Let's say I have a matrix
A = [2 3 2 5 6 7 2;
1 2 5 4 5 6 7;
7 5 3 9 8 1 2];
How do I remove 2s and keep one 2 in the first row and keep only one 5 in the second row?
The result can't be a matrix anymore, because each row will have a different length. You can obtain the result as a cell array of row vectors as follows:
B = mat2cell(A, ones(size(A,1),1)); %// convert matrix to cell array of its rows
B = cellfun(#(x) unique(x,'stable'), B, 'uniformoutput', 0); %// stably remove duplicates
For your example matrix
A = [2 3 2 5 6 7 2;
1 2 5 4 5 6 7;
7 5 3 9 8 1 2];
this gives
B{1} =
2 3 5 6 7
B{2} =
1 2 5 4 6 7
B{3} =
7 5 3 9 8 1 2
If you want to find out which values are duplicates within the row, you can do something like this:
[vals, col_idx] = sort(A,2);
idx = bsxfun(#plus,(col_idx-1)*size(A,1), (1:size(A,1))');
is_duplicate(idx(:,2:end)) = vals(:,1:end-1) == vals(:,2:end);
is_duplicate = reshape(is_duplicate, size(A));
is_duplicate =
0 0 1 0 0 0 1
0 0 0 0 1 0 0
0 0 0 0 0 0 0
From there, it depends what outcome you are looking for. You could set the duplicates to NaN or some other value, or you could set them to NaN, but then shift them to the end of the row, using something like the following:
col_idx = cumsum(~is_duplicate, 2);
idx = bsxfun(#plus,(col_idx-1)*size(A,1), (1:size(A,1))');
A_new = nan(size(A));
A_new(idx(~is_duplicate)) = A(~is_duplicate);
A_new =
2 3 5 6 7 NaN NaN
1 2 5 4 6 7 NaN
7 5 3 9 8 1 2
Related
I have an IxS matrix called numerateur and I would like to set one element per row to zero. However, the column to which this element belongs changes for each row. I can do it with a loop but would like to use a more efficient method.
You can find my loop below:
choice is an Ix1 vector containing, for each row, the column number of the element I want to set to zero in numerateur.
for i = 1:I
rank1 = choice(i);
numerateur(i,rank1) = 0;
end
You need to convert the column indices to linear indices:
ind = (1:size(numerateur,1)) + (choice(:).'-1)*size(numerateur,1);
numerateur(ind) = 0;
Example:
>> numerateur = randi(9,5,7)
numerateur =
8 7 8 3 2 8 4
7 8 6 3 6 1 3
8 9 5 5 5 9 5
4 4 2 7 4 9 9
1 6 7 3 8 7 9
>> choice = [2; 3; 6; 5; 1];
>> ind = (1:size(numerateur,1)) + (choice(:).'-1)*size(numerateur,1);
>> numerateur(ind) = 0;
>> numerateur
numerateur =
8 0 8 3 2 8 4
7 8 0 3 6 1 3
8 9 5 5 5 0 5
4 4 2 7 0 9 9
0 6 7 3 8 7 9
Use implicit expansion to find logical indices:
numerateur(choice == 1:size(numerator,2)) = 0;
Alternatively you can use sparse matrix to generate logical indices:
numerateur(sparse(1:I,choice,true,I,S)) = 0;
I will explain my question using an example. Imagine you have a 2D matrix like below:
5 4 3 8 0 0
5 4 2 9 1 0
5 6 2 7 2 0
5 4 7 9 0 0
5 6 7 1 2 0
By islands I mean column groups of same elements (except zeros).
I would like to find the histogram of length of islands except those consisting of zero elements.
This matrix has
island-length occurrence
5 1
2 3
1 12
How can I realize this task using Matlab ?
