I got a vector of numbers with length 6 like this a = [1 2 3 4 5 6] and I want to reconstruct the corresponding 4-by-4 matrix A like this where all the element from the diagonal to the other lower diagonal are all zero.
A = [0 1 2 3
0 0 4 5
0 0 0 6
0 0 0 0]
The relationship between the vector a and the corresponding matrix A is that if the dimension of the matrix is K then the length of vector a is K(K-1)/2. In this case the length of a is 6 which mean K = 4.
Another example case would be a = [1 2 3] then
A = [0 1 2
0 0 3
0 0 0
How can I do this?
If you have the Statistics Toolbox, just use squareform and triu:
a = [1 2 3 4 5 6];
A = triu(squareform(a, 'tomatrix'));
Without the toolbox:
a = [1 2 3 4 5 6];
n = (1 + sqrt(1+8*numel(a)))/2; % size of matrix
A = zeros(n); % initiallize
A((1:n).'>(1:n)) = a; % build logical mask using implicit expansion, and fill the
% lower half of the matrix with the desired values in column-major order
A = A.'; % transpose to put the values into the upper half in row-major order
I'm having trouble in looking for an element in a specific.
I have the array
A = [ 1 2 3 7 2 ; 2 8 5 7 2; 1 9 8 4 1; 8 7 2 10 9; 10 9 4 3 8]
I just want to get the index of A(3,4) for the element of 4. However my code spits out the two locations of the element 4, which is A(5,3) and A(3,4).
I used [row, col] = find(E==4)
Use
[row, col] = find(E==4, 1)
The second parameter is the number of elements you want to find. Find more details at https://www.mathworks.com/help/matlab/ref/find.html
Matlab searches a matrix in column-by-column order. If you want to find the first element by rows, you could transpose E before calling find. But you need to swap the resulting indices then:
[col, row] = find(E'==4, 1)
Here's a lengthier, iterative way to find the first index:
A = [ 1 2 3 7 2 ;...
2 8 5 7 2;...
1 9 8 4 1;...
8 7 2 10 9;...
10 9 4 3 8];
[a,b] = size(A);
for i = 1:a
for j = 1:b
if A(i,j) == 4
break
end
end
if A(i,j) == 4
break
end
end
index = [i,j]
It returned [3, 4] for me.
I have an array of doubles like this:
C = [1 2 3 4 0 3 2 5 6 7 1 2 3 4 150 30]
i want to find the pattern [1 2 3 4] within the array and then store the 2 values after that pattern with it like:
A = [1 2 3 4 0 3]
B = [1 2 3 4 150 30]
i can find the pattern like this but i don't know how to get and store 2 values after that with the previous one.
And after finding A, B if i want to find the number of occurrences of each arrays within array C how can i do that?
indices = cellfun(#(c) strfind(c,pattern), C, 'UniformOutput', false);
Thanks!
Assuming you're fine with a cell array output, this works fine:
C = [1 2 3 4 0 3 2 5 6 7 1 2 3 4 150 30 42 1 2 3 4 0 3]
p = [1 2 3 4]
n = 2
% full patttern length - 1
dn = numel(p) + n - 1
%// find indices
ind = strfind(C,p)
%// pre check if pattern at end of array
if ind(end)+ dn > numel(C), k = -1; else k = 0; end
%// extracting
temp = arrayfun(#(x) C(x:x+dn), ind(1:end+k) , 'uni', 0)
%// post processing
[out, ~, idx] = unique(vertcat(temp{:}),'rows','stable')
occ = histcounts(idx).'
If the array C ends with at least n elements after the last occurrence of the pattern p, you can use the short form:
out = arrayfun(#(x) C(x:x+n+numel(p)-1), strfind(C,p) , 'uni', 0)
out =
1 2 3 4 0 3
1 2 3 4 150 30
occ =
2
1
A simple solution can be:
C = [1 2 3 4 0 3 2 5 6 7 1 2 3 4 150 30];
pattern = [1 2 3 4];
numberOfAddition = 2;
outputs = zeros(length(A),length(pattern)+ numberOfAddition); % preallocation
numberOfFoundPattern = 1;
lengthOfConsider = length(C) - length(pattern) - numberOfAddition;
for i = 1:lengthOfConsider
if(sum(C(i:i+length(pattern)) - pattern) == 0) % find pattern
outputs(numberOfFoundPattern,:) = C(i:i+length(pattern)+numberOfAddition);
numberOfFoundPattern = numberOfFoundPattern + 1;
end
end
outputs = outputs(1:numberOfFoundPattern - 1,:);
How can I find all the cells that have the same values in a multi-dimensional array?
