I want to construct a function that accepts input n and gives the vector
[n n-1 n-2 ... n-n, n-1 n-2 ... n-n, ..., n-n]
//Example
input : n=3
output : [3 2 1 0 2 1 0 1 0 0]
I know how to do this using loops, but I'm looking for a clever way to do it in MATLAB
You can use repmat to repeat the matrix a few times, and then select only the triangular part by means of tril. Like this:
n=3;
x=repmat(n:-1:0,1,n+1);
result=x(tril(ones(n+1))>0)
Or in one line:
n=3;
getfield(repmat(n:-1:0,1,n+1),{reshape(tril(ones(n+1))>0,1,(n+1)^2)})
The result of this function is the desired output:
result =
3 2 1 0 2 1 0 1 0 0
Since you haven't gotten any answers, here's a way to do it:
N = 3;
x = repmat(N:-1:0,1,N+1)-cumsum(repmat([1 zeros(1,N)],1,N+1))+1
x = x(x>=0)
x =
3 2 1 0 2 1 0 1 0 0
Related
I hope you can help with a little problem I am having.
I want to upsample and downsample a vector with zeros and ones. We have the functions upsample and downsample for that, however, the upsample function in Matlab only adds zeros to the vector. I would like to repeat the value, instead of just putting in zeros.
Unfortunately the upsample function does not do that. Thus, I tried to use repmat (in the third dimension) and then reshape to get back to the old format. I know it must be possible with these functions, but if I simply use them, the vector just gets duplicated and added to the end.
An example:
The input vector is: [1 0 0 1 0 1 0 1 1 1 0 0] (these should be random).
Now I want to upsample (say) by a factor of 2. Then I want to get:
[1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 0 0 0 0].
Thanks in advance for any help!
You can use repelem:
>> repelem([1 0 1],2)
ans =
1 1 0 0 1 1
Or using repmat and reshape when input is a column vector:
>> input = [1 0 1];
>> reshape(repmat(input, 2, 1), 1, [])
ans =
1 1 0 0 1 1
I have a matrix with some zero values I want to erase.
a=[ 1 2 3 0 0; 1 0 1 3 2; 0 1 2 5 0]
>>a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
However, I want to erase only the ones after the last non-zero value of each line.
This means that I want to retain 1 2 3 from the first line, 1 0 1 3 2 from the second and 0 1 2 5 from the third.
I want to then store the remaining values in a vector. In the case of the example this would result in the vector
b=[1 2 3 1 0 1 3 2 0 1 2 5]
The only way I figured out involves a for loop that I would like to avoid:
b=[];
for ii=1:size(a,1)
l=max(find(a(ii,:)));
b=[b a(ii,1:l)];
end
Is there a way to vectorize this code?
There are many possible ways to do this, here is my approach:
arotate = a' %//rotate the matrix a by 90 degrees
b=flipud(arotate) %//flips the matrix up and down
c= flipud(cumsum(b,1)) %//cumulative sum the matrix rows -and then flip it back.
arotate(c==0)=[]
arotate =
1 2 3 1 0 1 3 2 0 1 2 5
=========================EDIT=====================
just realized cumsum can have direction parameter so this should do:
arotate = a'
b = cumsum(arotate,1,'reverse')
arotate(b==0)=[]
This direction parameter was not available on my 2010b version, but should be there for you if you are using 2013a or above.
Here's an approach using bsxfun's masking capability -
M = size(a,2); %// Save size parameter
at = a.'; %// Transpose input array, to be used for masked extraction
%// Index IDs of last non-zero for each row when looking from right side
[~,idx] = max(fliplr(a~=0),[],2);
%// Create a mask of elements that are to be picked up in a
%// transposed version of the input array using BSXFUN's broadcasting
out = at(bsxfun(#le,(1:M)',M+1-idx'))
Sample run (to showcase mask usage) -
>> a
a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
>> M = size(a,2);
>> at = a.';
>> [~,idx] = max(fliplr(a~=0),[],2);
>> bsxfun(#le,(1:M)',M+1-idx') %// mask to be used on transposed version
ans =
1 1 1
1 1 1
1 1 1
0 1 1
0 1 0
>> at(bsxfun(#le,(1:M)',M+1-idx')).'
ans =
1 2 3 1 0 1 3 2 0 1 2 5
can I know how can I replace values in specific matrix position without using for loop in MATLAB? I initialize matrix a that I would like to replace its value on specified row and column for each no. This has to be done a few time within num for loop. The num for loop is important here because I would want the update the value in the original code.
The real code is more complicated, I am simplifying the code for this question.
I have the code as follow:
a = zeros(2,10,15);
for num = 1:10
b = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
c = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
d = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
for no = 1:15
a(b(no),d(no),no) = c(1,no,:)
end
end
A sample output for, say no 13 is as follows:
a(:,:,13) =
Columns 1 through 8
0 0 0 0 0 7.3160 0 0
0 0 0 0 0 0 0 0
Columns 9 through 10
0 0
0 0
Thank you so much for any help I could get.
