How to call a variable with user string input

조회 수: 214 (최근 30일)
Chris
Chris 2014년 5월 13일
댓글: Walter Roberson 2022년 7월 30일
Hi all,
I know this is a likely a simple problem, but am having lots of trouble...
I am trying to use an array of data (say 1x50) in a for loop from a user input that is different than the name of the array.
In other words, the user inputs something like 'example' and I need to use the data in the existing array that is named 'exampleD'
exampleD = ones(1,50); % this is an existing array
inp = input('Which dataset are you working with? ','s');
% at this point user enters the string 'example'
Missing code here (some command that will concatenate user input and the letter 'D' to get 'exampleD' which is the name of the existing variable)
for i = 1:50
fprintf(fid,'%f',exampleD(i));
fprintf(fid,',');
end
Any help is greatly appreciated!
  댓글 수: 2
Jorge Ignacio Cisneros Saldana
Jorge Ignacio Cisneros Saldana 2022년 7월 29일
편집: Jorge Ignacio Cisneros Saldana 2022년 7월 30일
Do you know how to do same approach without eval command for a parfor loop?.
eval(['dataset = ',variableName,';']);
Walter Roberson
Walter Roberson 2022년 7월 30일
If you were able to do an assignment by eval() inside the body of a parfor, there would be no way for matlab to know that the variable had to be returned to the client. If it were possible at all, then it would only be useful for assigning to local variables. And if you are going to be creating local variables you might as well use dynamic field names inside a struct instead of using eval()
Remember that you can assign a struct into a cell array, and then after the parfor merge the struct. Or output a cell of names and a cell of values and then cell2struct after the parfor

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채택된 답변

Benjamin Avants
Benjamin Avants 2014년 5월 13일
Although it is often advised not to use this method, it would suit your purpose.
Once you have the user's input string and have created the variable name, you can use the eval function to access that variable.
eval(['dataset = ',variableName,';']);
the chosen dataset will now be stored in the variable 'dataset' and you can use it as per normal.
You could also store all the datasets as fields of a struct and use the variable name string to access the correct field of the struct.
  댓글 수: 2
Roberto
Roberto 2014년 5월 13일
편집: Roberto 2014년 5월 13일
it's highly NOT recommended to use eval to assign variable names...
% Instead of this
eval('myVar = myfunc(2)');
% try this
myVar = eval(myfunc(2));
Chris
Chris 2014년 5월 13일
Thanks a ton Benjamin and Roberto. Between these last two comments I am able to write just a few lines of code to figure out my issue.

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추가 답변 (2개)

Roberto
Roberto 2014년 5월 13일
편집: Roberto 2014년 5월 13일
exampleD = ones(1,50); % this is an existing array
inp = input('Which dataset are you working with? ');
for i = 1:numel(inp)
% whatever
end
  댓글 수: 1
Chris
Chris 2014년 5월 13일
Roberto, Thanks for your answer, but this doesn't actually seem to help. My problem lies in the fact that my script needs to call a variable with an 'incomplete' user input. -Chris

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Roberto
Roberto 2014년 5월 13일
exampleD = ones(1,50); % this is an existing array
inp = input('Which dataset are you working with? ','s');
% at this point user enters the string 'example'
inp = eval([inp 'D']);
for i = 1:50
fprintf(fid,'%f',inp);
fprintf(fid,',');
end

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