make vectors same length using min function

dear community let me ask your support.

my data:

x = [2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016];

y = [0, 0.05, 0.1, 0.15, 0.2];

a = [2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017,2018,2019];

b = [0, 0.05, 0.1, 0.15, 0.2, 0.4];

I'm trying to make x & a vector lenght same as y & b vector length so Im using min fucntion

minlen = min(length(y), length(x));

minlen = min(length(b), length(a));

I'm trying to get x & a vector updated instead of getting this ans

x(1:minlen);

how can I make x & a equal to ans respectively? like this:

x = [2008, 2009, 2010, 2011, 2012];

a = [2008, 2009, 2010, 2011, 2012, 2013];

any feedback will be highly appreciated

Answer 1

[Image Analyst]

How about

x = [2008,  2009,    2010,   2011,   2012,  2013,  2014,  2015,  2016];
y = [0,  0.05,   0.1,   0.15,  0.2];
a = [2008,  2009,    2010,   2011,   2012,  2013,  2014,  2015,  2016, 2017,2018,2019];
b = [0,  0.05,   0.1,   0.15,  0.2, 0.4];
x = x(1 : length(y))

x = 1x5

2008 2009 2010 2011 2012

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a = a(1 : length(b))

a = 1x6

2008 2009 2010 2011 2012 2013

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[dpb]

If a and b are hardcoded as the example, then the above is a simple as it gets and all that is needed.

However, if it is possible OP has other sets of data for which it is possible the shorter were to be y instead of b, then the use of the min() function to determne which would be the appropriate generalization as began.

The answer to the Q? then is to assign the result for both (or write the expression as the subscript (1:nMin) for both when using the result). The latter would avoid "throwing away" the other data which the direct substitution back into the original variable does.

[A-Rod]

nice, thanks a lor

[Image Analyst]

Yeah, I don't know the whole situation and just did what was asked for. It's possible that it might not work for other situations (such as y being shorter than x) and this is certainly not a robust solution for all possible caes. Originally I thought of using linspace to go from the min x to the max x

x = linspace(min(x), max(x), numel(y))

but then I saw he wanted it truncated instead of interpolated/rescaled, so then I just went with indexing to basically crop the x.

[dpb]

I was just trying to forestall the followup Q? when the situation changed... :)

Answer 2

[Steven Lord]

x = [2008,  2009,    2010,   2011,   2012,  2013,  2014,  2015,  2016];
y = [0,  0.05,   0.1,   0.15,  0.2];
a = [2008,  2009,    2010,   2011,   2012,  2013,  2014,  2015,  2016, 2017,2018,2019];
b = [0,  0.05,   0.1,   0.15,  0.2, 0.4];
minlen = min(length(y), length(x)); 

You could use indexing and assignment:

xbackup = x; % for use in the next block of code
x = x(1:minlen) % Note the added "x = " to assign the indexing result back to x

x = 1x5

2008 2009 2010 2011 2012

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Or if you're using a sufficiently recent release of MATLAB you could use trimdata.

x = xbackup % restoring the original x since the previous line overwrote it

x = 1x9

2008 2009 2010 2011 2012 2013 2014 2015 2016

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x = trimdata(x, minlen)

x = 1x5

2008 2009 2010 2011 2012

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[A-Rod]

this alos works, thank you for sharing