], [matrix.shape[1]]*(len(seq)-1))
neighbours = [check(*i) for i in params]
count = sum(neighbours)
return neighbours, count
seq = [7,8,6,3]
matrix = np.arange(1,10).reshape((3,3))
neighbours, count = get_neighbors(seq, matrix)
print('Matrix:')
print(matrix)
print('')
print('Sequence:', seq)
print('')
print('Count of neighbors:',count)
Matrix:
[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]]
Sequence: [10, 6, 7, 3]
Count of neighbors: 3
Another example -
seq = [7,8,6,3]
matrix = np.arange(1,10).reshape((3,3))
neighbours, count = get_neighbors(seq, matrix)
Matrix:
[[1 2 3]
[4 5 6]
[7 8 9]]
Sequence: [7, 8, 6, 3]
Count of neighbors: 2
So your input is the width of a table, the height of a table, and a list of numbers.
W = 4, H = 3, list = [10,6,7,3]
There are two steps:
Convert the list of numbers into a list of row/column coordinates (1 to [1,1], 5 to [2,1], 12 to [3,4]).
In the new list of coordinates, find consequent pairs, which have one coordinate identical, and the other one has a difference of 1.
Both steps are quite simple ("for" loops). Do you have problems with 1 or 2?
Working on a dicegame for school and I have trouble figuring out how to do automatic calculation of the result. (we don't have to do it automatically, so I could just let the player choose which dice to use and then just check that the user choices are valid) but now that I have started to think about it I can't stop...
the problem is as follows:
I have six dice, the dice are normal dice with the value of 1-6.
In this example I have already roled the dice and they have the following values:
[2, 2, 2, 1, 1, 1]
But I don't know how to calulate all combinations so that as many dicecombinations as possible whose value combined(addition) are 3 (in this example) are used.
The values should be added together (for example a die with value 1 and another die with the value 2 are together 3) then there are different rounds in the game where the aim is to get different values (which can be a combination(addition) of die-values for example
dicevalues: [2, 2, 2, 2, 2, 2]
could give the user a total of 12 points if 4 is the goal for the current round)
2 + 2 = 4
2 + 2 = 4
2 + 2 = 4
if the goal of the round instead where 6 then the it would be
2 + 2 + 2 = 6
2 + 2 + 2 = 6
instead which would give the player 12 points (6 + 6)
[1, 3, 6, 6, 6, 6]
with the goal of 3 would only use the dice with value 3 and discard the rest since there is no way to add them up to get three.
2 + 1 = 3
2 + 1 = 3
2 + 1 = 3
would give the user 9 points.
but if it where calculated the wrong way and the ones where used up together instead of each 1 getting apierd with a two 1 + 1 + 1 which would only give the player 3 points och the twos couldn't be used.
Another example is:
[1, 2, 3, 4, 5, 6]
and all combinations that are equal to 6 gives the user points
[6], [5, 1], [4 ,2]
user gets 18 points (3 * 6)
[1 ,2 ,3], [6]
user gets 12 points (2 * 6) (Here the user gets six points less due to adding upp 1 + 2 + 3 instead of doing like in the example above)
A dice can have a value between 1 and 6.
I haven't really done much more than think about it and I'm pretty sure that I could do it right now, but it would be a solution that would scale really bad if I for example wanted to use 8 dices instead and every time I start programming on it I start to think that have to be a better/easier way of doing it... Anyone have any suggestion on where to start? I tried searching for an answer and I'm sure it's out there but I have problem forumulating a query that gives me relevant result...
With problems that look confusing like this, it is a really good idea to start with some working and examples. We have 6 die, with range [1 to 6]. The possible combinations we could make therefore are:
target = 2
1 combination: 2
2 combination: 1+1
target = 3
1 combination: 3
2 combination: 2+1
3 combination: 1+1+1
target = 4
1 combination: 4
2 combination: 3+1
2+2
3 combination: 2+1+1
4 combination: 1+1+1+1
target = 5
1 combination: 5
2 combination: 4+1
3+2
3 combination: 2+2+1
4 combination: 2+1+1+1
5 combination: 1+1+1+1+1
See the pattern? Hint, we go backwards from target to 1 for the first number we can add, and then given this first number, and the size of the combination, there is a limit to how big subsequent numbers can be!
