Adding zeros to a column vector to match a larger column vector

I have x where the dimension is (100x1) and y (108x1)

I use:

new_x= [x,zeros(1,length(y)-length(x))];

but I got an error saying (Error using horzcat. Dimensions of arrays being concatenated are not consistent).

Answer 1

[Les Beckham]

new_x= [x; zeros(length(y)-length(x), 1)];
%        ^                          ^ switch the arguments to zeros  
%        use a semicolon instead of a comma

Answer 2

[Matt J]

This gracefully handles the case when length(x)=length(y), and requires no consideration of whether x,y are row or column vectors.

x(end+1:numel(y))=0;

Answer 3

[VBBV]

x = rand(100,1)  % column vector

x = 100×1

0.8303 0.5510 0.2123 0.2285 0.4678 0.7940 0.1815 0.2479 0.1557 0.1806

y = rand(108,1) % another column vector of different size

y = 108×1

0.4289 0.9965 0.6628 0.3147 0.2230 0.9301 0.1332 0.8500 0.8675 0.4001

zeros(1,length(y)-length(x)) % row vector

ans = 1×8

0 0 0 0 0 0 0 0

new_x= [x',zeros(1,length(y)-length(x))]

new_x = 1×108

0.8303 0.5510 0.2123 0.2285 0.4678 0.7940 0.1815 0.2479 0.1557 0.1806 0.1557 0.2500 0.1559 0.4368 0.1118 0.3869 0.5771 0.1627 0.6922 0.1240 0.2174 0.2439 0.4261 0.1834 0.3268 0.1863 0.2051 0.1857 0.8732 0.1211

%        ^transpose the x vector

Transpose the x vector since it is column vector whereas zeros(1,length(y)-length(x)) is the row vector. Both of them being concatenated using [ ] operator

[VBBV]

Alternately, you could transpose the row vector zeros(1,length(y)-length(x)) keeping the x vector same (without transpose)

Answer 4

[Steven Lord]

If you're using release R2023b or later, you could use the paddata function. Let's create some sample data:

% I have x where the dimension is (100x1) and y (108x1)
x = (1:100).';
y = (101:208).';

What sizes are the two vectors?

szX = size(x);
szY = size(y);

What is the size of the larger? I used my knowledge that they had the same number of dimensions here.

M = max(szX, szY);

Now pad and look at the sizes of the results.

x2 = paddata(x, M);
y2 = paddata(y, M);
whos x x2 y y2

  Name        Size            Bytes  Class     Attributes

  x         100x1               800  double              
  x2        108x1               864  double              
  y         108x1               864  double              
  y2        108x1               864  double              

Since y was already the correct size, paddata didn't change it.

isequal(y, y2)

ans = logical
   1

Let's look at the last dozen rows of x and x2.

tail(x, 12)

    89
    90
    91
    92
    93
    94
    95
    96
    97
    98
    99
   100

tail(x2, 12)

    97
    98
    99
   100
     0
     0
     0
     0
     0
     0
     0
     0

There are 8 copies of 0 at the end of x2 to make it the same size as y and y2.