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Closed 10 years ago.
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Find rows in matrix where entries match certain constraints?
In Matlab, i have a matrix (MxN) and I want to find the rows whose entry in a specific column is equal to a specified value. For example, I have a matrix as follow:
0 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 1 0 0 1
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 1 0
0 0 0 0 1 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
0 1 1 0 0 0 1 0 1 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
I want to find the rows whose 7th element us equal to 1. In this example, the result matrix should contains rows 2 and 7.
if your matrix is called A, just do:
A(A(:,7)==1,:)
PS: I think that this question has already been answered a million times...
Related
This question already has answers here:
Creating Indicator Matrix
(6 answers)
Closed 5 years ago.
I have vector like V[5000,1] with values in range 1-10
How to get matrix W[5000,10]
V[1] W[1 0 0 0 0 0 0 0 0 0]
[2] W[0 1 0 0 0 0 0 0 0 0]
[10] W[0 0 0 0 0 0 0 0 0 1]
[7] W[0 0 0 0 0 0 1 0 0 0]
... ...
its like W have "1" in column indicated by vector V
Solution
You can use the following approach:
W = zeros(length(v),10);
W(sub2ind(size(W),1:length(v),v'))=1;
Results
v = [10,4,10,4,9]
W =
0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
This question already has answers here:
How can I index a MATLAB array returned by a function without first assigning it to a local variable?
(9 answers)
Matlab Error: ()-indexing must appear last in an index expression
(1 answer)
Closed 5 years ago.
I'm trying to realise the following Octave command in MATLAB:
M = eye(x)(y,:);
x is just a number (in my example 10) and y is a vector (here 8x1):
y = [1 3 4 5 7 10 9 10];
The Octave command would generate:
M =
1 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
The ones are kept very near to the diagonal.
The nearest I came with MATLAB is with the following commands:
n = size(y,1);
Y = eye(n, x);
but it would generate something still different. If the difference between rows and columns gets bigger, it would be very different.
M =
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
How could I get the first matrix with MATLAB?
First you should find what this expression eye(x)(y,:) means. First create an identity matrix with the size of x by x, and then select rows with index in y. Therefore, the equivalent syntax would be:
E = eye(x);
M = E(y,:);
I have this matrix
X= [2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 250;
3 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 250;
2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 250;
3 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 250;
4 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 250;
3 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 250;
2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 250;
4 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 250;
3 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 250;
3 1 1 1 0 0 1 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 400]
I need to do three different sequence things in this matrix:
1- Search in this matrix to the following sequence 1 1 0 0 0 and write those rows that have this characteristic in new matrix (like row 1).
2- Use the matrix that generate in the first step and remove from it to the rows that have the same number in the same digits (like row 1,3,7) but at the same time keep only one row of each one (in the case of row 1,3,7 keep row 1 and remove other rows) .
3- use the matrix that generate in the second step and remove from this matrix any row that have following sequence 1 1 1 (like row 8) and put the other rows in this matrix in new matrix.
%Step-1
% Converting the matrix into a string, appending a semi-colon for similarity and removing the brackets from the string
req=mat2str(X); req(end)=';' ; req=req(2:end);
% Searching the sequence: 1 1 0 0 0
sp1=strfind(req, '1 1 0 0 0');
% Storing those rows of X in req matrix which contain the sequence
req=X(unique(ceil([sp1]/(size(req,2)/size(X,1)))),:);
%Step-2
req= unique(req,'rows');
%Step-3
% Converting the matrix into a string, appending a semi-colon for similarity and removing the brackets from the string
reqtemp=mat2str(req); reqtemp(end)=';' ; reqtemp=reqtemp(2:end);
% Searching the sequence: 1 1 1
sp1=strfind(reqtemp, '1 1 1');
% Removing those rows which contain the sequence
req(unique(ceil([sp1]/(size(reqtemp,2)/size(req,1)))),:)=[];
Here is an example matrix (but the result shouldn't be constrained to only working on this):
a=zeros(7,7);
a(5,3:6)=1;
a(2,2)=1;
a(2,4)=1;
a(7,1:2)=1
a=
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
I want to get rid of all the 1's that are alone (the noise), such that I only have the line of 1's on the fifth row.
rules:
-the 1's are in 'connected lines' if there are adjacent 1's (including diagonally) e.g.:
0 0 0 1 0 0 1 0 1
1 1 1 0 1 0 0 1 0
0 0 0 0 0 1 0 0 0
(The connected lines are what I want to keep. I want to get rid of all the 1's that are not in connected lines, the connected lines can intersect each other)
the 'connected lines need to be at least 3 elements long. So in the 7x7 example, there would only be one line that matches this criteria. If a(7,3) was set to 1, then there would be a connected line at the bottom left also
I am currently looking at this through a column by column approach, and here is the first draft of my code so far:
for nnn=2:6
rowPoss=find(a(:,nnn)==1);
rowPoss2=find(a(:,nnn+1)==1);
for nn=1:length(rowPoss)
if myResult(rowPoss(nn)-1:rowPoss(nn)+1,n-1)==0 %
%then?
end
end
end
My difficulty is, during this column by column process, I'd have to enable a way to recognise the beginning of the connected line, the middle of the connected line, and when a connected line ends. The same rules for this, when applied to noise (the lone 1's), would just ignore the lone 1's.
The output I want is basically:
b=
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 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
If you have image processing toolbox, try bwareaopen
b = bwareaopen(a, 3);
Sample Run #1:
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
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 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Sample Run #2:
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
I have one binary image so it has only 2 value like 0 and 1. After, I convert this into a padded image of different values, like the image will have curve shape. I took a 3 X 3 matrix of value and if i get curve shape then I padded the image with 1, or any number. I use 15 different types shape values like junction point, end point etc.
After, I give the values 1 to 15 - or the appropriate number according its shape. As such, I am getting an image like:
Figure
0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 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 0 0 0 0 0 0 0 0 0 0
I would like to count how many 1s there are in the image, followed by 2s, 3s, etc. up to 15. For example,
as shown in the figure, if the pad number was 5, the total number of pixels would be 3. If the pad number was 1, the total number of pixels would be 6.
Use histc:
>> im = [ 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 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 0 0 0 0 0 0 0 0 0 0 ]; %// data
>> values = 1:15; %// possible values
>> count = histc(im(:), values)
count =
6 %// number of 1's
0 %// number of 2's, etc
0
0
3
0
0
0
0
0
0
0
0
0
0
Or compute it manually with bsxfun:
>> count = sum(bsxfun(#eq, im(:), values(:).'), 1)
>> count =
6 0 0 0 3 0 0 0 0 0 0 0 0 0 0
I can also suggest using accumarray:
im = [0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 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 0 0 0 0 0 0 0 0 0 0 ]; %// data - borrowed from Luis Mendo
counts = accumarray(im(:) + 1, 1);
counts(1) = []
counts =
6
0
0
0
3
Note we have to offset by 1 as accumarray starts indexing the output array at 1. Because you want to disregard the 0s, I simply take the counts result and remove the first entry. This result agrees with what you are seeking. The first element is how many 1s we have encountered, which is 6. The last element is how many 5s you have encountered, which is 3. Because 5 is the largest number encountered in your image, we can say that all symbols after 5 (6, 7, 8, ..., 15) have a count of 0.