Re: sorting a data frame

[David L Lorenz <lorenz@usgs.gov>]

Walt,
  The reason that the data appear
not to be sorted correctly is that the year column is not an integer value.
There are very small differences in the value for yr because of minor differences
in the storage of the values. This is what I get when I subtract 2007 from
x$yr
x$yr - 2007
 [1]  1.000000e+000  3.183231e-012
 3.183231e-012  3.183231e-012  3.183231e-012 -5.684342e-012

-5.684342e-012 -5.684342e-012
 [9] -5.684342e-012 -5.684342e-012
  As you can see, the values that
started out greater than 7 are all slightly less than 2007 and the values
that started out less than 8 are all slightly larger than 2007, thus they
are sorted correctly. You need to round them when you do your computations.
Dave

"Data Analytics Corp." <dataanalytics@earthlink.net>
Sent by: s-news-owner@lists.biostat.wustl.edu

01/29/2008 08:56 AM

To	s-news@lists.biostat.wustl.edu
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Subject	[S] sorting a data frame

Good morning,

I posted a question yesterday regarding sorting a data frame.  The
problem is that the sort order is not correct when I use sort.col.  I
received many responses, all saying basically the same thing - I must 
have a factor.  I actually checked this before posting the query and
all 
was numeric.  So let me try again but be more explicit this time,
because I'm very puzzled.

I have a client file that has the dates as 1.2008, 4.2007, etc.  The
commands and example I used along with the as.numeric() checks are 
below.  As you can see, the sort is not correct.  I still don't
understand why.  Any hints?

 > y <- c(1.2008, 4.2007, 5.2007, 6.2007, 7.2007, 8.2007, 9.2007,
10.2007, 11.2007, 12.2007)
 > y
 [1]  1.2008  4.2007  5.2007  6.2007  7.2007  8.2007
 9.2007 10.2007 
11.2007 12.2007
 > is.numeric(y)
[1] T
 > x <- data.frame(yr = (y - floor(y))*10000, month = floor(y))
 > x
     yr month
 1 2008     1
 2 2007     4
 3 2007     5
 4 2007     6
 5 2007     7
 6 2007     8
 7 2007     9
 8 2007    10
 9 2007    11
10 2007    12
 > is.numeric(x$month)
[1] T
 > sort.col(x, "@ALL", 1:2)
     yr month
 6 2007     8
 7 2007     9
 8 2007    10
 9 2007    11
10 2007    12
 2 2007     4
 3 2007     5
 4 2007     6
 5 2007     7
 1 2008     1
 >


Any help is appreciated.

Walt Paczkowski
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