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From: Jan Simon on 11 Feb 2010 07:24
Dear Chris!

> spaces = (str == ' ') % returns 1 at all blank spaces
> str(spaces) = (str(spaces))+'a'-'A'

Another idea:
afterspaces = strfind(str, ' ') + 1;
str(afterspaces) = str(afterspaces) + ('a' - 'A');

The brackets accelerate the method: Without them this is calculated:
(str(afterspaces) + 'a') - 'A'
This might cause an overflow - but not for usual strings - and performs two vector operations, while the brackets around 'a' - 'A' yield to a single vector operation only.

Kind regards, Jan
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