On Apr 6, 2010, at 3:30 PM, David Winsemius wrote:
On Apr 6, 2010, at 9:56 AM, moleps islon wrote:
OK... next question.. Which is still a data manipulation problem so I
believe the heading is still OK.
##So now I read my population data from excel.
No, you read it from a text file and providing the first ten lines
of that text file should have been really easy. Read the Posting
Guide for advice about offering datasets either as structure()
objects with dput or dump or as attached files with "*.txt"
extension (not .csv). Just change the file name with your file
browser.
pop<-read.csv("pop.csv")
typeof(pop) ## yields a list
Really? I would have guessed it to yield just "list".
where I have age-specific population rows
and a yearly column population, where the years are suffixed by X
And had you used class(pop) you would have learned it was a
dataframe and even more informative would have been str(pop).
c<-(1953:2008)
No, no, no. Do not use variable names that are important function
names. The R interpreter can (usually) keep things straight but it
is our brains that experience problems. Other function names to
avoid: data, df, cut, mean, sd, list, vector, matrix
names(pop)<-c
c.div<-cut(c,break=seq(1950,2010,by=5)
(You should have gotten an error here.) After fixing the error, did
you you notice that there were only 3 of the first level???
Watch out for cut(). It uses the default convention of ( , ] , i.e.
open interval at right
er, ^left^
which is backwards to what some (most?) of us think natural. Because
of that the lowest level gets dropped unless you take special
precautions. That is undoubtedly why Harrell set up his Hmisc::cut2
to have the default be [ , )
Aggregating across columns? Certainly possible, but maybe not as
natural a fit to functions like split as would occur with working
across rows. I suppose you could use something like this untested
(because _still_ no sample dataset provided) code:
apply(pop, 1, # this works a row a time
function(x) tapply(x, list(c.div), sum) ) ) # or use aggregate
which uses tapply
I'm not sure it will work, since I don't know if the column names
would get carried over into "x" by apply(). You might need to create
a separate index that used the numeric positions of the columns
rather than their names. Perhaps use c.div <- seq(0,(2008-1953)) %/
% 5 or some such inside tapply.
Now I'd like to sum the agespecific population over the individual
levels of -c.div- and generate a new table for this with agespecific
rows and columns containing the 5-year bins instead of the original
yearly data. Do I have to program this from scratch or is it possible
to use an already existing function?
I think you ought to read more introductory material (and the
Posting Guide regarding how to offer example datasets). In this case
there are many functions that do data aggregation and most of them
should be illustrated in a good introductory text.
--
David.
//M
qta<- table(cut(age,breaks = seq(0, 100, by = 10),include.lowest =
TRUE),cut(year,breaks=seq(1950,2010,by=5),include.lowest=TRUE
On Mon, Apr 5, 2010 at 10:11 PM, moleps <mole...@gmail.com> wrote:
Thx Erik,
I have no idea what went wrong with the other code snippet, but
this one works.. Appreciate it.
qta<- table(cut(age,breaks = seq(0, 100, by = 10),include.lowest =
TRUE),cut(year,breaks=seq(1950,2010,by=5),include.lowest=TRUE))
M
On 5. apr. 2010, at 21.45, Erik Iverson wrote:
I don't know what your data are like, since you haven't given a
reproducible example. I was imagining something like:
## generate fake data
age <- sample(20:90, 100, replace = TRUE)
year <- sample(1950:2000, 100, replace = TRUE)
##look at big table
table(age, year)
## categorize data
## see include.lowest and right arguments to cut
age.factor <- cut(age, breaks = seq(20, 90, by = 10),
include.lowest = TRUE)
year.factor <- cut(year, breaks = seq(1950, 2000, by = 10),
include.lowest = TRUE)
table(age.factor, year.factor)
moleps wrote:
I already did try the regression modeling approach. However the
epidemiologists (referee) turns out to be quite fond of
comparing the incidence rates to different standard populations,
hence the need for this labourius approach. And trying the
"cutting" approach I ended up with :
table (age5)
age5
(0,5] (5,10] (10,15] (15,20] (20,25] (25,30] (30,35]
(35,40] (40,45] (45,50] (50,55] (55,60] (60,65] (65,70]
(70,75] (75,80] (80,85] (85,100] 35 34
33 47 51 109 157 231 362
511 745 926 1002 866 547 247
82 18
table (yr5)
yr5
(1950,1955] (1955,1960] (1960,1965] (1965,1970] (1970,1975]
(1975,1980] (1980,1985] (1985,1990] (1990,1995] (1995,2000]
(2000,2005] (2005,2009] 3 5
5 5 5 5 5
5 5 5 5 3
table (yr5,age5)
Error in table(yr5, age5) : all arguments must have the same
length
Sincerely,
M
On 5. apr. 2010, at 20.59, Bert Gunter wrote:
You have tempted, and being weak, I yield to temptation:
"Any good ideas?"
Yes. Don't do this.
(what you probably really want to do is fit a model with age as
a factor,
which can be done statistically e.g. by logistic regression; or
graphically
using conditioning plots, e.g. via trellis graphics (the
lattice package).
This avoids the arbitrariness and discontinuities of binning by
age range.)
Bert Gunter
Genentech Nonclinical Biostatistics
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
] On
Behalf Of moleps
Sent: Monday, April 05, 2010 11:46 AM
To: r-help@r-project.org
Subject: [R] Data manipulation problem
Dear R´ers.
I´ve got a dataset with age and year of diagnosis. In order to
age-standardize the incidence I need to transform the data into
a matrix
with age-groups (divided in 5 or 10 years) along one axis and
year divided
into 5 years along the other axis. Each cell should contain the
number of
cases for that age group and for that period.
I.e.
My data format now is
ID-age (to one decimal)-year(yearly data).
What I´d like is
age 1960-1965 1966-1970 etc...
0-5 3 8 10 15
6-10 2 5 8 13
etc..
Any good ideas?
Regards,
M
David Winsemius, MD
West Hartford, CT
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