Hi,
I am using aggregate to compute means for later plotting.
There are two factors involved and the problem is that the values of the
second factor ( Age ) in the means are not in the right order because
10 comes inbetween 1 and 2
What I really want is the numeric value of Age but as.numeric
: [R] aggregate factor
Hi,
I am using aggregate to compute means for later plotting.
There are two factors involved and the problem is that the
values of the second factor ( Age ) in the means are not in
the right order because 10 comes inbetween 1 and 2
What I really want is the numeric
Dear Lest,
I have a two-variable data frame as follows (the time peirod of the
actual data set is 10 years):
Date Amount
1 6/1/2007 1
2 6/1/2007 1
3 6/4/2007 2
4 6/5/2007 2
5 6/11/2007 3
6 6/12/2007 3
7 6/12/2007 3
8 6/13/2007 3
9
Hi,
Perhaps you can try:
df
Date Amount
1 2007-06-01 1
2 2007-06-01 1
3 2007-06-04 2
4 2007-06-05 2
5 2007-06-11 3
6 2007-06-12 3
7 2007-06-12 3
8 2007-06-13 3
9 2007-06-13 3
10 2007-06-18 4
11 2007-06-18 4
12 2007-06-25
Or,
z-mydata #zoo object
new.time - as.Date(7 * floor(as.numeric(time(z))/7) + 7)
z2 - aggregate(z, new.time, mean)
Henrique Dallazuanna escribió:
Hi,
Perhaps you can try:
df
Date Amount
1 2007-06-01 1
2 2007-06-01 1
3 2007-06-04 2
4 2007-06-05
Try this. I have changed output format to yyy/mm/Weekw so its ordered.
Lines - Date Amount
6/1/2007 1
6/1/2007 1
6/4/2007 2
6/5/2007 2
6/11/2007 3
6/12/2007 3
6/12/2007 3
6/13/2007 3
6/13/2007 3
6/18/2007 4
6/18/2007 4
6/25/2007
Dear Fellow Rers,
I have a table looks like this:
ca, la, 12
ca, sd, 22
ca, la, 33
nm, al, 9
ma, lx, 18
ma, bs, 90
ma, lx, 22
I want to sum the 3rd column grouped by the first and
the second column, so the result look like this table:
ca, la, 45
ca, sd, 22
nm, al, 9
ma, lx, 40
ma, bs, 90
This seems to work fine:
x - ca, la, 12
+ ca, sd, 22
+ ca, la, 33
+ nm, al, 9
+ ma, lx, 18
+ ma, bs, 90
+ ma, lx, 22
+
table - read.csv(textConnection(x), header=FALSE)
aggregate(table$V3,list(table$V1,table$V2),mean)
Group.1 Group.2x
1 nm al 9.0
2 ma bs 90.0
3
On 6/1/07, Alfonso Sammassimo [EMAIL PROTECTED] wrote:
Hi R-experts,
Thanks very much to Jim Holtman and Gabor on my previous question.
I am having another problem with data manipulation in zoo. The following is
data (Z) for first business day of every month in zoo format. I am trying to
I want to use the aggregate function to summarize data by a factor (my
field plots), but I want the summary to be the majority level of another
factor.
For example, given the dataframe:
Plot1 big
Plot1 big
Plot1 small
Plot2 big
Plot2 small
Plot2 small
Plot3
How about tapply?
plot - gl(2,3); plot
type - letters[c(1,2,2,1,1,1)]; type
tapply(type, list(plot), function(x) {tabl - table(x)
names(tabl[tabl==max
(tabl)])})
Hank
On May 31, 2007, at 3:25 PM, Thompson, Jonathan wrote:
I want to use the
On Thu, 2007-05-31 at 12:25 -0700, Thompson, Jonathan wrote:
I want to use the aggregate function to summarize data by a factor (my
field plots), but I want the summary to be the majority level of another
factor.
For example, given the dataframe:
Plot1 big
Plot1 big
Plot1
PROTECTED] On Behalf Of
Thompson, Jonathan
Sent: Friday, 1 June 2007 7:26 a.m.