Maybe there are shorter possibilities, but this will do - and it is fully vectorized:
A = [5 4 3 8 0 0
5 4 2 9 1 0
5 6 2 7 2 0
5 4 7 9 0 0
5 6 7 1 2 0]
%// pad zeros to first line of A
X(2:size(A,1)+1,:) = A;
%// differences of X
dX = diff(X)
%// cumulative sum of "logicalized" differences
cs = cumsum(logical(dX(:)))
%// filter out zeros
cs = cs(logical(A(:)))
%// count occurances
aa = accumarray(cs,1)
%// unique occurances
uaa = unique(aa)
%// count unique occurances
occ = hist(aa,uaa).'
%// accumarray may introduce new zeros, filter out
mask = logical(uaa)
%// output
out = [occ(mask) uaa(mask)]
out =
12 1
3 2
1 5
Needed a slight modification to one of my old snippets to filter the zeros. Here you go:
% Your Matrix
A = [ 5 4 3 8 0 0;
5 4 2 9 1 0;
5 6 2 7 2 0;
5 4 7 9 0 0;
5 6 7 1 2 0];
% Find Edges (Ends of Islands)
B = diff(A);
B = [ones(1,size(A,2));B~=0;ones(1,size(A,2))];
% At each column, find distances between island edges, filter out zero islands.
R = cell(size(A,2),1);
for i = 1:size(A,2)
[C ~] = find(B(:,i));
Ac = A(C(1:end-1),i);
D = diff(C);
D(Ac==0)=[];
R{i} = D;
end
% Find histogram of island lengths
R = R(find(~cellfun(#isempty,R)),1);
R = cell2mat(R);
[a,~,c] = unique(R);
out = [a, accumarray(c,ones(size(R)))];
I have a matrix of unsorted numbers and I want to keep the n largest (not necessarily unique) values per column and set the rest to zero.
I figured out how to do it with a loop:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4]
k = 2
for n = 1:4
[y, ind] = sort(a(:,n), 'descend');
a(ind(k+1:end),n) = 0;
end
a
which gives me:
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
k =
2
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
However when I try to eliminate the loop, I can't seem to get the indexing right, because this:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4]
k = 2
[y, ind] = sort(a, 'descend');
b = ind(k+1:end,:)
a(b) = 0
which gives me this: (which is not what I wanted to do)
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
k =
2
b =
4 3 3 1
1 2 2 4
a =
0 8 12 5
0 2 6 18
0 3 9 7
0 9 12 4
Am I indexing this wrong? Do I have to use the loop?
I referenced this question to get started but it wasn't exactly what I was trying to do: How to find n largest elements in an array and make the other elements zero in matlab?
You're very close. ind in the sort function gives you the row locations for each column where that particular value would appear in the sorted output. You need to do some additional work if you want to index into the matrix properly and eliminate the entries. You know that for each column of I, that tells you that we need to eliminate those entries from that particular column. Therefore, what I would do is generate column-major linear indices using each column of I to be the rows we need to eliminate.
Try doing this:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4];
k = 2;
[y, ind] = sort(a, 'descend');
%// Change here
b = sub2ind(size(a), ind(k+1:end,:), repmat(1:size(a,2), size(a,1)-k, 1));
a(b) = 0;
We use sub2ind to help us generate our column major indices where the rows are denoted by the values in ind after the kth element and the columns we need are for each column in this matrix. There are size(a,1)-k rows remaining after you truncate out the k values after sorting, and so we generate column values that go from 1 up to as many columns as we have in a and as many rows as there are remaining.
We get this output:
>> a
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
Here's one using bsxfun -
%// Get descending sorting indices per column
[~, ind] = sort(a,1, 'descend')
%// Get linear indices that are to be set to zeros and set those in a to 0s
rem_ind = bsxfun(#plus,ind(n+1:end,:),[0:size(a,2)-1]*size(a,1))
a(rem_ind) = 0
Sample run -
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
n =
2
ind =
3 4 1 2
2 1 4 3
4 3 3 1
1 2 2 4
rem_ind =
4 7 11 13
1 6 10 16
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
I have a variable like this that is all one row:
1 2 3 4 5 6 7 8 9 2 4 5 6 5
I want to write a for loop that will find where a number is less than the previous one and put the rest of the numbers in a new row, like this
1 2 3 4 5 6 7 8 9
2 4 5 6
5
I have tried this:
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
m = zeros(size(test));
for i=1:numel(test)-1;
for rows=1:size(m,1)
if test(i) > test(i+1);
m(i+1, rows+1) = test(i+1:end)
end % for rows
end % for
But it's clearly not right and just hangs.
