rounding up (always)

On Oct 20, 2010, at 8:38 PM, Dimitri Liakhovitski wrote:

Thank you for your help, everyone.
Actually, I am building a lot of graphs (in a loop) but the values on
the y axes from graph to graph could range from [-5; 5] to [-10,000;
10,000].
So, I am trying to create ylim ranging from ymin to ymax such that
they look appropriate for the range.
For example, if we are taking the actual range from -4.28 to 6.45, I'd
like the range to be -5 to 7.
But if the range is from -1225 to 2248, then I'd like it to be from
-1500 to 2500 or from -2000 to 3000.
Hence, my original question.

Except you did not express that desire correctly. And you still have not
been very clear about the degree of rounding-ness needed. The first example
above suggests to the next highest absolute value if the minimum is less
than 10. (I didn't acheive that.) ?You initially wanted the rounding to the
next regular interval on the usual meaning of "greater than" when you wrote
... "always up" ?... and ... "higher" which would not lead most people to
deliver an answer that satisfies your more recent specification.

In particular any log()-ging would need to first apply an abs() function.
See if this is any more applicable to the (new) problem statement. This
rounds out on both the negative and positive sides to the next highest power
of 10.

roundout3 <- function(x){k=floor(log(abs(x),10)+1);

? ? ? ? ?sign(x)*ifelse(x==0, 0, ?(floor( abs(x/10^k)+0.499999999999 )
+1)*10^k)}
# zero is a problem since log(0) and log(abs(0)) both return Inf.

roundout3(c(-1250,-250, -3, 0, 4, 250, 1250))

[1] -10000 ?-1000 ? ?-10 ? ? ?0 ? ? 10 ? 1000 ?10000

--
David.

Dimitri

On Wed, Oct 20, 2010 at 5:55 PM, Ted Harding <ted.harding at wlandres.net>
wrote:

On 20-Oct-10 21:27:46, Duncan Murdoch wrote:

On 20/10/2010 5:16 PM, Dimitri Liakhovitski wrote:

Hello!

I am trying to round the number always up - i.e., whatever the
positive number is, I would like it to round it to the closest 10 that
is higher than this number, the closest 100 that is higher than this
number, etc.

For example:
x<-3241.388

signif(x,1) rounds to the closest thousand, i.e., to 3,000, but I'd
like to get 4,000 instead.
signif(x,2) rounds to the closest hundred, i.e., to 3,200, but I'd
like to get 3,300 instead.
signif(x,3) rounds to the closest ten, i.e., to 3,240, but I'd like to
get 3,250 instead.

Of course, I could do:
floor(signif(x,1)+1000)
floor(signif(x,2)+100)
floor(signif(x,3)+10)

But it's very manual - because in the problem I am facing the numbers
sometimes have to be rounded to a 1000, sometimes to a 100, etc.

Write a function. ?You have very particular needs, so it's unlikely
there's already one out there that matches them.
Duncan Murdoch

As Duncan and Clint suggest, writing a function is straightforward:
for the problem as you have stated it, on the lines of

?function(x,k){floor(signif(x,k-as.integer(log(x,10)-1))) + 10^k}

However, what do you *really* want to happen to 3000?

Ted.

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--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

David Winsemius, MD
West Hartford, CT