To: r-help@stat.math.ethz.ch
Subject: [R] Aggregate to find majority level of a factor
I want to use the aggregate function to summarize data by a
factor (my field plots), but I want the summary to be the
majority
This should do the trick. Also labels ties with NA.
a=as.data.frame(cbind(c(1,1,1,2,2,2,3,3,3,4,4),c
('big','big','small','big','small','small','small','small','small','big'
,'small')))
a$V2=factor(a$V2)
maj=function(x){
y=table(x)
z=which.max(y)
if(sum(y==max(y))==1){
Hi R-experts,
Thanks very much to Jim Holtman and Gabor on my previous question.
I am having another problem with data manipulation in zoo. The following is
data (Z) for first business day of every month in zoo format. I am trying to
get mean of open for each year. I subset Z - Z[,2] then
Hi,
Does anyone know if: with R can you take a set of numbers and aggregate
them like you can in SPSS? For example, could you calculate the percentage
of people who smoke based on a dataset like the following:
smoke = 1
non-smoke = 2
variable
1
1
1
2
2
1
1
1
2
2
2
2
2
2
When aggregated, SPSS
?table
On Wednesday 25 April 2007 14:32, Natalie O'Toole wrote:
Hi,
Does anyone know if: with R can you take a set of numbers and aggregate
them like you can in SPSS? For example, could you calculate the percentage
of people who smoke based on a dataset like the following:
smoke = 1
Hi Nat,
can I suggest, without offending, that you purchase and read Peter
Dalgaard's Introductory Statistics with R or Michael Crawley's
Statistics: An Introduction using R or Venables and Ripley's Modern
Applied Statistics with S or Maindonald and Braun's Data Analysis
and Graphics Using R: An
Andrew Robinson [EMAIL PROTECTED] wrote:
can I suggest, without offending, that you purchase and read Peter
Dalgaard's Introductory Statistics with R or Michael Crawley's
Statistics: An Introduction using R or Venables and Ripley's Modern
Applied Statistics with S or Maindonald and Braun's
Hello,
is there a way to use the aggregate function to calculate monthly mean
in case i have one row in data frame that holds the date like
-mm-dd? i know that it works for daily means. i also like to do it
for monthly and yearly means. maybe there is something like aggregate(x,
try this. The first group of lines recreates your data frame, DF, and
the last line is the aggregate:
Input - DateTimez
2006-01-01 21:00 6,2
2006-01-01 22:00 5,7
2006-01-01 23:00 3,2
2006-01-02 00:00 7,8
2006-01-02 01:00 6,8
2006-01-02
it works. thanks a lot.
Gabor Grothendieck wrote:
try this. The first group of lines recreates your data frame, DF, and
the last line is the aggregate:
Input - DateTimez
2006-01-01 21:00 6,2
2006-01-01 22:00 5,7
2006-01-01 23:00 3,2
2006-01-02
If monthly should aggregate per -mm combination, you could try
something like
aggregate(x$z,list(cut(as.Date(x$Date),m)),mean)
for monthly aggregation and
aggregate(x$z,list(cut(as.Date(x$Date),y)),mean)
for yearly means.
If monthly aggregation should aggregate over different years
Dear list members,
I am facing some problems using the aggregate() function.
I want to calculate a sum and a mean of one variable over the
combination of 12 factors with the aggregate() function to avoid loops
but it doesn't work (or the job is far too long, it exceeds 2 hours). It
works with
Joachim Claudet wrote:
Dear list members,
I am facing some problems using the aggregate() function.
I want to calculate a sum and a mean of one variable over the
combination of 12 factors with the aggregate() function to avoid loops
but it doesn't work (or the job is far too long, it
Peter Dalgaard wrote:
Alternatively, rewrite aggregate() and send us a patch ;-)
It is not necessarily all that hard. Here's a rough idea
IX - as.data.frame(by)
OO - do.call(order,IX)
Y - x[OO,]
g - cumsum(!duplicated(IX))
FF - unique(IX)
cbind(FF, sapply(split(x,g),FUN))
(completely
Hi All,
I think i'm failing to undersatnd how aggregate() is supposed to work.
example:
test1-sample(c(0,1),100,replace=T)
test2-sample(letters,100,replace=T)
aggregate(test1,list(test2),sum)
Error in data.frame(w, lapply(y, unlist, use.names = FALSE)) :
arguments imply differing
It does that for me without errors ...
(R 2.3.1 on Mac OSX 10.4.8)
Best, Ingmar
From: Gustaf Rydevik [EMAIL PROTECTED]
Date: Fri, 8 Dec 2006 12:58:01 +0300
To: r-help@stat.math.ethz.ch
Subject: [R] Aggregate?
