Loading in a for loop

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Tim Johansson
Tim Johansson am 15 Okt. 2020
Kommentiert: Tim Johansson am 16 Okt. 2020
im in a situation on which i have to load a lot of files (about 400) and was wondering if there were a smart way to do it.
I was thinking that a for loop was the best way to do it, but the for loop does not recognise the 'i' as the variable.
for i=1:400
run_i='C:\Users\unknown\run_i'
end
is there a simple way to solve this?
  2 Kommentare
Stephen23
Stephen23 am 15 Okt. 2020
Tim Johansson
Tim Johansson am 16 Okt. 2020
Thanks, it worked.

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Antworten (1)

Sudhakar Shinde
Sudhakar Shinde am 15 Okt. 2020
Bearbeitet: Sudhakar Shinde am 15 Okt. 2020
You can use 'int2str'
for i=1:400
run_i=['C:\Users\unknown\run_',int2str(i)];
end
%or
for i=1:400
run_i=strcat('C:\Users\unknown\run_',num2str(i));
end
  2 Kommentare
Stephen23
Stephen23 am 15 Okt. 2020
Or use the recommended and more versatile sprintf.
Tim Johansson
Tim Johansson am 16 Okt. 2020
Thanks for the help

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