Extract the value of a array from a function

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]

Does the function period only accept integer inputs? Is it possible to write its inverse based on its definition? Can you show us the code of the period function?

[Ganesh Kini]

Yes, it will be only integers.

how do i write the inverse of it ?

Please let me know

Answer 1

[Ameer Hamza]

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]

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]

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]

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]

Ganesh, based on the information provided in your question, it is impossible to get the 5th element of vector a. That information about vector a was not used by the function period, and therefore lost. If there is some other information available about how to construct vector a from the value of t1 then it might be possible.

[Ganesh Kini]

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]

Can you show your actual code. This code does not make much sense. what is the value of variable p? what is the value of variable time(p)? How thw vectors a, b, and c are loaded?

[Ganesh Kini]

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]

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]

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]

If it is in a text file, then you cannot access it directly in MATLAB. If it is saved in .mat file, then you can directly access in on your disk without loading into MATLAB.

[Ganesh Kini]

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]

See matfile: https://www.mathworks.com/help/matlab/ref/matlab.io.matfile.html

It will directly access variables from the disk without loading them into memory. For matrices, it can be used with a slight modification. Is p stored as matrix? If yes, what are its dimensions?

[Ganesh Kini]

Hi,

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

[Ameer Hamza]

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

[Ganesh Kini]

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]

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]

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]

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]

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]

62988 is the linear index. To convert into a,b, and c you need to use ind2sub. See the difference between linear indexing and Subscripts indexing on this page: https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

[Ganesh Kini]

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]

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]

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

[Ameer Hamza]

Use round() function.

[Ganesh Kini]

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]

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]

Hi

The above code is not working

I have tried this before

Can you suggest some other alternative ?

[Ameer Hamza]

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

[Ganesh Kini]

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]

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]

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]

Is this the complete error message shown by MATLAB?

[Ganesh Kini]

I am actually doing it Octave, which is similar to matlab

[Ameer Hamza]

The solution I suggest works in MATLAB but not I octave, so there are some fundamental differences between two. You should try to convert my method to octave according to its function.

[Ameer Hamza]

I have already posted an answer to that question.

[Ganesh Kini]

Sorry i posted the wrong link

Can you check this

https://de.mathworks.com/matlabcentral/answers/518903-finding-an-closest-possible-element-in-an-7-dimensional-matrix-array

[Ganesh Kini]

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]

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]

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]

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

[Ganesh Kini]

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 ?

[Ameer Hamza]

How is nw_vec defined?

[Ganesh Kini]

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]

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

[Ganesh Kini]

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

[Ameer Hamza]

You will need to share period_fun values to diagnose the issue.

[Ganesh Kini]

Yes I have shared it earlier on this link

https://de.mathworks.com/matlabcentral/answers/518903-finding-an-closest-possible-element-in-an-7-dimensional-matrix-array

its the same code

Attaching the file again

[Ameer Hamza]

No value in period_fun is 1.1. How can it find this value?

[Ganesh Kini]

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]

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]

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

can you try this combination ??

Just to cross verify

[Ameer Hamza]

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]

But I am getting that way, what can be the potential reason ?

[Ameer Hamza]

are you running it in MATLAB?

[Ganesh Kini]

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]

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]

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