Hi All,
I think i'm failing to undersatnd how aggregate() is supposed to work
: Gustaf Rydevik [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Aggregate?
Hi All,
I think i'm failing to undersatnd how aggregate() is supposed to work.
example:
test1-sample(c(0,1),100,replace=T)
test2-sample(letters,100,replace=T)
aggregate
Try the summary function, which pretty much does exactly that.
-Alex
On 20 Oct 2006, at 23:44, Jonathan Greenberg wrote:
Is there a way to calculate, say, the mean, min and max using
aggregate
using one line of code? Or do I need to call them separately (e.g.
aggregate(...,mean);
Is there a way to calculate, say, the mean, min and max using aggregate
using one line of code? Or do I need to call them separately (e.g.
aggregate(...,mean); aggregate(...,min)) and then merge the data back
together?
--j
--
Jonathan A. Greenberg, PhD
NRC Research Associate
NASA Ames Research
Try summaryBy in package doBy. e.g. using the built in dataset CO2:
summaryBy(uptake ~ Plant, CO2, FUN = c(mean, min, max))
On 10/20/06, Jonathan Greenberg [EMAIL PROTECTED] wrote:
Is there a way to calculate, say, the mean, min and max using aggregate
using one line of code? Or do I need to
Try summaryBy in package doBy. e.g. using the built in dataset CO2:
summaryBy(uptake ~ Plant, CO2, FUN = c(mean, min, max))
Or with reshape with a little more work:
cm - melt(CO2, id=1:4)
cast(cm, Type ~ Treatment, c(min,mean,max))
but you get some extra flexibility:
cast(cm,
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hello,
I'm a novice user trying to figure out how to retain NA aggregate
values. For example, given a data frame with data for 3 of the 4
possible factor colors(orange is omitted from the data frame), I want
to calculate the average height by color,
On Thu, 2006-10-05 at 15:44 -0700, Kaom Te wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hello,
I'm a novice user trying to figure out how to retain NA aggregate
values. For example, given a data frame with data for 3 of the 4
possible factor colors(orange is omitted from the data
Dear r-help reader,
I have some problems with the aggregate function.
My datframe looks like
frame
Day Time V1 V2
1 M0 3 NA
2 M0 4 NA
3 M0 5 2
4 M1 NA 4
5 M1 10 6
6 T0 4 45
7 T1 4 3
8 T1 3 2
9 T1 6 1
I used the aggegate
Frank [EMAIL PROTECTED] writes:
aggregate(frame[,c(-1)],list(frame$Day,frame$Time),mean)
My problem is now that I do not obtain a 'mean' for Day=M/Time=0 and
Day=M/Time=1,
because aggregate ignores all values for a grouping variable if NA
occurs.
No. But mean() will give an NA
aggregate(frame[,c(-1)],list(frame$Day,frame$Time),mean, na.rm=T)
2006/10/1, Frank [EMAIL PROTECTED]:
Dear r-help reader,
I have some problems with the aggregate function.
My datframe looks like
frame
Day Time V1 V2
1 M0 3 NA
2 M0 4 NA
3 M0 5 2
4 M1 NA
See ?mean and note the na.rm= argument:
aggregate(frame[-1], frame[1:2], mean, na.rm = TRUE)
On 10/1/06, Frank [EMAIL PROTECTED] wrote:
Dear r-help reader,
I have some problems with the aggregate function.
My datframe looks like
frame
Day Time V1 V2
1 M0 3 NA
2 M0 4
?
- Original Message -
From: John Kane [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: R R-help r-help@stat.math.ethz.ch
Sent: Monday, August 21, 2006 6:59 PM
Subject: Re: [R] aggregate example : where is the
state.region variable?
--- Gabor Grothendieck [EMAIL PROTECTED
Gabor == Gabor Grothendieck [EMAIL PROTECTED]
on Mon, 21 Aug 2006 21:03:49 -0400 writes:
Gabor It is worthwhile to note that what is being
Gabor illustrated here is aggregating a numeric matrix by a
Gabor factor using the aggregate.default method and, of
Gabor course, a
there is
no factor in the dataset but why there is not one and
why a call to another dataset is totally opaque.
