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Creating variables consisting of many elements

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Gratitude Kim
Gratitude Kim on 4 Aug 2017
Commented: Andrei Bobrov on 4 Aug 2017
I want to create a variable consisting of X1 to X10 . Is there an easy way instead of typing X1 to X10?
V1={'X1','X2','X3','X4','X5','X6','X7','X8','X9','X10'}
Thanks

  3 Comments

the cyclist
the cyclist on 4 Aug 2017
Your problem is upstream, and is a great illustration of why it is almost always a terrible idea to name variables X1, X2, etc.
Can you fix that prior code, to use a vector with elements X(1), X(2), or a cell array with elements X{1}, X{2}, etc?
If that is impossible, there are still solutions to your awkward problem, via even more awkward methods. What is the class and size of X1, etc?
Stephen Cobeldick
Stephen Cobeldick on 4 Aug 2017
@Gratitude Kim: what do you need these string for? The best solution may well depend on your usecase.
>> arrayfun(@(n)sprintf('X%d',n),1:10,'uni',0)
ans =
'X1'
'X2'
'X3'
'X4'
'X5'
'X6'
'X7'
'X8'
'X9'
'X10'
or perhaps
>> regexp(sprintf('X%d ',1:10),'\S+','match')
If you are planning on accessing variables with those names then DON'T. Read this to know why:

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Answers (3)

Daniel Burke
Daniel Burke on 4 Aug 2017
From your question it looks like you want V1 to be a cell array of strings, ranging from 'X1' to 'X10', this code should do the trick if that is your goal:
V1 = cell(1,10)
for i = 1:10
V1{i} = ['X' num2str(i)]
end

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Philip Borghesani
Philip Borghesani on 4 Aug 2017
If you don't mind strings and have R2017a I like:
"x"+(1:10)
Or if you want a cell array or have 2016b:
cellstr(string('x')+(1:10))

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