Re: [R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work
Hello R-users The help I received from Petr helped me created this solution to my problems. t1-with(mydata ,aggregate(mydata$Y, list(mydata$time,mydata$treatment, mydata$expREP, mydata$techREP) , median, na.rm=T)) ### find median by factors colnames(t1)-c(time,treatment,expREP,techREP,Y50) ### column name ## newdata-merge(mydata, t1, by.x= names(mydata)[2:5], by.y=names(t1)[1:4], all=T) Thank you, Alan ### Message: 97 Date: Thu, 08 Mar 2007 08:00:53 +0100 From: Petr Pikal [EMAIL PROTECTED] Subject: Re: [R] how to apply functions to unbalanced data in long format by factors..cant get by or aggregate to work To: ALAN SMITH [EMAIL PROTECTED], r-help@stat.math.ethz.ch Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=US-ASCII Hi you can use aggregate to create table of medians with(mydata, aggregate(Y, list(time, tratment, expRep,), median) repeats of unique factors either by rle or aggregate with length function Then you can do replication by norep - rep(your.median, each = your replicates) Regards Petr submitted question abrigded Hello R users, #Example data frame## mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 23, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47), time = structure(as.integer(c(1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2)), .Label = c(120hr, 24hr), class = factor), treatment = structure(as.integer(c(1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 2, 1)), .Label = c(control, trt), class = factor), expREP = structure(as.integer(c(1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 3, 1, 3, 3, 3, 1, 2, 1, 2, 2, 2, 2, 3, 3, 2, 2, 1, 2, 3, 3, 1, 1, 2, 3, 1, 3, 3, 3, 3, 1, 3, 1, 1, 2, 1, 1, 2, 3, 2, 2, 1, 3, 2, 2, 2, 3, 2, 1, 2, 2, 2, 2, 1, 1, 1, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2, 3, 2, 3, 1, 2, 3, 3, 1, 1, 1, 3, 3, 1, 1, 3, 1, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 2, 3, 2, 2, 3, 2, 2, 2, 1, 1, 1, 3, 3, 2, 2, 2, 1, 3, 1, 2, 3, 1, 3, 3, 1, 2, 3, 1, 2, 1, 3, 1, 3, 3, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 3, 3, 3, 1, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 3, 2, 1, 1, 2, 1, 3, 2, 1, 2, 1, 3, 1, 1, 2, 3)), .Label = c(expREP1, expREP2, expREP3), class = factor), techREP = structure(as.integer(c(3, 2, 1, 1, 1, 3, 1, 3, 3, 2, 2, 1, 1, 3, 2, 3, 3, 1, 2, 1, 2, 1, 3, 1, 3, 2, 2, 3, 1, 1, 3, 3, 2, 3, 3, 3, 2, 2, 2, 2, 1, 1, 2, 3, 1, 2, 3, 1, 3, 2, 1, 1, 2, 2, 3, 3, 3, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 1, 3, 2, 2, 1, 2, 1, 3, 2, 2, 1, 1, 3, 3, 3, 1, 3, 1, 3, 2, 2, 2, 1, 2, 1, 3, 1, 3, 2, 1, 3, 1, 3, 2, 3, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 1, 2, 3, 2, 2, 3, 2, 2, 1, 2, 2, 3, 3, 3, 1, 1, 1, 3, 1, 1, 2, 3, 2, 3, 3, 1, 1, 3, 2, 3, 3, 3, 3, 1, 3, 2, 1, 3, 3, 1, 3, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 3, 3, 1, 2, 3, 2, 3, 3, 3, 1, 2, 2, 1, 3, 2, 3, 3, 2, 2, 2, 3, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 2, 2, 3)), .Label = c(techREP1, techREP2, techREP3), class = factor), log2Abun =
Re: [R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work
Hello R-users The help I received from Petr helped me created this solution to my problems. t1-with(mydata ,aggregate(mydata$Y, list(mydata$time,mydata$treatment, mydata$expREP, mydata$techREP) , median, na.rm=T)) ### find median by factors colnames(t1)-c(time,treatment,expREP,techREP,Y50) ### column name ## newdata-merge(mydata, t1, by.x= names(mydata)[2:5], by.y=names(t1)[1:4], all=T) Another way is to use the reshape package, http://had.co.nz/reshape library(reshape) molten - melt(mydata, m=log2Abun) cast(molten, time + treatment + expREP + techREP ~ ., median) # You can also create many other shapes easily: cast(molten, expREP + techREP ~ time + treatment , median) cast(molten, expREP + techREP ~ time + treatment , median, margins=TRUE) Hadley __ R-help@stat.math.ethz.ch 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.