The reason is purely historical. The state dataset is about
10 years older than the data.frame concept. At the time the
state.* variables were constructed it was not possible to
--- Richard M. Heiberger [EMAIL PROTECTED] wrote:
there is
no factor in the dataset but why there is not one
and
why a call to another dataset is totally opaque.
The reason is purely historical. The state dataset
is about
10 years older than the data.frame concept. At the
time
I was looking ?aggregate and ran the first example
aggregate(state.x77, list(Region = state.region),
mean)
The variables in state.x77 appear to be :
state.x77
Population Income Illiteracy Life Exp Murder HS Grad
Frost Area
Where is the state.region variable coming from?
Its not part of state.x77. Its a completely separate variable.
Try ls(package:datasets) and notice its in the list
or try ?state.region and note that its a variable in datasets.
On 8/21/06, John Kane [EMAIL PROTECTED] wrote:
I was looking ?aggregate and ran the first example
On Mon, 21 Aug 2006, John Kane wrote:
I was looking ?aggregate and ran the first example
aggregate(state.x77, list(Region = state.region),
mean)
The variables in state.x77 appear to be :
state.x77
Population Income Illiteracy Life Exp Murder HS Grad
Frost Area
Where is the
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Its not part of state.x77. Its a completely
separate variable.
Try ls(package:datasets) and notice its in the
list
or try ?state.region and note that its a variable in
datasets.
Thanks. I was wondering if it was going something like
that.
, August 21, 2006 6:59 PM
Subject: Re: [R] aggregate example : where is the state.region variable?
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Its not part of state.x77. Its a completely
separate variable.
Try ls(package:datasets) and notice its in the
list
or try ?state.region
It is worthwhile to note that what is being illustrated here is aggregating a
numeric matrix by a factor using the aggregate.default method and, of course,
a factor can't be part of a numeric matrix.
Of course, that is not say that the examples could not be improved in
terms of clarity,
Hi, everyone,
I have a data.frame named eva like this:
IND PARTNO VC1 EO1 EO2 EO3 EO4 EO5
114 114001 2 5 4 4 5 4
114 114001 2 4 4 4 4 4
114 114001 2 4 NA NA NA NA
112 112002 2 3 3 6 2 6
112 112002 2 1 1 3 4 4
112 112003 2 6 6 6 5
Hi Wei-Wei,
try this:
eva.agg - aggregate(x = list(
VC1=eva$VC1,
EO1=eva$EO1,
EO2=eva$EO2,
EO3=eva$EO3,
EO4=eva$EO4,
EO5=eva$EO5
),
Hi Andrew,
Thank you very much! It works so well than I can expect.
All the best,
Wei-Wei
2006/6/30, Andrew Robinson [EMAIL PROTECTED]:
Hi Wei-Wei,
try this:
eva.agg - aggregate(x = list(
VC1=eva$VC1,
EO1=eva$EO1,
Hello,
I have a data set with a grouping variable (TRIPID) and several other
variables. TRIPID is repeated in some areas and I would like to use a
function like aggregate to sum the variable UNITS according to TRIPID.
However I would also like to retain the other variables as they are in
the
Suppose we want to sum C over levels of A and that B is constant
within levels of A. Then:
DF - data.frame(A = gl(2,2), B = gl(2,2), C = 1:4) # test data
do.call(rbind, by(DF, DF$A, function(x) replace(x[1,], C, sum(x$C
On 5/3/06, Guenther, Cameron [EMAIL PROTECTED] wrote:
Hello,
I
Nice trick, thx...
Stéphane.
On Wed, 2006-03-29 at 11:17 -0500, jim holtman wrote:
try 'by':
x
S_id AF_Class count... R_gc_percent S_length
5 82644971 30 0.4835678
6 826449737 0.4835678
8 82645541 31
Dear R users,
I have some trouble with the aggregate function. Here are my data
daf
S_id AF_Class count... R_gc_percent S_length
5 82644971 30 0.4835678
6 826449737 0.4835678
8 82645541 31 0.5138894
9
try 'by':
x
S_id AF_Class count... R_gc_percent S_length
5 82644971 30 0.4835678
6 826449737 0.4835678
8 82645541 31 0.5138894
9 82645542 11 0.5138894
10 826455431
Dear Colleagues,
does anybody know how to aggregate a data.frame using different functions for
different columns?
