I think this is the shortest (and my favorite) way: w = ["col$i" for i = 1:100_000]
- Dahua On Wednesday, January 1, 2014 9:22:12 AM UTC-6, John Myles White wrote: > > Since we’re throwing out ways to do this, here’s another one: > > w = Array(UTF8String, 100_000) > > for i in 1:100_000 > w[i] = @sprintf "col%d" i > end > > I’ve come to really love using the *printf macros for anything that even > vaguely resembles formatting. > > — John > > On Jan 1, 2014, at 6:27 AM, [email protected] <javascript:> wrote: > > > Hi, > > > > First of all you initialize "w" as Float64 which you do not probably > want. > > > > For creating vector of strings you colud write: > > > > w = ["col" * string(i) for i in 1:100000] > > > > or: > > > > w = Array(ASCIIString, 100000) > > > > for i = 1:100000 > > w [i] = "col" * string(i) > > end > > > > Bogumil > > > > On Wednesday, January 1, 2014 11:07:06 AM UTC+1, paul analyst wrote: > > I need a vector of string of subtitles: col1, col2, col3, ..., col100000 > > How to join "col" phrase with numbers of var(i) ( in the loop) > > > > w = zeros (100000); > > > > for i = 1:100000 > > w [i] =??? > > end > > > > Paul > >