Let x be your data vector. What you want can be done quite simply as follows:
ind = [find(diff(x)<0) numel(x)]; %// find ends of increasing subsequences
ind(2:end) = diff(ind); %// compute lengths of those subsequences
y = mat2cell(x, 1, ind); %// split data vector according to those lenghts
This produces the desired result in cell array y. A cell array is used so that each "row" can have a different number of columns.
Example:
x = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
gives
y{1} =
1 2 3 4 5 6 7 8 9
y{2} =
2 4 5 6
y{3} =
5
If you are looking for a numeric array output, you would need to fill the "gaps" with something and filling with zeros seem like a good option as you seem to be doing in your code as well.
So, here's a bsxfun based approach to achieve the same -
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5] %// Input
idx = [find(diff(test)<0) numel(test)] %// positions of row shifts
lens = [idx(1) diff(idx)] %// lengths of each row in the proposed output
m = zeros(max(lens),numel(lens)) %// setup output matrix
m(bsxfun(#le,[1:max(lens)]',lens)) = test; %//'# put values from input array
m = m.' %//'# Output that is a transposed version after putting the values
Output -
m =
1 2 3 4 5 6 7 8 9
2 4 5 6 0 0 0 0 0
5 0 0 0 0 0 0 0 0
can any one help me to Vectorized this loop.
i have large Matrix and i want to replace all the pixel values whose length is less then some threshold Value For simplicity lets say
a = randi([1 5],10,10);
for i = 1:length(a)
someMat=a(a==i);
if length(someMat)<20
a(a==i)=0;
end
end
but its killing me.
Example:
a = randi([1 5],10,10)
a =
5 2 1 5 5 5 2 2 3 2
3 3 5 4 4 4 3 1 1 5
5 1 3 5 3 3 4 1 3 1
3 1 5 3 2 5 1 1 5 1
1 1 4 3 4 3 4 4 5 1
1 4 3 5 1 1 2 2 2 1
3 3 5 2 4 1 1 3 2 4
4 1 5 3 4 5 3 4 3 3
5 3 5 5 4 3 1 3 4 1
4 1 1 3 5 5 1 3 3 5
Result for Thresold 20
5 0 1 5 5 5 0 0 3 0
3 3 5 0 0 0 3 1 1 5
5 1 3 5 3 3 0 1 3 1
3 1 5 3 0 5 1 1 5 1
1 1 0 3 0 3 0 0 5 1
1 0 3 5 1 1 0 0 0 1
3 3 5 0 0 1 1 3 0 0
0 1 5 3 0 5 3 0 3 3
5 3 5 5 0 3 1 3 0 1
0 1 1 3 5 5 1 3 3 5
length of pixel 4 was 17
length of pixel 2 was 10
i try it by some thing like
[nVal Index] = histc(a(:),unique(a)); %
nVal(nVal>20) = 1; % just some threshold value and assigning by some Number may be zero as well
But I dont Know how to replace the Index Values of the corresponding Pixal and apply reshape to get it in original form. Here Even i am not sure that i will get the same Matrix With Reshape . Please Help me.....
thanks
I think this does what you want:
threshold_length = 20;
replace_value = 0;
u = unique(a); %// values of a
h = histc(a(:), u); %// count for each value
r = u(h<threshold_length); %// values to be removed
a(ismember(a,r)) = replace_value; %// remove those values
I see #LuisMendo arrived at mostly the same solution quicker than I did, but an alternative to using ismember is to use more of what unique gives you:
threshold = 20;
[vals, ~, ix] = unique(a); % capture the values and their indices
counts = histc(a(:), vals); % count the occurrences of each value
vals(counts<threshold) = 0; % zero the values that aren't common enough
a(:) = vals(ix); % recreate the matrix with updated values