Sincerely
___
Markus Preisetanz
Consultant
Client Vela GmbH
Albert-Roßhaupter-Str. 32
81369 München
fon: +49 (0) 89 742 17-113
fax:
you can use mapply()...
z - as.data.frame(matrix(1:3,3,3,T))
mapply(function(x,y) x(y), c(sum,prod,sum), z)
Markus Preisetanz a écrit :
Dear Colleagues,
does anybody know how to aggregate a data.frame using different functions for
different columns?
Sincerely
___
Hi
have you tried
?aggregate
eg df1-aggregate(mydata, list(mean1=x1,mean2=x2),mean)
-
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
Thanks Peter!
I had a feeling that there must be a simpler, better, more elegant
solution.
/Hans
Peter Dalgaard wrote:
hadley wickham [EMAIL PROTECTED] writes:
I faced a similar problem. Here's what I did
tmp -
Dear all,
I'm wanting to do a series of comparisons among 4 categorical variables:
a - aggregate(y, list(var1, var2, var3, var4), sum)
This gets me a very nice 2-dimensional data frame with one column per
variable, BUT, as help for aggregate says, empty subsets are
removed. I don't see in
I faced a similar problem. Here's what I did
tmp -
data.frame(A=sample(LETTERS[1:5],10,replace=T),B=sample(letters[1:5],10,replace=T),C=rnorm(10))
tmp1 - with(tmp,aggregate(C,list(A=A,B=B),sum))
tmp2 - expand.grid(A=sort(unique(tmp$A)),B=sort(unique(tmp$B)))
merge(tmp2,tmp1,all.x=T)
At least
I faced a similar problem. Here's what I did
tmp -
data.frame(A=sample(LETTERS[1:5],10,replace=T),B=sample(letters[1:5],10,replace=T),C=rnorm(10))
tmp1 - with(tmp,aggregate(C,list(A=A,B=B),sum))
tmp2 - expand.grid(A=sort(unique(tmp$A)),B=sort(unique(tmp$B)))
merge(tmp2,tmp1,all.x=T)
At
Thanks, Phil! I've literally spent two hours on my own trying to find
something that does exactly that. Thanks for another pair of functions
added to my (slowly!) growing R vocabulary.
-jlb
Phil Spector wrote:
Joseph -
I'm sure there are clearer and more efficient ways to do it, but
hadley wickham [EMAIL PROTECTED] writes:
I faced a similar problem. Here's what I did
tmp -
data.frame(A=sample(LETTERS[1:5],10,replace=T),B=sample(letters[1:5],10,replace=T),C=rnorm(10))
tmp1 - with(tmp,aggregate(C,list(A=A,B=B),sum))
tmp2 -
Hi,
aggregate() does not preserve the order of levels for
ordered factors, e.g.,
levs - c(Low, Med, Hi)
d - data.frame(x = 1:30, fac = ordered(rep(levs, 10), levels = levs))
out - aggregate(d[,x], by = list(fac=d$f), FUN = mean)
cat(Original ordered levels:, levels(d$fac), \n)
Hi,
Yesterday, I have analysed data with 16 rows and 10 columns.
Aggregation would be impossible with a data frame format, but when converting
it to a matrix with *numeric* entries (check, if the variables are of class
numeric!) the computation needs only 7 seconds on a Pentium III. I´m
Here is the way that I would do it. Using 'lapply' to process the list and
create a matrix; take less than 1 second:
dat - data.frame(D=sample(32000:33000, 33000, T),
+ Fid=sample(1:10,33000,T), A=sample(1:5,33000,T))
system.time({
+ result - lapply(split(seq(nrow(dat)), dat$D), function(.d){ #
Many thanks for all your answers. Converting to a matrix didn't help,
I tried with Hmisc but didn't get anywhere (different summary
functions, multiple levels).
2005/10/14, jim holtman [EMAIL PROTECTED]:
Here is the way that I would do it. Using 'lapply' to process the list and
create a
Hi,
I use the code below to aggregate / cnt my test data. It works fine,
but the problem is with my real data (33'000 rows) where the function
is really slow (nothing happened in half an hour).
Does anybody know of other functions that I could use?
Thanks,
Hans-Peter
--
dat -
Convert dat to a matrix and see if working with the
matrix instead of a data frame speeds things up
enough.
