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period (t1, t2, t3) = time (p)

where t1 , t2, t3 are n dimensional arrays say

a = (1,3,4,6,8,9,0,5,4,)

b = (3)

c = (4,56,7,8,5,1)

t1 = length(a)

t2 = length(b)

t1 = length(c)

Since I have a function called period it gives some random value based on the parameters that have been passed to t1, t2, t3.

output -- period (5,1,2) = 12.

what I'm trying is to do trace it back

i have time as 12, just by looking at the output i want to trace the value of t1 that has been passed.

I want the 5 element of t1 as the result, thats 8

How do i get it ?

kindly help

Ameer Hamza
on 5 Apr 2020

In case the function period is a block-box or cannot be inverted, then you can an optimization-based approach to find its inverse. For example, if the input to the function period can be any real number

obf_fun = @(x) (period(x(1),x(2),x(3)) - 12).^2;

sol = fmincon(obf_fun, rand(1,3));

a = sol(1);

b = sol(2);

c = sol(3);

If the function period can only accept integer inputs, then you will need global optimization toolbox to find the solution

obf_fun = @(x) (period(x(1),x(2),x(3)) - 12).^2;

sol = ga(obf_fun, 3, [], [], [], [], [], [], [], 1:3)

a = sol(1);

b = sol(2);

c = sol(3);

Ganesh Kini
on 5 Apr 2020

Is it possible to explain in a few words?

the time can be in decimal too for example time 12.18, but the inputs to the period will be integers.

If the time is a decimal number, can i trace it back?

Can you please let me know

Ameer Hamza
on 5 Apr 2020

Ganesh, if the inputs to period are integers, then you will need to use the second code, i.e., ga() algorithm. Yes, the time can be fractional too. You can replace 12 with a decimal number.

Explanation: We have defined objective functions like this

Which will search for the value of a, b, and c which will minimize the value of the . As you can see, when this value is minimized, the output of the period will be the same as time. The algorithm will output the value of a,b, and c, which brings the output of the period closest to the time value.

Ganesh Kini
on 5 Apr 2020

Thank you for the help

But my query was regarding getting the value of the 5th element of t1 based on the time.

I guess the above code minimises and then gets the value closer to the time = 12

Ameer Hamza
on 5 Apr 2020

Ganesh Kini
on 5 Apr 2020

a = (1,3,4,6,8,9,0,5,4,)

b = (3)

c = (4,56,7,8,5,1)

a, b,c are saved in .vec extension

for t1 =1:1:length(a)

for t2=1:1:length(b)

for t3=1:1:length(c)

period_fun(a,b,c)=time(p);

p=p+1;

end

end

end

So this code will give me the time

period fun(5,1,3) which means it will fetch

5th element of a = 8, 1st element of b = 3, 3rd element of c = 7, and the output is time = 1.1

Now just by looking the output time 1.1 i want to extract the 5th element of a or the 1st element of b or 3rd element of c and get that element as output

is it possible ?

Ameer Hamza
on 5 Apr 2020

Ganesh Kini
on 5 Apr 2020

a =load('numbers.vec');

b =load('numbers1.vec');

c =load('numbers2.vec');

the file number.vec = (1,3,4,6,8,9,0,5,4) and so on

time (p) is just the index

this is a part of the output for time(p)

12.7271

6.0866

6.0278

3.7492

9.181100000000001

5.2952

13.8514

6.918

6.1957

4.113

11.8323

5.4656

7.3774

5.3561

13.4326

6.7934

6.3158

4.1806

9.674099999999999

6.0949

14.502

1.1

So i will choose a random time i.e 1.1 and i need to get the 5 th element of a

I hope you understood, Please let me know

Ameer Hamza
on 5 Apr 2020

If you have time vector like give above, then you can do something like this

p = find(time == 1.1);

it will give you the value of p. But you didn't explain how the value of p increase with the values of t1, t2, and t3 inside the for loops.

