array = [2 1 3 2 1]
for i = 2:length(array)
value = array(i);
j = i - 1;
array_j=array(1:j);
array_j_indices=cumsum(array_j>value);
[~,n]=find(array_j_indices==1);
newArray=array;
array(n+1:i)=array_j(array_j>value);
j=j-max(array_j_indices);
array(j+1) = value;
end %forLoop
disp(array);
Hello,
I saw this code for vectorising while loop insertion code but i cannot seem to understand how it works.
How does cumsum(array_j>value) work? I understand and tested cumsum functions but i can't seem to understand how the rational operator of (array_j>value) works in the within a cumsum function under the for loop.
Also, i dont understand how [~,n]=find(array_j_indices==1) stores value for the matrix of n. Does it store it only in columns because there is a not (~) in the rows?
cumsum(array_j>value)?
array_j>value: due to the sorted nature of array_j, the result is always some zeros followed by some ones, e.g. [0 0 0 0 1 1 1 1]
cumsum(array_j>value) = [0 0 0 0 1 2 3 4]: at most one element will be equal to 1.
[~,n]=find(array_j_indices==1); ?
Because there is only one row, this is equal to n=find(array_j_indices==1);.
Fastest implementation?
Note that this 'vectorised' code is slower the following (easier) implementation:
for i = 2:length(array)
value = array(i);
j = i - 1;
n=find(array(1:j)>value,1);
array(n+1:i)=array(n:j);
array(n) = value;
end
and much slower than the built-in matlab sort method.
Related
I am trying to create a single column vector (out), which is comprised of a sequence of ones and zeros. These should occur in sets of length B and C respectively, which are repeated A number of times. For example:
out=[1
0
0
1
0
0
1
0
0]
It is currently set up as:
out=[0]; %not ideal, but used to initially define 'out'
A=3;
B=1;
C=2;
for i = 1:length(A)
for ii = 1:length(B)
out(end+1,1) = ones(ii,1);
end
for iii = 1:length(C)
out(end+1,1) = zeros(iii,1);
end
end
This is not working - current output:
out=[0
1
0]
How can I correct these loops to get the desired output? Also, is there a better way of achieving this with the given the inputs?
Many thanks.
1) You do not need to use length as this returns the length of an array type, so A,B,C will all be length of 1.
2) Just directly use the values as shown below. Also you can initialize an empty array with empty brackets []
3) If you're using the zeros and ones commands, these generate whole arrays/matrices and do not need to be in a loop. If you want to keep your loop version, just use =1 or =0
out=[]; %--> you can use this instead
A=3;
B=1;
C=2;
for i = 1:A
out(end+1:end+B,1) = ones(B,1);
out(end+1:end+C,1) = zeros(C,1);
end
... or of course to be more "Matlaby" just do what David said in the comments repmat([ones(B,1);zeros(C,1)],A,1), but the above is there to help you on your way.
How about some modulo arithmetic?
result = double(mod(0:(B+C)*A-1, B+C)<B).';
Example:
>> B = 2; %// number of ones in each period
>> C = 4; %// number of zeros in each period
>> A = 3; %// number of periods
>> result = double(mod(0:(B+C)*A-1, B+C)<B).'
result =
1
1
0
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
I can suggest 2 ways:
a)Using for loop-
A=3;
B=2;
C=3;
OneVector=ones(1,B); % B is the length of ones.
zeroVector=zeros(1,C); % C is the length of zeros.
combinedVector=cat(2,OneVector,zeroVector);
Warehouse=[]; % to save data
for(i=1:A)
Warehouse=cat(2,Warehouse,combinedVector);
end
b)using repmat:
OneVector=ones(1,B); % B is the length of ones.
zeroVector=zeros(1,C); % C is the length of zeros.
combinedVector=cat(2,OneVector,zeroVector);
Warehouse=repmat(combinedVector, [A,1]);
I hope, this will solve your problem.