On 10/13/05, Hans-Peter [EMAIL PROTECTED] wrote:
Hi,
I use the code below to aggregate / cnt my test data. It works fine,
but the problem is with my real data (33'000 rows) where the
Gabor Grothendieck wrote:
Convert dat to a matrix and see if working with the
matrix instead of a data frame speeds things up
enough.
In the Hmisc package the asNumericMatrix and matrix2dataFrame functions
facilite this.
Also look at the summarize and mApply functions in Hmisc, which can be
How can I aggregate this data.frame to list the min and max date for
each unique id?
From this :
r = data.frame(id=rep(seq(1:3), 3), date= as.Date(c(rep(2005-08-25,3),
rep(2005-08-26,3), rep(2005-08-29, 3)), %Y-%m-%d))
r
id date
1 2005-08-25
2 2005-08-25
3 2005-08-25
1
://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Omar Lakkis [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Tuesday, August 30, 2005 4:36 PM
Subject: [R] aggregate
How can I aggregate this data.frame to list the min and max date for
each unique id?
From this :
r
Dear all:
Here is my problem:
Example data:
dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17))
If I wanted to aggregate each level of column dat$x I
could use:
aggregate(dat$y,list(x=dat$x),sum)
But I just want to aggregate two levels (?c? and ?d?)
to obtain a new level ?e?
I am expecting
On 6/17/05, alex diaz [EMAIL PROTECTED] wrote:
Dear all:
Here is my problem:
Example data:
dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17))
If I wanted to aggregate each level of column dat$x I
could use:
aggregate(dat$y,list(x=dat$x),sum)
But I just want to aggregate two levels (c
Dear All,
I have tried to calculate tree mean growth but I think the structure I used
below (growthresumo) is not the most elegant, even though it worked. The only
problem I had in this first part was that I cannot use 'summary', just 'mean'
(sorry but 'R' is pretty new for me).
growthresumo
Dear Paulo,
On May 25, 2005, at 8:01 PM, Paulo Brando wrote:
Dear All,
I have tried to calculate tree mean growth but I think the structure I
used below (growthresumo) is not the most elegant, even though it
worked. The only problem I had in this first part was that I cannot
use
I have a data frame of daily open, high, low and settle prices. How
can I aggregate this data weekly?
The data frame has five columns, the first is the date column and the
rest are the prices.
__
R-help@stat.math.ethz.ch mailing list
: Wednesday, May 11, 2005 3:45 PM
To: r-help@stat.math.ethz.ch
Subject: [R] aggregate
I have a data frame of daily open, high, low and settle prices. How
can I aggregate this data weekly?
The data frame has five columns, the first is the date column and the
rest are the prices
$c),list(wk),mean)
colnames(aggr) - etc
-Original Message-
From: Omar Lakkis [mailto:[EMAIL PROTECTED]
Sent: Wednesday, May 11, 2005 3:45 PM
To: r-help@stat.math.ethz.ch
Subject: [R] aggregate
I have a data frame of daily open, high, low and settle prices. How
can I aggregate
hello,
Does anybody know how to aggregate a lag series ?
when I try to use aggregate I get the following message
try-ts(1:100,start=c(1985,1),freq=12)
aggregate(try,4,mean,na.rm=T)
Qtr1 Qtr2 Qtr3 Qtr4
1985258 11
1986 14 17 20 23
1987 26 29 32 35
1988 38 41
.
Example:
R x - ts(1:20,start=c(1990,1),freq=12)
R aggregate(window(x, start = c(1990, 1), end = c(1991, 9),
extend = TRUE), 4, mean, na.rm = TRUE)
Qtr1 Qtr2 Qtr3 Qtr4
1990 2.0 5.0 8.0 11.0
1991 14.0 17.0 19.5
R aggregate(window(lag(x, k = -1), start = c(1990, 1),
end = c(1991
Dear all
I use aggregate with variables of type numeric and dates. For type numeric
functions, such as sum() are very fast, but similar simple functions, such
as min() are much slower for the variables of type 'dates'. The difference
gets bigger the larger the 'id' var is - but see this sample
On 4/15/05, Christoph Lehmann [EMAIL PROTECTED] wrote:
Dear all
I use aggregate with variables of type numeric and dates. For type numeric
functions, such as sum() are very fast, but similar simple functions, such
as min() are much slower for the variables of type 'dates'. The difference
gets
On 4/15/05, Christoph Lehmann [EMAIL PROTECTED] wrote:
Dear all
I use aggregate with variables of type numeric and dates. For type numeric
functions, such as sum() are very fast, but similar simple functions, such
as min() are much slower for the variables of type 'dates'. The difference
gets
R-folks,
Is there a function, like aggregate, that allows users to bin values?