Ganesh Kini
on 5 Apr 2020

p = p+1

it increases by one step after the computation of one set.

this will work if the data is arranged in a matrix. I need to access the file directly

Ameer Hamza
on 5 Apr 2020

Ganesh Kini
on 5 Apr 2020

it is saved in .mat

how do i access it ? Does the above solution work to extract data which is in a array and not a matrix

Ameer Hamza
on 5 Apr 2020

Ganesh Kini
on 6 Apr 2020

Hi,

Could you please help me with other source. I tried it is not working

Ameer Hamza
on 6 Apr 2020

Ganesh, you will need to share your files and current code so that I can easily suggest a solution.

Ganesh Kini
on 7 Apr 2020

Hi Ameer,

after looking at the existing code and analysis.

period_fun(a,b,c)=time(p);

period_fun is not a function, its just an array.

Could you please let me know how to access the elements passed to it ?

Ameer Hamza
on 7 Apr 2020

From the limited information I have, i can only suggest this

idx = find(period_fun == 1.1); % suppose your are looking for 1.1 in period_fun

[a,b,c] = ind2sub(size(period_fun), idx);

Ganesh Kini
on 7 Apr 2020

Hi Ameer,

idx = find(period_fun == 1.1)

output of idx -- 1

this will find the index of the time, which does not serve the purpose tracing back the elements that are passed

[a,b,c] = ind2sub(size(period_fun), idx);

And finding 1.1 in period_fun does not make sense

please correct me if im wrong

Ameer Hamza
on 7 Apr 2020

The line

idx = find(period_fun == 1.1)

does not give an index in time. It gives a linear index in the matrix period_fun. The line

[a,b,c] = ind2sub(size(period_fun), idx)

convert the linear index into the values of subscripts a, b, and c

Ganesh Kini
on 7 Apr 2020

Sorry for the last comment. It was my mistake

so my code

value = period_fun(2,1,1,10,10,15,3);

Disp(value) gives me 1.1

now when i run

idx = find(period_fun == 1.1)

disp(idx)-- 62988

i tried looking for what it is, didnt get a clue

could you please tell what 62988 indicate ?

Ameer Hamza
on 7 Apr 2020

Ganesh Kini
on 7 Apr 2020

Hi,

i have passed something like

value = period_fun(2,1,1,10,10,15,3); i.e. value(a,b,c,d,e,f,g,h) it should be a 7D matrix.

but here its 62988. Can you help me here ?

Ameer Hamza
on 7 Apr 2020

You can get the value of a,b,...h back using

[a,b,c,d,e,f,g,h] = ind2sub(size(period_fun), 62988);

Ganesh Kini
on 16 Apr 2020

Hi Ameer,

Sorry for the late reply, the solution works.

But suppose the output is 1.0 instead 1.1

the closest value of 1.0 is obviously 1.1 ?

But how do i do that ? How do i map the closest value based on a random output ?

Can you please guide me

Ganesh Kini
on 16 Apr 2020

Hi,

I guess you didnt get my query.

I have the output as 1. I have to find the nearest value of 1 (i.e 1.1)

How do i do that ?

Ameer Hamza
on 16 Apr 2020

You can try something like

[~,idx] = min(abs(time-1));

idx is the index of 1.1 in time

time(idx) % output should be 1.1

Ganesh Kini
on 16 Apr 2020

Hi

The above code is not working

I have tried this before

Can you suggest some other alternative ?

Ameer Hamza
on 17 Apr 2020

Can you attach variable time in a .mat file? It will make it easier to sugges a solution.

Ganesh Kini
on 17 Apr 2020

Hi, Please forget the previous part. I will explain here

period = period_temp(2,1,1,4,5,6,1)+0.3;

the output is period = 11.5

I have an array period_temp (a,b,c,d,e,f,g) which gives a lot of values in 2 * 1 * 1 * 10 * 10 *10* 8 matrix

but 11.5 doesnt exist in the matrix since we i have added 0.3 as constant to period_temp.

Now How do I find the closest value to 11.5?

