help? @sprintf
INFO: Loading help data...
Base.@sprintf(%Fmt, args...)
Return @printf formatted output as string.
julia s = @sprintf this is a %s test
this is a test
julia println(s)
this is a test
That is because `@sprintf` returns a string which you haven't printed.
El miércoles, 22 de
Thanks, it makes sense now. I actually read the documentation earlier, but
didn't understand it. Now with the example it's very clear.
On Thursday, July 23, 2015 at 3:56:39 PM UTC+3, Ismael VC wrote:
help? @sprintf
INFO: Loading help data...
Base.@sprintf(%Fmt, args...)
Return @printf
https://github.com/JuliaLang/julia/pull/12279
On Thursday, July 23, 2015 at 9:55:48 PM UTC+3, Ismael VC wrote:
You are welcome, if you think this needs some clarification you could try
to edit the manual for the good of everyone else!
Cheers.
On Thu, Jul 23, 2015 at 8:26 AM, Tero
You are welcome, if you think this needs some clarification you could try
to edit the manual for the good of everyone else!
Cheers.
On Thu, Jul 23, 2015 at 8:26 AM, Tero Frondelius tero.frondel...@gmail.com
wrote:
Thanks, it makes sense now. I actually read the documentation earlier, but
I'm trying to learn macros. Can you help me to get this working? Currently
the error is that arr is not defined. Probably an obvious mistake, but I
just don't get hang of it.
macro Write(arr)
@eval begin
for i in arr
@sprintf(%12.6f\n,i)
end
end
end
a =
Thanks. My real error was to use @sprintf macro, thus a follow up question,
why this isn't printing anything:
macro Write(arr)
quote
for i in $arr
@sprintf(%12.6f\n,i)
end
end
end
a = 1e5*rand(10)
@Write a
This is purely for learning purposes. This was simple enough
You forgot to interpolate the expression with `$`:
julia macro write(arr)
quote
for i in $arr
@printf(%12.6f\n,i)
end
end
end
julia a = 1e5*rand(10)
10-element Array{Float64,1}:
46310.6
25130.5
30710.8
82089.6
I wrote this some time ago for my own projects. It's reasonably fast:
https://gist.github.com/dpo/11000433
Thanks for the info. Actually my question comes from old fortran style, where I
can write something of the form
Write(1,'1000f12.6') a
where a is an array. The string inside the write function says I can print 1000
doubkes in 12 characters with 6 decimals. So the string is a constant literal,
Would this work for you:
julia a = 1e5*rand(1000)
julia for i in a
@printf(%12.6f\n, i)
end
74708.038385
71244.774457
5057.229038
3761.297034
...
Remember that loops are fast in Julia...
Kaj
On Tuesday, July 14, 2015 at 9:14:37 AM UTC+3, Ferran Mazzanti wrote:
Thanks for the
Yes thanks, I knew already looped solutions :)
I was looking for somethin' compact as in the fortran statement above,
though. It makes things more *neat*, if there's any such thing.
On Tuesday, July 14, 2015 at 12:08:59 PM UTC+2, Kaj Wiik wrote:
Would this work for you:
julia a =
You could just use a macro to take the format and the array and let it
write the messy loop for you.
On Tuesday, July 14, 2015 at 8:39:44 PM UTC+10, Ferran Mazzanti wrote:
Yes thanks, I knew already looped solutions :)
I was looking for somethin' compact as in the fortran statement above,
Hi,
I have a simple question: is it possible to feed printf() a list with a
variable number of arguments? Specifically I want to print a formatted list of
numbers contained in an arrary, but I can't tell in advance how many elements
does the array contain. Is there an easy way to do that, or
On Mon, Jul 6, 2015 at 6:47 PM, Ferran Mazzanti
ferran.mazza...@gmail.com wrote:
Hi,
I have a simple question: is it possible to feed printf() a list with a
variable number of arguments? Specifically I want to print a formatted list
of numbers contained in an arrary, but I can't tell in
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