There is a finite list of possible combinations. You can by looking for 1 combination scores, and remove these from the die available. Then move on to look for 2 combination scores, etc.
If you want to read more about this sub-field of mathematics, the term you need to look for is "Combinatorics". Have fun!
Let's say I have 5 connected components (labelled objects) in an image called labelledImage from bwlabel. How can I manipulate labelledImage so that the objects that are labelled as 1 and 4 only display, while removing the objects that are labelled as 2, 3 and 5. Then, how can I manipulate the original RGB image so that the connected components that are labelled as 1 and 4 only display.
I know how to retain a single connected component by using this line of code below. However, I don't know how to do this for multiple labelled regions.
Works.
connectedComponent1 = (labelledImage == 1);
imshow(connectedComponent1)
Doesn't work.
connectedComponent1and4 = (labelledImage == [1 4]);
imshow(connectedComponent1and4)
You can't do logical indexing that way. The simplest way is to perhaps use Boolean statements to combine things.
connectedCompoonent1and4 = labelledImage == 1 | labelledImage == 4;
In general, supposing you had a vector of elements that denote which components you want to keep, you could use bsxfun, permute and any to help you with that. Something like this should work:
components = [1 4];
connected = any(bsxfun(#eq, labelledImage, permute(components, [1 3 2])), 3);
The above code uses matrix broadcasting to create a temporary 3D matrix where each slice i contains the ith value of the vector components which contain the desired labels you want to keep. labelledImage is also replicated in the third dimension so the result using bsxfun creates a 3D matrix where each slice i segments out the ith object you want to keep. We then combine all of the objects together using any and looking in the third dimension.
If you don't like one-liners, you could even use a simple for loop:
components = [1 4];
connected = false(size(labelledImage, 1), size(labelledImage, 2));
for ind = 1 : numel(components)
connected = connected | labelledImage == components(ind);
end
This creates an output image that is all false, then we loop through each value in the vector of components you want to keep and append those results on top of the result. The end will give you all of the components you want to keep.
Lastly, you could use also use ismember and determine those values in your matrix that can be found between the label matrix and the components vector and simply create your mask that way:
connected = ismember(labelledImage, components);
Now that you have a mask of objects you want to extract out, to use this on the original image, simply multiply each channel with the mask. Another use of bsxfun can do that for you. Assuming your image in RGB is called img, simply do the following:
outImg = bsxfun(#times, img, cast(connected, class(img)));
To perform element-wise multiplication, you must ensure that both matrices that are being multiplied have the same type. I convert the mask into the same class as whatever the input image is and perform the multiplication.
Use ismember.
Ex:
A = randi(5,5); % your connected component matrix
B = [1 4] % list of components you want to keep
A =
4 2 1 3 5
2 4 2 5 1
3 4 5 1 4
1 4 1 3 5
4 3 5 1 5
A(~ismember(A,B)) = 0
A =
4 0 1 0 0
0 4 0 0 1
0 4 0 1 4
1 4 1 0 0
4 0 0 1 0
I have a matrix M=[4 3 2 1;1 2 3 4]. I want to append different size matrices at each iteration:
M=[4 3 2 1;1 2 3 4];
for i=1:t
newM=createNewMatrix;
M=[M;newM];
end
newM can be [] or a Nx4 matrix. This is very slow though. What is the fastest way to do this?
Update
Pre-allocating would look like this?
M=zeros(200000,4)
start=1
M(1:2,:)=M=[4 3 2 1;1 2 3 4];
for i=1:t
newM=createNewMatrix;
size_of_newM=size(newM,1);
finish=start+size_of_newM-1;
M(start:finish,:)=newM;
start=finish;
end
Like suggested, preallocation gives the most boost.
Using cell arrays is another good approach and could be implemented like this:
M = cell(200000, 1);
M{1} = [4 3 2 1; 1 2 3 4];
for t=2:200000
i = randi(3)-1;
M{t}=rand(i,4);
end
MC = vertcat(M{:});
In principle you generate a cell array with arbitrary long arrays in each cell and then concatenate them afterwards.
This worked for me nearly twice as fast as your preallocation update. On the other hand, this still was only around one second for the example with 200k iterations...