I have tried with

idx = interp1(period_temp, 1: length (period_temp), period, 'nearest');

Output: error: y(168000,_): but y has size 15x1

Please suggest

Ameer Hamza
on 17 Apr 2020

See this example:

x = rand(2,2,2);

val = 0.5;

[~,idx] = min(abs(x-val), [], 'all', 'linear');

[i1,i2,i3] = ind2sub(size(x), idx); % return index in each dimension

closest_value = x(i1,i2,i3);

Ganesh Kini
on 18 Apr 2020

Hi

The solutuion is not working

Please suggest some other solution

error warning: print_usage: Texinfo formatting filter exited abnormally

warning: called from

print_usage at line 74 column 5

warning: print_usage: raw Texinfo source of help text follows...

error: Invalid call to min. Correct usage is:

error: called from

print_usage at line 91 column 5

Ameer Hamza
on 18 Apr 2020

Ganesh Kini
on 27 Apr 2020

Hi Ameer,

I am facing one issue here

idx = find(period_fun == 1.1); % suppose your are looking for 1.1 in period_fun

[a,b,c] = ind2sub(size(period_fun), idx);

So when i run it. sometimes a carries more than one index

Suppose for eg for the code when i run it it get

a = 1.000 4.0000

Expected answer is 1.000 but why did 4.0000 come ?

How can we eliminate the other values?

Please help me out

PS: FYI the array of a = (1,3,4,6,8,9,0,5,4,)

Ameer Hamza
on 27 Apr 2020

two values of a mean that 1.1 exist at two location in period_fun. You can get a single value like this

idx = find(period_fun == 1.1, 1);

Ganesh Kini
on 27 Apr 2020

1.1 is unique. a,b,c combination will be unique

but a is taking two or three values from its array.

Can you please help me in this ?

Ameer Hamza
on 27 Apr 2020

If 1.1 is uniuq, then a cannot take more then one value. Can you show an example where this happen?

Ganesh Kini
on 27 Apr 2020

abc = period_fun(2,1,2,5,5,13,8);

%finding the nearest possible value

dist = abs(period_fun - abc);

[min_dist, idx] = min(dist(:));

nearestvalue = period_fun(idx);

actualidx= find(period_fun ==nearestvalue);

[p1,p2,p3,p4,p5,p6,p7] = ind2sub(size(period_fun), actualidx);

v1 = nw_vec(p5);

First Case

v1 is displaying as follows

0.80000 0.65000 0.75000

all these are in the nw_vec array but 0.8 is expected answer. I should get only 0.8 as the answer

Second case

v1 is displaying as follows

0.40000 0.55000

0.2 is expected but its not giving it as output. I should get only 0.2 as the answer

How can i solve this ?

Ganesh Kini
on 27 Apr 2020

nw_vec is .vec file

the file is as follows

0.2, 0.36, 0.38, 0.40, 0.47, 0.5, 0.55, 0.65, 0.75, 0.8

Ameer Hamza
on 27 Apr 2020

You are just using p5, why are you not using all the indexes: p1,p2,p3,p4,p5,p6,p7?

Ganesh Kini
on 27 Apr 2020

Because i want only p5.

abc = period_fun(2,1,2,5,5,13,8) takes the form of abc = period_fun(ioc,ipc,irc,inw,ivw,ivdp,itp);

where ivw=1:1:length(nw_vec) and so on for all the ioc,ipc,irc,inw,ivw,ivdp,itp.

We have 7 arrays that has been passed to period_fun

and from abc we are tracing it back

I want only values from nw_vec

Ganesh Kini
on 27 Apr 2020

Yes I have shared it earlier on this link

its the same code

Attaching the file again

Ganesh Kini
on 27 Apr 2020

Yes, i just gave it as example.

let me give you the results from the output from my screen

abc = period_fun(2,1,2,5,5,13,8)

abc= 2.3918

and after the above code i should get

v1 = nw_vec(p5);

v1 should be 0.47 ( Only for this combination)

But at the output i am getting 0.47, 0.65

I should get only 0.47 as the answer

Ameer Hamza
on 27 Apr 2020

When I run this in MATLAB. It just get single value (0.8)

fid = fopen('Mat file.txt');

data = textscan(fid, '%f', 'HeaderLines', 6);

period_arr = data{1};

fclose(fid);

period_arr = padarray(period_arr, 168000 - numel(period_arr), 0, 'post');

period_arr = reshape(period_arr, [2 7 1 10 10 15 8]);

abc = 2.3918;

nw_vec = [0.2, 0.36, 0.38, 0.40, 0.47, 0.5, 0.55, 0.65, 0.75, 0.8];

dist = abs(period_arr - abc);

