husain_javan 9 months ago updated by Pavel Holoborodko 9 months ago 3

is there a way to use arrayfun for mp? I need to define a function in integral form. but integral only uses scalar values in its limits. If I use loops it would be a little slower.

Under review

Toolbox provides arrayfun and it works with mp variables. Do you have any issues using it? Please provide minimal and reproducible example of the issue.

>> S(1).f1 = rand(1,5,'mp');
>> S(2).f1 = rand(1,10,'mp');
>> S(3).f1 = rand(1,15,'mp');
>> A = arrayfun(@(x) mean(x.f1),S)

A =

0.5690631027410031794744327271473594 0.6602111831687765719500760042137699 0.4687452915777556405885206913808361

>> A = arrayfun(@(x) mean(x.f1),S,'UniformOutput',false)

A =

1×3 cell array

{1×1 mp} {1×1 mp} {1×1 mp}

hello Mr. Holoborodko. for example this function I can create:

>>myfun = @(x)arrayfun( @(x) integral(@(t) t,eps, x, 'RelTol', eps), x);
>>myfun([1 2]) ans =
0.500000000000000 2.000000000000000

but with your toolbox:

>>mufun = @(x)arrayfun( @(x) quadgk(@(t) t,100*eps('mp'), x, 'RelTol', 100*eps('mp')), x);
>>mufun(mp([1 2])) ans =
[1x1 mp] [1x1 mp]

however when I try to give scalar without arrayfun, it works:

mufun = @(x) quadgk(@(t) t,100*eps('mp'), x, 'RelTol', 100*eps('mp'));
mufun(mp(2)) ans =

what does it mean when it displays [1x1 mp] ?


This means that function returned array of cells, each cell is a "mp" number. 
There is nothing wrong with it, you can use/see the actual number by  ans{1:2} or other indexing.

Why arrayfun returns the cell array in case of mp([1 2]) is a mystery, we will investigate it in more detail. I guess it is specific for non-standard types (like "mp").