I've got to break down a data frame into classes of 5cm (or something like
it), and I only know how to do it using code like,
signif - symnum( stems$dbh,
corr = FALSE,
na = FALSE,
On Thu, 2005-03-31 at 09:17 -0800, Jeff D. Hamann wrote:
R-folks,
Is there a function, like aggregate, that allows users to bin values?
I've got to break down a data frame into classes of 5cm (or something like
it), and I only know how to do it using code like,
signif - symnum(
On 6 Jan 2005 at 16:55, Karla Meurk wrote:
Hi, some time ago I asked R-help about aggregating data as a result I
was able to put together some code which includes the line
rain.ag - aggregate(newdata, list(hod6=cut(mindata,6 hours)), mean,
na.rm=T)
I also want to aggregate daily, and
Hi, some time ago I asked R-help about aggregating data as a result I
was able to put together some code which includes the line
rain.ag - aggregate(newdata, list(hod6=cut(mindata,6 hours)), mean,
na.rm=T)
I also want to aggregate daily, and 30 minutely etc.
My question is why is it that I get
Karla Meurk ksm32 at student.canterbury.ac.nz writes:
:
: Hi, some time ago I asked R-help about aggregating data as a result I
: was able to put together some code which includes the line
:
: rain.ag - aggregate(newdata, list(hod6=cut(mindata,6 hours)), mean,
: na.rm=T)
:
: I also want to
I am trying to use the function aggregate with the median function but I
get the following error:
Error in FUN(X[[1]], ...) : Argument INDEX
When I replace median by mean, it works perfectly
Can someone tell me where the problem comes from?
Thx
I am running R 2.0.0 on SunOS 5.9
--
Philippe Hupé
On 21 Dec 2004 at 12:46, Philippe Hup wrote:
I am trying to use the function aggregate with the median function but
I get the following error:
Error in FUN(X[[1]], ...) : Argument INDEX
When I replace median by mean, it works perfectly
Hi Philippe
I suppose that you have some typo in
Hi all,
I have the folowing frame(there are more columns than shown),
1 2 34 5
Year Total TusWhi Norw
1994 1.00 1830 0 355
1995 1.00 0 00
1995 1.00 0
Hi,
# x ... your frame
attach(x)
sum(Total[Year==1997 Tus 0])
I hope this helps
Best,
Matthias
-Ursprüngliche Nachricht-
Von: Luis Rideau Cruz [mailto:[EMAIL PROTECTED]
Gesendet: Montag, 26. Juli 2004 14:52
An: [EMAIL PROTECTED]
Betreff: [R] aggregate function
Hi all,
I
Hi,
# x ... your frame
attach(x)
sum(Total[Year==1997 Tus 0])
I hope this helps
Best,
Matthias Templ
-Ursprüngliche Nachricht-
Von: Luis Rideau Cruz [mailto:[EMAIL PROTECTED]
Gesendet: Montag, 26. Juli 2004 14:52
An: [EMAIL PROTECTED]
Betreff: [R] aggregate function
I would try something like:
lapply(frame[3:5], function(i) tapply(frame$Total[i0], frame$Year[i0],
sum))
$Tus
1994 1995 1997 1999
1121
$Whi
1995 1997 1999
1.00 4.00 2.04
$Norw
1994 1995 1997 1998 1999
11512
HTH,
Andy
From: Luis Rideau Cruz
Hi
[Sorry if this gets posted twice. I have been having some
problems with gmane posting.]
We can use rowsum like this:
rowsum(frame$Total * (frame[,3:5]0), frame$Year)
Tus Whi Norw
1994 1 0.00 1
1995 1 1.00 1
1997 2 4.00 5
1998 0 0.00 1
Hi,
I have a set of data like the following:
[,1] [,2]
[1,] 102
[2,]70
[3,]10
[4,]10
[5,] 150
[6,] 174
[7,]40
[8,] 198
[9,] 102
[10,] 195
I'd like to aggregate it in order to obtain the frequency (the number of
occurences)
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