how to Compare Values in Array on Specific Indexes in Matlab?
Suppose:
A= [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8]
so i want to Find that
on index 2,3,4,5,6 values or same or not ?
Note: Index can be Dynamically Entered.
Number(length) of Values in Array also can be changed..
To check if they are all equal: use diff to subtract pairs of values, and then check if all those differences are 0.
A = [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8];
ind = [2 3 4 5 6];
result = ~any(diff(A(ind)));
This is faster than using unique. With A and ind as in your example,
>> tic
for cont = 1:1e5
result = ~any(diff(A(ind)));
end
toc
tic
for cont = 1:1e5
result=numel(unique(A(ind)))==1;
end
toc
Elapsed time is 0.371142 seconds.
Elapsed time is 4.754007 seconds.
Hey this should do the trick:
A= [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8];
B= [1,3,5];
C=A(B);
result=numel(unique(C))==1;
Here A is your data.
B is the index vector.
C contains the elements corresponding to the index vector.
result is 1 if all values were the same and 0 otherwise.
You can even "shorten" the code further by joining the two line:
result=numel(unique(A(B)))==1;
There are some ways, it depends on your taste.
For example, if the variable indexing contain the corresponding indexes:
unique(A(indexing));
will give you a vector with the unique elements in the sub-vector A(indexing). Then you just need to check the length:
length(unique(A(indexing))) == 1
I would avoid the use of numel when the function length is available (it is much more clearer what you are trying to achieve).
Other option is to compare the first element to the rest of the element in the sub-vector:
sub_vector = A(indexing);
all(sub_vector == sub_vector(1));
The second option assumes that the sub-vector will never be empty!
Related
I am trying to generate random numbers between 1 and 6 using Matlab's randperm and calling randperm = 6.
Each time this gives me a different array let's say for example:
x = randperm(6)
x = [3 2 4 1 5 6]
I was wondering if it was possible to create pairs of random numbers such that you end up with x like:
x = [3 4 1 2 5 6]
I need the vector to be arranged such that 1 and 2 are always next to each other, 3 and 4 next to each other and 5 and 6 next to each other. As I'm doing something in Psychtoolbox and this order is important.
Is it possible to have "blocks" of random order? I can't figure out how to do it.
Thanks
x=1:block:t ; %Numbers
req = bsxfun(#plus, x(randperm(t/block)),(0:block-1).'); %generating random blocks of #
%or req=x(randperm(t/block))+(0:block-1).' ; if you have MATLAB R2016b or later
req=req(:); %reshape
where,
t = total numbers
block = numbers in one block
%Sample run with t=12 and block=3
>> req.'
ans =
10 11 12 4 5 6 1 2 3 7 8 9
Edit:
If you also want the numbers within each block in random order, add the following 3 lines before the last line of above code:
[~, idx] = sort(rand(block,t/block)); %generating indices for shuffling
idx=bsxfun(#plus,idx,0:block:(t/block-1)*block); %shuffled linear indices
req=req(idx); %shuffled matrix
%Sample run with t=12 and block=3
req.'
ans =
9 8 7 2 3 1 12 10 11 5 6 4
I can see a simple 3 step process to get your desired output:
Produce 2*randperm(3)
Double up the values
Add randperm(2)-2 (randomly ordered pair of (-1,0)) to each pair.
In code:
x = randperm(3)
y = 2*x([1 1 2 2 3 3])
z = y + ([randperm(2),randperm(2),randperm(2)]-2)
with result
x = 3 1 2
y = 6 6 2 2 4 4
z = 6 5 2 1 3 4
I have for example a=[1 2 3 4 5 6 7 8 9 10]; and I have to delete each 2 following numbers from 3.
like at the end it should be a=[1 4 7 10];
How to do this without a for loop.
And also if there is a way to guarantee that at the end the resulting array will have exact number of entries, like here it should be a with 4 entries at the end.
But for example we have b=[1 2 3 4 5 6 7 8 9 ]; and if I want make sure that at the end I still have 4 entries in the rest array, so that b can't be equal to [1 4 7] because I need 4 entries for sure.
