CODEWITHSUNDEEP
matlabmatrix
claushojmark
claushojmark

Reputation: 27

Repmat function in matlab

I have been through a bunch of questions about the Repeat function in MatLab, but I can't figure out how this process work.

I am trying to translate it into R, but my problem is that I do not know how the function manipulates the data.

The code is part of a process to make a pairs trading strategy, where the code takes in a vector of FALSE/TRUE expressions.

The code is:

% initialize positions array 
positions=NaN(length(tday), 2);

% long entries
positions(shorts, :)=repmat([-1 1], [length(find(shorts)) 1]);

where shorts is the vector of TRUE/FALSE expressions.

Hope you can help.

Upvotes: 0

Views: 1376

Answers (2)

hbaderts
hbaderts

Reputation: 14336

The MATLAB repmat function replicates and tiles the array. The syntax is

B = repmat(A,n)

where A is the input array and n specifies how to tile the array. If n is a vector [n1,n2] - as in your case - then A is replicated n1 times in rows and n2 times in columns. E.g.

A = [ 1 2 ; 3 4]
B = repmat(A,[2,3])

B =           |           |
     1     2     1     2     1     2
     3     4     3     4     3     4   __
     1     2     1     2     1     2
     3     4     3     4     3     4

(the lines are only to illustrate how A gets tiled)

In your case, repmat replicates the vector [-1, 1] for each non-zero element of shorts. You thus set each row of positions, whos corresponding entry in shorts is not zero, to [-1,1]. All other rows will stay NaN.

For example if

shorts = [1; 0; 1; 1; 0];

then your code will create

positions = 
          -1    1
          NaN   NaN
          -1    1
          -1    1
          NaN   NaN

I hope this helps you to clarify the effect of repmat. If not, feel free to ask.

Upvotes: 1

Ander Biguri
Ander Biguri

Reputation: 35525

repmat repeats the matrix you give him [dim1 dim2 dim3,...] times. What your code does is:

1.-length(find(shorts)): gets the amount of "trues" in shorts.

e.g: 

shorts=[1 0 0 0 1 0]
length(find(shorts))
ans = 2

2.-repmat([-1 1], [length(find(shorts)) 1]); repeats the [-1 1] [length(find(shorts)) 1] times.

continuation of e.g.:

repmat([-1 1], [length(find(shorts)) 1]);
ans=[-1 1
     -1 1];

3.- positions(shorts, :)= saves the given matrix in the given indexes. (NOTE!: only works if shorts is of type logical).

continuation of e.g.:

At this point, if you haven't omit anything, positions should be a 6x2 NaN matrix. the indexing will fill the true positions of shorts with the [-1 1] matrix. so after this, positions will be:

positions=[-1 1
           NaN NaN
           NaN NaN
           NaN NaN
           -1 1
           NaN NaN]

Hope it helps

Upvotes: 3

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