So you have to "rip" a number apart into individual digits... The end output is a cell.
That is if:
x = 12345678
out = {[1 2 3 4 5 6 7 8]}
x = [1 23 53 1 88]
out = {[1 2 3 5 3 1 8 8]}
x = [213 102 9 12;
11 323 34 22];
out = {[2 1 3 1 0 2 9 1 2];
[1 1 3 2 3 3 4 2 2]};
x = [120395;
2]
out = {[1 2 0 3 9 5];
[2]}
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1 Comment
Frequency Domain
on 24 Jul 2013
Nice 'for'! Sometimes I simply forget to use that!
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