How to assign values to a string? [duplicate]

Question

I have a loop running on multiple workstations, each loop is dealing with 5 different classes, when the results is acquired it is to save the result with the name of the classes it used.

For example on workstation 1 I am using:

class1 = 1;
class2 = 10;

%code that will run on all images associated with classes 1:10
% final_result from all 10 classes is calculated here

I want to save this now, with the name such as:

result_1_10 = final_result;
save result_1_10 result_1_10;

I can do this manually but its becoming very difficult to change the values on all machines after one value is changed, I would rather it save these and pick up the numbers from the two variables class1 and class2.

Here is what I tried:

['result_' num2str(class1) '_' num2str(class2)];

This would give me result_1_10. Which is what I wanted but it is a string, rather than a variable, so I cannot assign the result value to it

['result_' num2str(class1) '_' num2str(class2)] = final_result;

Would give the error:

Error: An array for multiple LHS assignment cannot contain LEX_TS_STRING.

I even tried str2num(num2str(class1)) but that would also give an error.

How do I actually do this?

Thank you

Suever · Accepted Answer · 2016-03-16 14:56:30Z

2

While you can do this, it is very much discouraged by The Mathworks themselves. Any time you are trying to store information about what a variable contains within the variable name itself, that's a sign that maybe things should be rearranged a bit. Maybe consider using a different data structure.

Consider for example using a struct where you keep the classes as fields and the result as a field.

S.class1 = 1;
S.class2 = 10;
S.result = final_result;

You could then even create an array of structs holding your data.

S = struct('class1', {1, 2, 1}, ...
           'class2', {10, 11, 10}, ...
           'result', {rand(10), rand(10), rand(10)});

Then you could grab all results when class1 was 1:

S([S.class1 == 1]);

Or all results when class1 as 1 and class2 was 10

S([S.class1 == 1] & [S.class2 == 10]);

If you insist on doing it the way that you've laid out, you'll have to use eval or assignin to do that. Also, sprintf is often more concise than string concatenations.

variable = sprintf('result_%d_%d', class1, class2);
eval([variable, '= final_result;']);

edited Mar 16, 2016 at 14:56

answered Mar 16, 2016 at 14:30

Suever

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4 Comments

StuckInPhDNoMore Over a year ago

Thank you for your detailed answer. After the eval([variable, '= final_result;']); statement I have achieved the end result, that is assign the value to the name generated. What is the purpose of the assignin function after that?

Suever Over a year ago

@StuckInPhD Did you read the entire post? I would recommend trying the other approach. That being said the assignin statement does the exact same thing so it's an alternative.

StuckInPhDNoMore Over a year ago

Yes, I did read the post and your alternative method, I was just curious. Regarding the alternative, I could save everything inside a struct upto the line: S.variable = sprintf('result_%d_%d', class1, class2); but this would be in string, and I am back to where I started, how do I assign this new name to my final_result to be saved? Sorry if I misunderstood anything

Suever Over a year ago

@StuckInPhD If you want to convert all of your existing result_x_y variables to the struct versions, you will have to use eval (like in the last part of the question) to accomplish this. Then once they are in the struct, you will never have to use eval again! eval(sprintf('S.variable = result_%d_%d;', class1, class2));

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