Say i have a 2D matrix A:
A = [ 1 1 0 0
1 0 0 0
1 1 1 0];
A is not necessarily binary, or even integer (i.e., floats are possible). I want to remove any column that contains uniform valued elements. In the above example, i would get:
1 0
0 0
1 1
To make this fully general, i'd like to allow the user to select the dimension along which rows/columns/slices are removed (i.e., with a DIM option).
Any ideas?
You could try using the min and max functions, which allow you to use the dim argument.
For example
index = min(A,[],1)==max(A,[],1);
A(:,index)=[];
will remove the columns you want. It is straightforward to do the same for rows
index = min(A,[],2)==max(A,[],2);
A(index,:)=[];
One-liner:
B = A(:,range(A)~=0); %//columns
The other one-liner is not that nice, and ugly one-liners should not be written down. :-) But is basically the same solution as S..'s, except is way more expensive (requires stats toolbox).
Please note that "generality" of subscript-based solutions doesn't extend to N-dimensional arrays as easily, because subscripting in ND arrays without checking beforehand the number of dimensions is difficult. Also, for the 1D arrays the notion of "uniformity" is a bit odd along the singleton dimension (the result is always empty).
Besides the neat solution provided by #S.. there is this simple hack also for your example:
for ii = 1:size(A,2)
T(ii) = all(A(:,ii) == sum(A(:,ii))/numel(A(:,ii)));
end
A(:,~T)
ans =
1 0
0 0
1 1
As suggested by #gariepy the right side of the equation can be replaced with mean function.
for ii = 1:size(A,2)
T(ii) = all( A(:,ii) == mean(A(:,ii)) );
end
A(:,~T)
A(:,~all(A == repmat(A(1,:),[size(A,1) 1])))
Inspired by #S.. but only checks if every element of the column equals the first element of the column. Seems like a little less work for the processor than finding the min and the max, and checking for equality.
I've written a simple function that takes a vector vec, iterates through it, performing an operation whose result is stored in another vector vecRes of same size at same index, and returns vecRes upon completing the loop. Below is function code:
function [ vecRes ] = squareTerms( vec )
vecSize = size(vec);
vecRes = zeros(vecSize);
for i = 1:vecSize
vecRes(i) = vec(i)^2;
end
end
Problem is that it seems to exit too early, after only one iteration in fact as the output appears as:
vecRes = 1 0 0 0 0 0 0 0 0 0
For input:
vec = 1 2 3 4 5 6 7 8 9 10
I can't figure out why it does so. Any help is greatly appreciated.
Size returns 2 values, rows and columns. Probably you are a having a 1xN vector. So size returns [1 N] and your loop runs 1 time.
>>> size ([1 2 3])
>
> ans =
>
> 1 3
>
>>> 1:size ([1 2 3])
>
> ans =
>
> 1
Others have pointed out the problem. My preferred solution in this sort of case is to use numel, i.e.
vecRes = zeros(size(vec));
for i = 1:numel(vec)
vecRes(i) = vec(i) ^ 2;
end
Of course, in this case, vectorisation is better still:
vecRes = vec .^ 2;
Replace
for i = 1:vecSize
with
for i = 1:vecSize(2)
vecSize is an array of numbers, not just a single value. For example, if vec is a 1 by 8 vector, then size(vec) will return [1, 8].
Therefore, your for-loop-statement,
for i = 1:vecSize
, is actually equivalent to something like:
for i = 1:[1, 8]
This doesn't make much sense. There are a number of ways to fix the problem. You could write:
for i = 1:length(vec)
or
for i = 1:numel(vec) % "numel" stands for "number of elements"
If the vector is 1xn instead of nx1, you can write:
for i = 1:size(vec, 2)
Yet another alternative is:
for i = 1:max(vecSize)
However, the most sensible option is not to write the squareTerms function at all and simply write
vecRes = vec.^2;
Note the dot before the caret. vec^2 and vec . ^2 are not the same thing.