You can use indexing for this:
A = 1:10;
B = A(1:3:end)
B =
[1 4 7 10]
Or, if you really want to remove elements:
A = 1:10;
A(2:3:end) = [];
A(3:3:end) = [];
For your second question regarding length checking, it's unclear what you're asking. Would an if comparison be enough ?
if numel(A) ~= 4
% ... Handle unexpected values here
end
Best,
As you mentioned in the question and in the comments that you need 4 elements at the end and if elements are less than 4 then you want to include the last element/s of b, the following should work:-
b=[1 2 3 4 5 6 7 8 9]
b_req=b(1:3:end);
temp=length(b_req);
if temp<4 b_req(end+1:4)=b(end-3+temp:end); % for including the elements of b so that total elements are 4 at the end
elseif temp>4 b_req=b_req(1:4); % for removing the extra elements
end
b_req
Output:-
b =
1 2 3 4 5 6 7 8 9
b_req =
1 4 7 9
and
if instead b=[1 2 3 4 5 6 7 8 9 10]; then the same code gives what you require, i.e. b_req = [1 4 7 10]
This code speaks for itself:
a = 1:15; % some vector
% returns every third element after the first one:
third_elemets = a(1:3:end);
% returns the missing elements for the vector to be in size 4 from the end of a
last_elements = a(end-3+length(third_elemets):end);
% take maximum 4 elements from a
logic_ind = true(min(4,length(third_elemets)),1);
% and concatanate them with last_elements (if there are any)
a = [third_elemets(logic_ind) last_elements]
and under the assumption that whenever there are less than 4 elements you simply take the last one(s) - it should always work.
Suppose A is a 3-D matrix as below (2 rows-2 columns-2 pages).
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
I want to have a vector, say "a", whose inputs are the average of diagonal elements of matrices on each page. So in this simple case, a=[(1+4)/2;(5+8)/2].
But I have difficulties in matlab to do so. I tried the codes below but failed.
mean(A(1,1,:),A(2,2,:))
You can use "partially linear indexing" in the two dimensions that define the diagonal, as follows:
Since partially linear indexing can only be applied on trailing dimensions, you first need to apply permute to rearrange dimensions, so that the first and second dimensions become second and third.
Now you leave the first dimension untouched, linearly-index the diagonals in the second and third dimensions (which effectly reduces those two dimensions to one), and apply mean along the (combined) second dimension.
Code:
B = permute(A, [3 1 2]); %// step 1: permute
result = mean(B(:,1:size(A,1)+1:size(A,1)*size(A,2)), 2); %// step 2: index and mean
In your example,
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
this gives
result =
2.5000
6.5000
You can use bsxfun for a generic solution -
[m,n,r] = size(A)
mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
Sample run -
>> A
A(:,:,1) =
8 4 1
7 6 3
1 5 8
A(:,:,2) =
1 7 6
8 5 2
1 2 7
A(:,:,3) =
6 2 8
1 1 6
1 4 5
A(:,:,4) =
8 1 6
1 5 1
9 2 7
>> [m,n,r] = size(A);
>> sum(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
22 13 12 20
>> mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
7.3333 4.3333 4 6.6667
How to find out all array element indices equal to several values (>2)
For example, I have an array a=[1 2 3 4 5 5 4 3 2 2 2 1], I want to know the indices of all elements equal to b=[2 5]
Remember, I cannot use style like a==b(1) | a==b(2) because number of elements in b is arbitrary.
Do I have to use a for loop to do this?
You can use ismember (as Daniel said just seconds before I hit enter...) ;-)
a=[1 2 3 4 5 5 4 3 2 2 2 1];
b=[2 5];
c=find(ismember(a,b))
Output:
c =
2 5 6 9 10 11
If you want to do it more manually, you can use bsxfun:
c = find(any(bsxfun(#eq, a(:).', b(:)), 1));
Given A = [3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 8]
Output B = [3 4 5 6 8]
Is there a Matlab function or command to get this result? I am new to Matlab. Just now I am doing it going through for each element and keeping a counter for it. I have very big array so this is taking too much time.
Use a combination of unique and histc:
uA = unique(A); %// find unique values
B = uA(histc(A, uA)>=2); %// select those that appear at least twice
The above code gives the values that appear at least twice. If you want values that appear exactly twice, replace >= by ==.