I've two matrices A is 2x9 and B is 6x3,
A= zeros(2,9)
A =
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
B=round(10*rand(6,3))
B =
7 6 9
6 7 8
4 1 7
1 1 1
8 5 1
3 5 1
by using these codes I want to add first three row of matrix B to first row of A and second three rows of matrix B to second row of matrix A by using these codes :::
for i=1:6
if i<=3
x=x(y,:)
else
end
end
I tried I don't know how inside if condition
j=1; k=1;
for i=1:6 % loop over rows of B
if (j > 9) % reset your index to point to the first entry of the next row of A
k=k+1;
j=1;
end
A(k,j:j+2)=A(k,j:j+2)+B(i,:);
j=j+3;
end
If the sizes of your Matrices change you'll have to adapt the conditions and strides respectively. This is just a quick solution. I wonder if there is a more elegant way to solve this problem - there always is, when using Matlab...
Related
I have a NxNx5 array T that I would like to convert into a Rx5 array TT such that the following condition is satisfied (where R is the number of non-zero entries of the array T(:,:,1)):
If T(i,j,1) == 0 then we ignore. If T(i,j,1) != 0 then I would like a row of TT whose entry is
[T(i,j,1) T(i,j,2) T(i,j,3) T(i,j,4) T(i,j,5)]
Note that T(i,j,k) (k = 2,3,4,5) could be zero. For example,
If
T(3,2,1) = 3
then I would like a row of TT to be
[3 0 2 1 5].
Some notes:
The entries of TT are all integers.
The entries accent in order column wise. i.e the first column of TT(:,:,1) maybe
[1 2 0 0 3 4 0 0 0 5 6]'
then the next column
[7 8 0 0 0 0 0 9 10 11 12]'
I think this does what you want:
ind = find(T(:,:,1));
ind = bsxfun(#plus, ind(:), (0:size(T,3)-1)*size(T,1)*size(T,2));
result = T(ind);
This will do it:
clear
rng(343)
N=7;
K=5;
T=randi([0,4],[N,N,K])
TT=reshape(T,[N*N,K])
TT(T(:,1)==0,:)=[] %delete rows with first col equal to 0
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
This question already has answers here:
How can I find indices of each row of a matrix which has a duplicate in matlab?
(3 answers)
Closed 8 years ago.
I have two matrices and I want to find the indices of rows in Matrix B which have the same row values in Matrix A. Let me give a simple example:
A=[1,2,3; 2,3,4; 3,5,7; 1,2,3; 1,2,3; 5,8,6];
B=[1,2,3; 29,3,4; 3,59,7; 1,29,3; 1,2,3; 5,8,6;1,2,3];
For example, for first row in matrix A, The row1, row5, and row7 in Matrix B are correspondences.
I have written below code but it doesn't return back all indices which have the same row value in matrix A and only one of them (row7) is backed !!
A_sorted = sort(A,2,'descend'); % sorting angles
B_sorted = sort(B,2,'descend'); % sorting angles
[~,indx]=ismember(A_sorted,B_sorted,'rows')
the result is
indx_2 =
7
0
0
7
7
6
It means for the first row in matrix A , only one row ( row 7) in Matrix B is available !! But as you can see for first row in matrix A there is three correspondent rows in matrix B (Row 1, row 5 and row 7)
I think the best strategy is to apply ismember to unique rows
%make matrix unique
[B_unique,B2,B3]=unique(B_sorted,'rows')
[~,indx]=ismember(A_sorted,B_unique,'rows')
%For each row in B_unique, get the corresponding indices in B_sorted
indx2=arrayfun(#(x)find(B3==x),indx,'uni',0)
If you want to compare all pairs of rows between A and B, use
E = squeeze(all(bsxfun(#eq, A, permute(B, [3 2 1])), 2));
or equivalently
E = pdist2(A,B)==0;
In your example, this gives
E =
1 0 0 0 1 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
1 0 0 0 1 0 1
1 0 0 0 1 0 1
0 0 0 0 0 1 0
The value E(ia,ib) tells you if the ia-th row of A equals the ib-th row of B.
I would line to find the number of the first consecutive zero elements. For example in [0 0 1 -5 3 0] we have two zero consecutive elements that appear first in the vector.
could you please suggest a way without using for loops?
V=[0 0 1 -5 3 0] ;
k=find(V);
Number_of_first_zeros=k(1)-1;
Or,
Number_of_first_zeros=find(V,1,'first')-1;
To solve #The minion comment (if that was the purpose):
Number_of_first_zeros=find(V(find(~V,1,'first'):end),1,'first')-find(~V,1,'first');
Use a logical array to find the zeros and then look at where the zeros and ones are alternating.
V=[1 2 0 0 0 3 5123];
diff(V==0)
ans =
0 1 0 0 -1 0
Create sample data
V=[1 2 0 0 0 3 5123];
Find the zeros. The result will be a logical array where 1 represents the location of the zeros
D=V==0
D =
0 0 1 1 1 0 0
Take the difference of that array. 1 would then represent the start and -1 would represent the end.
T= diff(D)
ans =
0 1 0 0 -1 0
find(T==1) would give you the start and find(T==-1) would give you the end. The first index+1 of T==1 would be the start of the first set of zeros and the first index of T==-1 would be the end of the first set of zeros.
You could find position the first nonzero element using find.
I=find(A, 1);
The number of leading zeros is then I-1.
My solution is quite complex yet it doesn't use the loops and it does the trick. I am pretty sure, that there is a more direct approach.
Just in case no one else posts a working solution here my idea.
x=[1 2 4 0 20 0 10 1 23 45];
x1=find(x==0);
if numel(x1)>1
x2=[x1(2:end), 0];
x3=x2-x1;
y=find(x3~=1);
y(1)
elseif numel(x1)==1
display(1)
else
display('No zero found')
end
x is the dataset. x1 contains the index of all zero elements. x2 contains all those indices except the first one (because matrix dimensions must agree, one zero is added. x3 is the difference between the index and the previous index of zeros in your dataset. Now I find all those differences which are not 1 (do not correspond to sequences of zeros) and the first index (of this data is the required result. The if case is needed in case you have only one or no zero at all.
I'm assuming your question is the following: for the following vector [0 0 1 -5 3 0], I would like to find the index of the first element of a pair of 0 values. Is this correct? Therefore, the desired output for your vector would be '1'?
To extend the other answers to find any such pairs, not just 0 0 (eg. 0 1, 0 2, 3 4 etc), then this might help.
% define the pattern
ptrn = [ 0 0 ];
difference = ptrn(2) - ptrn(1)
V = [0 0 1 -5 3 0 0 2 3 4 0 0 1 0 0 0]
x = diff(V) == difference
indices = find(x)
indices =
1 6 11 14 15
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