Problem 2159. A SUBSREF variant that accepts the 'end'-operator.

Created by J-G van der Toorn

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Unfortunately, Matlab's <tt>subsref</tt> function does not support the <tt>end</tt> -operator. The <tt>end</tt> -operator is a powerfull method to refer to the rear-end of an array,<pre> a = 1:10;
 a(end-1) % is 9
 a(end-2:end) % is [8 9 10]</pre>It works similar for cell arrays and struct-arrays.Alhough <tt>subsref</tt> is quite powerful, and accepts for example the <tt>:</tt> colon oparator, it fails to process more complex stings, including <tt>end</tt> -operators.This assignment is to create a function that accepts those more complex element definitions.For example<pre> s.a.b(3).c{2}.d = 'a':'z';
 subsref(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{1:3})) </pre>returns<pre> ans = </pre><pre> abc </pre>And<pre class="language-matlab">subsrefbetter(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{'[1 end]'})) 
</pre>should return<pre> ans = </pre><pre> az </pre>and<pre class="language-matlab">subsrefbetter(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{'end-2:end'})) 
</pre>should return<pre> ans =</pre><pre> xyz</pre>The function <tt>subsrefbetter</tt> should accept both structures like created with <tt>substruct</tt>, and list of arguments like accepted by <tt>substruct</tt>.That's all.

a = 1:10;
 a(end-1) % is 9
 a(end-2:end) % is [8 9 10]

It works similar for cell arrays and struct-arrays.

This assignment is to create a function that accepts those more complex element definitions.

For example

s.a.b(3).c{2}.d = 'a':'z';
 subsref(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{1:3}))

returns

ans =

abc

And

subsrefbetter(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{'[1 end]'}))

should return

ans =

and

subsrefbetter(s,substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{'end-2:end'}))

should return

ans =
  
 xyz

That's all.

Solve

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Last Solution submitted on Sep 23, 2021

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1 Comment

Andrew Newell on 20 May 2017

Pity that eval is forbidden - I had quite an elegant solution using it.

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Problem 2159. A SUBSREF variant that accepts the 'end'-operator.

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Problem 2159. A SUBSREF variant that accepts the 'end'-operator.

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