I can get it partially to work with result=A(:,:,1)==A(:,:,2) but I'm not sure how to also include A(:,:,3)
I tried result=A(:,:,1)==A(:,:,2)==A(:,:,3) but the results come back as all 0 when there should be 1 correct answer
which is where the number 8 is located in the same cell on all the pages of the array. Note: this is just a test the repeating number could be found multiple times and as different numbers.
PS: I'm using octave 3.8.1 which is like matlab
See code below:
clear all, tic
%graphics_toolkit gnuplot %use this for now it's older but allows zoom
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8]; A(:,:,2)=[9 1 7; 6 5 4; 7 2 8]; A(:,:,3)=[2 4 6; 8 9 1; 3 5 8]
[i j k]=size(A)
for ii=1:k
maxamp(ii)=max(max(A(:,:,ii)))
Ainv(:,:,ii)=abs(A(:,:,ii)-maxamp(ii));%the extra max will get the max value of all values in array
end
%result=A(:,:,1)==A(:,:,2)==A(:,:,3)
result=A(:,:,1)==A(:,:,2)
result=double(result); %turns logical index into double to do find
[row col page] = find(result) %gives me the col, row, page
This is the output it gives me:
>>>A =
ans(:,:,1) =
1 2 3
4 5 6
7 9 8
ans(:,:,2) =
9 1 7
6 5 4
7 2 8
ans(:,:,3) =
2 4 6
8 9 1
3 5 8
i = 3
j = 3
k = 3
maxamp = 9
maxamp =
9 9
maxamp =
9 9 9
result =
0 0 0
0 1 0
1 0 1
row =
3
2
3
col =
1
2
3
page =
1
1
1
Use bsxfun(MATLAB doc, Octave doc) and check to see if broadcasting the first slice is equal across all slices with a call to all(MATLAB doc, Octave doc):
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
If we're playing code golf, a one liner could be:
result = all(bsxfun(#eq, A, A(:,:,1)), 3);
The beauty of the above approach is that you can have as many slices as you want in the third dimension, other than just three.
Example
%// Your data
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8];
A(:,:,2)=[9 1 7; 6 5 4; 7 2 8];
A(:,:,3)=[2 4 6; 8 9 1; 3 5 8];
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
... gives us:
>> result
result =
0 0 0
0 0 0
0 0 1
The above makes sense since the third row and third column for all slices is the only value where every slice shares this same value (i.e. 8).
Here's another approach: compute differences along third dimension and detect when all those differences are zero:
result = ~any(diff(A,[],3),3);
You can do
result = A(:,:,1) == A(:,:,2) & A(:,:,1) == A(:,:,3);
sum the elements along the third dimension and divide it with the number of dimensions. We get back the original value if the values are the same in all dimension. Otherwise a different (e.g. a decimal) value. Then find the location where A and the summation are equal over the third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
or
You could also do
all(A==repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]),3)
where repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]) would highlight the implicit broadcasting of this when compared with A.
or
you skip the broadcasting altogether and just compare it with the first slice of A
A(:,:,1) == sum(A,3)./size(A,3)
Explanation
3 represents the third dimension .
sum(A,3) means that we are taking the sum over the third dimension.
Then we divide that sum by the number of dimensions.
It's basically the average value for that position in the third dimension.
If you add three values and then divide it by three then you get the original value back.
For example, A(3,3,:) is [8 8 8]. (8+8+8)/3 = 8.
If you take another example, i.e. the value above, A(2,3,:) = [6 4 1].