[min_dist, idx] = min(dist(:));

nearestvalue = period_arr(idx);

actualidx = find(period_arr == nearestvalue);

[p1,p2,p3,p4,p5,p6,p7] = ind2sub(size(period_arr), actualidx);

v1 = nw_vec(p5);

Result:

>> idx

idx =

37747

>> nearestvalue

nearestvalue =

2.3918

>> actualidx

actualidx =

37747

>> p5

p5 =

10

>> v1

v1 =

0.8000

Ganesh Kini
on 27 Apr 2020

period_arr(2,1,1,9,9,13,8);

can you try this combination ??

Just to cross verify

Ameer Hamza
on 27 Apr 2020

It get a single value (0.75) again

fid = fopen('Mat file.txt');

data = textscan(fid, '%f', 'HeaderLines', 6);

period_arr = data{1};

fclose(fid);

period_arr = padarray(period_arr, 168000 - numel(period_arr), 0, 'post');

period_arr = reshape(period_arr, [2 7 1 10 10 15 8]);

abc = period_arr(2,1,1,9,9,13,8);

nw_vec = [0.2, 0.36, 0.38, 0.40, 0.47, 0.5, 0.55, 0.65, 0.75, 0.8];

dist = abs(period_arr - abc);

[min_dist, idx] = min(dist(:));

nearestvalue = period_arr(idx);

actualidx = find(period_arr == nearestvalue);

[p1,p2,p3,p4,p5,p6,p7] = ind2sub(size(period_arr), actualidx);

v1 = nw_vec(p5);

Result:

>> v1

v1 =

0.7500

Ganesh Kini
on 27 Apr 2020

for ioc=1:1:length(ring_vec)

for ipc=1:1:length(opcon_vec)

for irc=1:1:length(rccorner_vec)

for inw=1:1:length(nw_vec)

for ivw=1:1:length(pw_vec)

for ivdp=1:1:length(vd_vec)

for itp=1:1:length(temp_vec)

abc = period_fun(ioc,ipc,irc,inw,ivw,ivdp,itp);

%finding the nearest possible value

dist = abs(period_fun - abc);

[min_dist, idx] = min(dist(:));

nearestvalue = period_fun(idx);

actualidx= find(period_fun ==nearestvalue);

[p1,p2,p3,p4,p5,p6,p7] = ind2sub(size(period_fun), actualidx);

v1 = nw_vec(p5);

end

end

end

end

end

end

This is the overall program, do you think i need to refresh or clear the variables ? Yes in MATLAB

or is it even possible to tweak/ change the code a little bit, so that we get the accurate results

Ameer Hamza
on 27 Apr 2020

I cannot see any reason why you are getting two values in v1. The most I can suggest us to use a breakpoint and run your code line by line. Also, have you tried to find() with second input as I mentioned in one of my previous comment

actualidx= find(period_fun ==nearestvalue, 1);

Ganesh Kini
on 28 May 2020

Hi Ameer,

Thanks for the help

t1 ´= 2.3

t2 = 5.6

%its is a 2*7*1*10*10*15*8 matrix

p = period_arr(1,:,:,:,:,:,:); % this is of the form p = period_arr(i1, i2, i3, i4, i5, i6, i7);

n = period_arr(2,:,:,:,:,:,:); % this is of the form n = period_arr(i1, i2, i3, i4, i5, i6, i7);

dist_p = abs(p - t1);

[min_dist_p, idx_p] = min(dist_p(:));

dist_n = abs(n - t2);

[min_dist_n, idx_n] = min(dist_n(:));

c_tp = p(idx_p);

c_tn = n(idx_n);

So based on the minimum distance i get the closest value, and it is working fine.

But, I have a problem here I have to get only the closest value where the indices i4 of p = i4 of n and i5 of p = i5 of n. It should regulate the same value for both p and n.

How do i do that? please help me out. I am stuck in this problem from 2 days

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