It can be done using sub2ind, which casts the subs to a linear index.
Following your vague variable names, it would look like this (omitting the useless loop over num):
a = zeros(2,10,15);
b = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
d = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
c = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
% // we vectorize the loop over no:
no = 1:15;
a(sub2ind(size(a), b, d, no)) = c;
Apart from the sub2ind based approach as suggested in Nras's solution, you can use a "raw version" of sub2ind to reduce a function call if performance is very critical. The related benchmarks comparing sub2ind and it's raw version is listed in another solution. Here's the implementation to solve your case -
no = 1:15
a = zeros(2,10,15);
[m,n,r] = size(a)
a((no-1)*m*n + (d-1)*m + b) = c
Also for pre-allocation, you can use a much faster approach as listed in Undocumented MATLAB blog post on Preallocation performance with -
a(2,10,15) = 0;
The function sub2ind is your friend here:
a = zeros(2,10,15);
x = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
y = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
z = 1:15;
dat = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
inds = sub2ind(size(a), x, y, z);
a(inds) = dat;
Matlab provides a function 'sub2ind' may do what you expected.
with variable as the same you posted:
idx = sub2ind(size(a),b,d,[1:15]); % return the index of row a column b and page [1:15]
a(idx) = c;
I want to create a variable that finds a pattern (let's say [1 1]) in different rows of a matrix (A). Of course there aren't an equal number of occurrences of this string in each row.
A = [ 0 0 0 1 1
1 1 1 0 0
0 1 0 1 1
1 1 1 0 0
0 1 0 0 1
1 0 1 1 1
0 1 0 1 0
1 1 1 0 1];
I could do:
for i = 1:n
var(i,:) = strfind(A(i,:),[1 1]);
end
but then both sides of the equation won't be equal.
ERROR: ??? Subscripted assignment dimension mismatch.
I try to preallocate. I create a matrix with what I think would be the maximum number of occurrences of this string in each row of matrix A (let's say 50).
for i = 1:n
var(i, :) = NaN(1,50)
end
That's followed by the previous bit of code and it's no good either.
I've also tried:
for i = 1:n
var(i,1:numel(strfind(A(i,:),[1 1])) = strfind(A(i,:),[1 1])
end
Error: The expression to the left of the equals sign is not a valid
target for an assignment.
How should I go about doing this?
The output I expect is a matrix var(i,:) that gives me the position in the matrix where each of these patterns occur. It works fine for just one row.
For example:
var(1,:) = [1 2 5 8 10 22 48]
var(2,:) = [2 3 4 7 34 45 NaN]
var(3,:) = [4 5 21 32 33 NaN]
Thanks!
In your first try: you tried to build a matrix with different length of rows.
In your second try: you pre-allocated, but then run it over by re-definning var(i,:), while you tried to put there your desired result.
In your third try: unfortunately you just missed one brackets- ) at the end of left expression.
This code suppose to work (what you did at 2nd and 3rd attempts, with pre-allocate and fixed brackets):
var=NaN(1,50);
for i = 1:n
var(i,1:numel(strfind(A(i,:),[1 1]))) = strfind(A(i,:),[1 1])
end
Say I have a vector containing only logical values, such as
V = [1 0 1 0 1 1 1 1 0 0]
I would like to write a function in MATLAB which returns a 'streak' vector S for V, where S(i) represents the number of consecutive 1s in V up to but not including V(i). For the example above, the streak vector would be
S = [0 1 0 1 0 1 2 3 4 0]
Given that I have to do this for a very large matrix, I would very much appreciate any solution that is vectorized / efficient.
This should do the trick:
S = zeros(size(V));
for i=2:length(V)
if(V(i-1)==1)
S(i) = 1 + S(i-1);
end
end
The complexity is only O(n), which I guess should be good enough.
For your sample input:
V = [1 0 1 0 1 1 1 1 0 0];
S = zeros(size(V));
for i=2:length(V)
if(V(i-1)==1)
S(i) = 1 + S(i-1);
end
end
display(V);
display(S);
The result would be:
V =
1 0 1 0 1 1 1 1 0 0
S =
0 1 0 1 0 1 2 3 4 0
You could also do it completely vectorized with a couple intermediate steps:
V = [1 0 1 0 1 1 1 1 0 0];
Sall = cumsum(V);
stopidx = find(diff(V)==-1)+1;
V2=V;
V2(stopidx) = -Sall(stopidx)+[0 Sall(stopidx(1:end-1))];
S2 = cumsum(V2);
S = [0 S2(1:end-1)];
Afaik the only thing that can take a while is the find call; you can't use logical indexing everywhere and bypass the find call, because you need the absolute indices.
It's outside the box - but have you considered using text functions? Since strings are just vectors for Matlab it should be easy to use them.
Regexp contains some nice functions for finding repeated values.