That code throws multiple errors. Can you at least test your code
before posting?
(And, again, please avoid using function names as names for your
objects.)
-- David.
On Apr 7, 2010, at 8:54 AM, moleps islon wrote:
So.. here we try again.
##generate dataset
age.cat<-seq(0,100,10)
year<-(1953:(1953+55))
data.vec<-sample(10000:10000,(age.cat*year))
data.matrix<-matrix(data.vec,c(length(age.cat),length(year))
rownames(data.matrix)<-age.cat
colnames(data.matrix)<-year
##divide into 5 year periods
age.div<-cut(year,seq(1950,2010,6),include.lowest=T) ##interval is
beyond my datainterval so I doubt the include.lowest matters
Now what I'd like to do is summarise the rows within the 5-year
intervals.
I did read about apply in its different variants and Dahlgaard, but I
do not know understand how it could be applied in this setting.
I tried making an array and summarise by that (used the vector and
applied it into a
length(age.cat)*max(vector(table(age.div)*length(age.div) array. It
worked but required a bit of tweaking (inserting null columns) and I
find myself in this situation quite often whereby I need to add
multiple columns based on another vector so I'd be very interested in
another more general approach.
//M
On Tue, Apr 6, 2010 at 9:41 PM, David Winsemius <dwinsem...@comcast.net
> wrote:
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
On Tue, Apr 6, 2010 at 9:41 PM, David Winsemius <dwinsem...@comcast.net
> wrote:
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
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@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.
