R help - Oct 2008 - Condition warning: has length > 1 and only the first element will be used

Hello,

I've been learning R functions recently and I've come across a problem
that
perhaps someone could help me with.

# I have a a chron() object of times
>
hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00"))
# I would like to subtract 12 hours from each time element, so I created a
vector of 12 hours (actually, more like a vector of noons)
> less.hours=chron(time=rep("12:00:00",4))
# If I just subtract the two vectors, I get some negative values, as
fractions of a day
> hours-less.hours
[1] -0.45833333  0.25000000  0.04166667 -0.08333333

# I would like those negative values to cycle around and subtract the amount
< 0 from midnight
# That is to say, 01:00:00 - 12:00:00 should be 13:00:00
# because 01:00:00-12:00:00 = -11:00:00, and 24:00:00-11:00:00 = 13:00:00
# It's sort of like going back to the previous day, but without actually
including information about which day it is

# This is what I tried
> test.function=function(x,y) {
+ sub = x-y
+ if(sub<0) x+y
+ }
> test.function(hours,less.hours)
Time in days:
[1] 0.5416667 1.2500000 1.0416667 0.9166667
Warning message:
In if (sub < 0) x + y :
  the condition has length > 1 and only the first element will be used


# My questions are, why does it only use the first element?? Why does it not
apply to all? Also, what happened to the elements where sub>= 0, it looks
like they followed the rules
# of if(sub<0).  I feel like I must not be understanding something rather
basic about how logical expressions operate within R
# Help would be appreciated...

	[[alternative HTML version deleted]]

The if statement is for program flow control (do this bunch of code if this
condition is true, else do this), the vectorized version is the ifelse function,
that is the one that you want to use.

Also note (this is for times in general, not sure about chron specifically)
subtracting noon from the times gives you time differences, subtracting a time
difference from a time will give you another time, so it may be simpler to
subtract just 12, rather than noon from your times.

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Joe Kaser
> Sent: Wednesday, October 15, 2008 2:19 PM
> To: r-help at r-project.org
> Subject: [R] Condition warning: has length > 1 and only the first
> element will be used
>
> Hello,
>
> I've been learning R functions recently and I've come across a
problem
> that
> perhaps someone could help me with.
>
> # I have a a chron() object of times
>
> >
hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00"))
>
> # I would like to subtract 12 hours from each time element, so I
> created a
> vector of 12 hours (actually, more like a vector of noons)
>
> > less.hours=chron(time=rep("12:00:00",4))
>
> # If I just subtract the two vectors, I get some negative values, as
> fractions of a day
>
> > hours-less.hours
> [1] -0.45833333  0.25000000  0.04166667 -0.08333333
>
> # I would like those negative values to cycle around and subtract the
> amount
> < 0 from midnight
> # That is to say, 01:00:00 - 12:00:00 should be 13:00:00
> # because 01:00:00-12:00:00 = -11:00:00, and 24:00:00-11:00:00 >
13:00:00
> # It's sort of like going back to the previous day, but without
> actually
> including information about which day it is
>
> # This is what I tried
>
> > test.function=function(x,y) {
> + sub = x-y
> + if(sub<0) x+y
> + }
>
> > test.function(hours,less.hours)
> Time in days:
> [1] 0.5416667 1.2500000 1.0416667 0.9166667
> Warning message:
> In if (sub < 0) x + y :
>   the condition has length > 1 and only the first element will be used
>
>
> # My questions are, why does it only use the first element?? Why does
> it not
> apply to all? Also, what happened to the elements where sub>= 0, it
> looks
> like they followed the rules
> # of if(sub<0).  I feel like I must not be understanding something
> rather
> basic about how logical expressions operate within R
> # Help would be appreciated...
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

Joe Kaser wrote:
> Hello,
> 
> I've been learning R functions recently and I've come across a
problem that
> perhaps someone could help me with.
> 
> # I have a a chron() object of times
> 
>>
hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00"))
> 
> # I would like to subtract 12 hours from each time element, so I created a
> vector of 12 hours (actually, more like a vector of noons)
> 
>> less.hours=chron(time=rep("12:00:00",4))
> 
> # If I just subtract the two vectors, I get some negative values, as
> fractions of a day
> 
>> hours-less.hours
> [1] -0.45833333  0.25000000  0.04166667 -0.08333333
> 
> # I would like those negative values to cycle around and subtract the
amount
> < 0 from midnight
> # That is to say, 01:00:00 - 12:00:00 should be 13:00:00
> # because 01:00:00-12:00:00 = -11:00:00, and 24:00:00-11:00:00 = 13:00:00
> # It's sort of like going back to the previous day, but without
actually
> including information about which day it is
> 
> # This is what I tried
> 
>> test.function=function(x,y) {
> + sub = x-y
> + if(sub<0) x+y
> + }
> 
>> test.function(hours,less.hours)
> Time in days:
> [1] 0.5416667 1.2500000 1.0416667 0.9166667
> Warning message:
> In if (sub < 0) x + y :
>   the condition has length > 1 and only the first element will be used
> 
> 
> # My questions are, why does it only use the first element?? Why does it
not
> apply to all? Also, what happened to the elements where sub>= 0, it
looks
> like they followed the rules
> # of if(sub<0).  I feel like I must not be understanding something
rather
> basic about how logical expressions operate within R
> # Help would be appreciated...
OK: help(ifelse), ordinary if doesn't vectorize.

Actually, ifelse does awkward things with classed objects. You might 
want something like

sub <- x - y
sub[sub<0] <- x+y

instead. And don't forget to return sub in all cases.

-- 
    O__  ---- Peter Dalgaard             ?ster Farimagsgade 5, Entr.B
   c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
  (*) \(*) -- University of Copenhagen   Denmark      Ph:  (+45) 35327918
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