Create matrix from two arrays using colon

I'm trying to avoid using for loops in my code as much as possible. I was advised that using vectorization would be quicker and so I though it would be good practice. The problem is that I have two column vectors and I want to create a matrix that contains the values in between the two values of each column for each row. For example if the in the first row, first column, I had the value 1 and in the first row, second column I had the value 10 I want a row vector of [1,2,3,4,5,6,7,8,9,10]. I want the code to do this for all rows.

I tried this:

indexsnips(:,1) = indexstamps_start; %89x2 double
indexsnips(:,2) = indexstamps_end;   %89x2 double

indexsnips = indexsnips(:,1):indexsnips(:,2); %all starts and ends have a consistent  
                                              %difference of 60. The result should be 89x6.

The result returns only the vector from the first row, but not the others.

[Image Analyst]

Do they have to be integers? What if you have [1,10] in the first row for 10 values, and you have [1,6] in the second row for 6 values? Or do all rows have 10 values? You can't have different numbers of columns in a matrix.

What do yo umean that all rows have a consistent difference of 60? Do you mean like this

[1 61;
100, 160]

If so, how does your example with 10 agree with that? And if it's 60, and the new vector changes by 1, why do you have 6 columns instead of 60 columns?

[Daniel]

What I have is something like this:

blah(:,1) =   [100 , 200 , 300 , 400 , 500]; 
blah(:,2) =   [105 , 205 , 305 , 405 , 505];

I tried

m = blah(:,1):blah(:,2);

what I got was only the first row

m =

     100   101   102   103   104   105

what I want is a matrix like this

m =

     100   101   102   103   104   105
     200   201   202   203   204   205
     300   301   302   303   304   305
     400   401   402   403   404   405
     500   501   502   503   504   505

Answer 1

[Roger Stafford]

Assuming 1) you want integer values in your matrix, 2) the two vectors have integers, and 3) that corresponding elements in the vectors have equal differences, then do this. Let the first column vector be u and the second one v.

 M = bsxfun(@plus,v,0:v(1)-u(1));

[Daniel]

I tried a test run on some arbitrary code:

v = [100,200,300,400,500];
u = [105,205,305,405,505];
M = bsxfun(@plus,v,0:v(1)-u(1))

answer was:

M =

     Empty matrix: 5-by-0

what I want is:

m =
     100   101   102   103   104   105
     200   201   202   203   204   205
     300   301   302   303   304   305
     400   401   402   403   404   405
     500   501   502   503   504   505

[Roger Stafford]

What you originally stated was that "I have two column vectors" but what you have used for u and v here are two row vectors. Change them to column vectors and try it again.

Answer 2

[David Sanchez]

A =   [100 , 200 , 300 , 400 , 500]; 
B =   [105 , 205 , 305 , 405 , 505]; 
C = zeros(numel(A),numel(B)+1);

for k=1:numel(A)
  C(k,:) = A(k):B(k);
end

C =

     100   101   102   103   104   105
     200   201   202   203   204   205
     300   301   302   303   304   305
     400   401   402   403   404   405
     500   501   502   503   504   505

Answer 3

[Jos (10584)]

I do not see why you have to specify the last values, since it is always the same number more than the starting values. If that is the case, you can use this

V = [100,200,300,400,500]
N = 5
% engine
  C = ones(numel(V), N+1)
  C(:,1) = V(:)
  C = cumsum(C,2)
% or in one big unreadable but not faster step
  C2 = cumsum([V(:) ones(numel(V),N)],2)

If N varies, you can use cell arrays:

V = [100,200,300,400,500]
U = [105,210,303,406, 508]
C = arrayfun(@(x) V(x):U(x),1:numel(V),'un',0)
% the cell C{k} holds the array [V(k):U(k)]

[Jos (10584)]

Note that you can replace N by (U(1)-V(1)) to avoid having a variable called N ...

Answer 4

[Andrei Bobrov]

v = [100,200,300,400,500];
u = [105,205,305,405,505];

M = bsxfun(@plus,min(v(:),u(:)),0:abs(v(1)-u(1)))

Answer 5

[Perry Liu]

indexsnips(:,1)+(0:60)

[Walter Roberson]

Note: this requires R2016b or later.