If you put a dot before an operator sign, the operation will be performed element-wise. For example,
C = A * B
performs matrix multiplication, but
C = A .* B
will cause the first element of A to by multiplied by the first element of B, and the result will be assigned to the first element of C. Then, the product of the second elements of A and B will be taken, and the result will be stuck in the second element of C, and so on.
vecRes = vec.^2;
I'm trying to get a logical matrix as a result of a condition that is specific for each column M(:,i) of the original matrix, based on the value of the same index i in vector N, that is, N(i).
I have looked this up online, but can't find anything quite like it. There must be a simple and clean way of doing this.
M =
3 -1 100 8
200 2 300 4
-10 0 0 400
N =
4 0 90 7
and my desired solution is, for each column of M(:,i), the values less than N(i):
1 1 0 0
0 0 0 1
1 0 1 0
It's a standard use-case for bsxfun:
O = bsxfun(#lt, M, N)
Here #lt is calling the "less than" function, i.e. it is the function handle to the < operator. bsxfun will then "expand" N along its singleton dimension by applying the function #lt to each row of M and the whole of N.
Note that you can easily achieve the same thing using a for-loop:
O = zeros(size(M));
for row = 1:size(M,1)
O(row,:) = M(row,:) < N;
end
Or by using repmat:
O = M < repmat(N, size(M,1), 1);
but in MATLAB the bsxfun is usually the most efficient.
Possible two-line solution using arrayfun to apply the comparison to each column and index pair:
T = arrayfun(#(jj)M(:,jj) < N(jj), 1:numel(N), 'UniformOutput', false);
result = cat(2,T{:});
Edit: Of course, the bsxfun solution is much more efficient.
Is is possible to put two for statements into one statement. Something like
A = [ 0 0 0 5
0 2 0 0
1 3 0 0
0 0 4 0];
a=size(A);
b=size(A);
ind=0;
c=0;
for ({i=1:a},{j=1:b})
end
Your question is very broad, but one thing to consider is that in MATLAB you can often take advantage of linear indexing (instead of subscripting), without actually having to reshape the array. For example,
>> A = [ 0 0 0 5
0 2 0 0
1 3 0 0
0 0 4 0];
>> A(3,2)
ans =
3
>> A(7) % A(3+(2-1)*size(A,1))
ans =
3
You can often use this to your advantage in a for loop over all the elements:
for ii=1:numel(A),
A(ii) = A(ii) + 1; % or something more useful
end
Is the same as:
for ii=1:size(A,2),
for jj=1:size(A,1),
A(jj,ii) = A(jj,ii) + 1;
end
end
But to address your specific goal in this problem, as you stated in the comments ("I am storing the non zero elements in another matrix; with elements like the index number, value, row number and column number."), of making sparse matrix representation, it comes to this:
>> [i,j,s] = find(A);
>> [m,n] = size(A);
>> S = sparse(i,j,s,m,n)
S =
(3,1) 1
(2,2) 2
(3,2) 3
(4,3) 4
(1,4) 5
But that's not really relevant to the broader question.
Actually you can combine multiple loops into one for, however it would require you to loop over a vector containing all elements rather than the individual elements.
Here is a way to do it:
iRange = 1:2;
jRange = 1:3;
[iL jL] = ndgrid(iRange,jRange);
ijRange = [iL(:) jL(:)]';
for ij = ijRange
i = ij(1); j = ij(2);
end
Note that looping over the variables may be simpler, but perhaps this method has some advantages as well.
No
read this http://www.mathworks.com/help/matlab/matlab_prog/loop-control-statements.html
i also don't see any added value even if it was possible
No I don't think you can put two for loops in one line.
Depends on your operation, you may be able to reshape it and use one for loop. If you are doing something as simple as just printing out all elements,
B = reshape(A,a*b,1);
for i=1:a*b
c = B(i);
...
end