I guess "little" means different things to different people:
x = sample(1:100,650000,replace=TRUE)
system.time(a<-formatC(x,digits=10,flag='0'))
user system elapsed
32.854 0.444 34.813
system.time(b<-sprintf("%011d",x))
user system elapsed
0.352 0.012 0.363
If you look at the definitions of the functions, you'll see
that formatC is written in R, and sprintf uses a single call
to an .Internal function. I
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
[EMAIL PROTECTED]
On Mon, 12 May 2008, Anh Tran wrote:
Yea, thanks all. I checked back and I got a few things mistyped.
The array is 650,000 and it took 25 seconds :p. It's acceptable. Just that I
had too many variable at the time I ran it.
Also, seems like sprintf is a little faster.
Thanks all.
Anh Tran
On Mon, May 12, 2008 at 2:55 PM, Uwe Ligges <[EMAIL PROTECTED]>
wrote:
Anh Tran wrote:
Thanks. formatC(flag) works.
But it's awefully slow. I try to do that for 65000 numbers (generating
ID
for each item) and it seems like forever.
On my not that recent laptop:
system.time(formatC(1:65000, width=10, flag="0"))
user system elapsed
1.92 0.00 1.94
I think 2 seconds is less than "forever".
Uwe Ligges
Is there any faster way?
Thank all.
Anh Tran
On Mon, May 12, 2008 at 2:36 PM, Uwe Ligges <
[EMAIL PROTECTED]> wrote:
Anh Tran wrote:
Hi,
What's one way to convert an integer to a string with preceding 0's?
such that
'13' becomes '00000000013'
to be put into a string
I've tried formatC, but they removes all the zeros and replace it
with
blanks
Not so for me:
formatC(13, digits=10, flag="0")
Uwe LIgges
Thanks
--
Regards,
Anh Tran
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