Then (6+4+1)/3=3.667. This is not equal to A(2,3,:).
sum(A,3)./size(A,3)
ans =
4.0000 2.3333 5.3333
6.0000 6.3333 3.6667
5.6667 5.3333 8.0000
Therefore, we know that the elements are not the same
throughout the third dimension. This is just a trick I use
to determine that. You also have to remember that
sum(A,3)./size(A,3) is originally a 3x3x1 matrix
that will be automatically expanded (i.e. broadcasted) to a
3x3x3 matrix when we do the comparison with A (A == sum(A,3)./size(A,3)).
The result of that comparison will be a logical array with 1 for the positions that are the same throughout the third dimension.
A == sum(A,3)./size(A,3)
ans =
ans(:,:,1) =
0 0 0
0 0 0
0 0 1
ans(:,:,2) =
0 0 0
1 0 0
0 0 1
ans(:,:,3) =
0 0 0
0 0 0
0 0 1
Then use all(....,3) to get those. The result is a 3x3x1
matrix where a 1 indicates that the value is the same in the
third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
This question already has answers here:
Repeat copies of array elements: Run-length decoding in MATLAB
(5 answers)
Closed 8 years ago.
My question is similar to this one, but I would like to replicate each element according to a count specified in a second array of the same size.
An example of this, say I had an array v = [3 1 9 4], I want to use rep = [2 3 1 5] to replicate the first element 2 times, the second three times, and so on to get [3 3 1 1 1 9 4 4 4 4 4].
So far I'm using a simple loop to get the job done. This is what I started with:
vv = [];
for i=1:numel(v)
vv = [vv repmat(v(i),1,rep(i))];
end
I managed to improve by preallocating space:
vv = zeros(1,sum(rep));
c = cumsum([1 rep]);
for i=1:numel(v)
vv(c(i):c(i)+rep(i)-1) = repmat(v(i),1,rep(i));
end
However I still feel there has to be a more clever way to do this... Thanks
Here's one way I like to accomplish this:
>> index = zeros(1,sum(rep));
>> index(cumsum([1 rep(1:end-1)])) = 1;
index =
1 0 1 0 0 1 1 0 0 0 0
>> index = cumsum(index)
index =
1 1 2 2 2 3 4 4 4 4 4
>> vv = v(index)
vv =
3 3 1 1 1 9 4 4 4 4 4
This works by first creating an index vector of zeroes the same length as the final count of all the values. By performing a cumulative sum of the rep vector with the last element removed and a 1 placed at the start, I get a vector of indices into index showing where the groups of replicated values will begin. These points are marked with ones. When a cumulative sum is performed on index, I get a final index vector that I can use to index into v to create the vector of heterogeneously-replicated values.
To add to the list of possible solutions, consider this one:
vv = cellfun(#(a,b)repmat(a,1,b), num2cell(v), num2cell(rep), 'UniformOutput',0);
vv = [vv{:}];
This is much slower than the one by gnovice..
What you are trying to do is to run-length decode. A high level reliable/vectorized utility is the FEX submission rude():
% example inputs
counts = [2, 3, 1];
values = [24,3,30];
the result
rude(counts, values)
ans =
24 24 3 3 3 30
Note that this function performs the opposite operation as well, i.e. run-length encodes a vector or in other words returns values and the corresponding counts.
accumarray function can be used to make the code work if zeros exit in rep array
function vv = repeatElements(v, rep)
index = accumarray(cumsum(rep)'+1, 1);
vv = v(cumsum(index(1:end-1))+1);
end
This works similar to solution of gnovice, except that indices are accumulated instead being assigned to 1. This allows to skip some indices (3 and 6 in the example below) and remove corresponding elements from the output.
>> v = [3 1 42 9 4 42];
>> rep = [2 3 0 1 5 0];
>> index = accumarray(cumsum(rep)'+1, 1)'
index =
0 0 1 0 0 2 1 0 0 0 0 2
>> cumsum(index(1:end-1))+1
ans =
1 1 2 2 2 4 5 5 5 5 5
>> vv = v(cumsum(index(1:end-1))+1)
vv =
3 3 1 1 1 9 4 4 4 4 4