[R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work
Hello R users, Problem...I do not understand how to use aggregate,by, or the appropriate apply to perform a function on data with more than one factor on unbalanced data... I have a data frame in the long format that does not contain balanced data. The ID is a unique identifier corresponding to the experimental unit that will later be examined by ANOVA, T-tests etc. Y is the data generated from the experiment. The factors represent the differences between each sample or run measured. str(mydata) ### sample of table at bottom of email ### 'data.frame': 129982 obs. of 6 variables: $ ID: num 7 7 7 7 7 7 8 8 8 8 ... $ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ... $ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ... $ expREP : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2 2 ... $ techREP : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3 1 3 3 2 ... $ Y : num 14.4 14.1 14.2 13.8 14.1 ... Could someone please help with doing something like the following 1. I would like to find the median for each unique combination of factors using the data in the long format (like finding the median of a single column of data). 2. Create a new column where the median is repeated for the number of rows of the unique factor combination 3. I would like to learn the most efficient way to do this because I want to avoid recreating the table from scratch with many commands like the series below. I will have to perform this operation on many different data sets some, with many more factors then this example. ### help me learn to use an apply or other command that will do the following # m0-mydata$cpdID[mydata$time==24hr mydata$treatment==control mydata$expREP==expREP1 mydata$techREP==techREP1] m1-mydata$Y[mydata$time==24hr mydata$treatment==control mydata$expREP==expREP1 mydata$techREP==techREP1] m2-median(m1) m3-cbind(ID=m0,time=rep(24hr,length(m1)), treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)), techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1))) # I would like to avoid writing the above hundreds of times ## I am able to reshape into wide format and then find the column medians. However restacking the data and regenerating the factors becomes very very messy on data sets with 150 columns. I am able to preform this analysis is SAS easily using BY, but I would like to know how to do it in R. I have tried these commands in a number of different variations with no luck and similar error messages test1-aggregate(mydata[,-1], list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP) ,median, na.rm=T) Error in median.default(X[[1]], ...) : need numeric data ### Y in numeric test1-by(mydata[,-1], list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP) ,median, na.rm=T) Error in median.default(data[x, ], ...) : need numeric data Thanks Alan winXP R 2.4.1 #Example data frame## mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 23, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47), time = structure(as.integer(c(1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2)), .Label = c(120hr, 24hr), class = factor), treatment = structure(as.integer(c(1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2,
Re: [R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work
Here is one way of doing it:
# create the rows for each unique combination
x.split - split(seq(nrow(mydata)), list(mydata$time, mydata$treatment,
+ mydata$expREP, mydata$techREP), drop=TRUE)
# now go through the list of indices and add the median
mydata$Y50 - 0 # add the dummy median column
for (i in x.split){
+ mydata$Y50[i] - median(mydata$Y[i]) # median for each group
+ }
head(mydata,20)
cpdID time treatment expREP techREPY Y50
1 7 120hr control expREP1 techREP3 14.42331 15.74599
2 7 120hr control expREP1 techREP2 14.05282 15.10810
3 7 120hr control expREP1 techREP1 14.22814 14.63248
4 7 120hr trt expREP3 techREP1 13.84921 15.08641
5 7 24hr trt expREP1 techREP1 14.07648 15.17235
6 7 24hr control expREP1 techREP3 14.21394 14.63314
7 8 24hr control expREP1 techREP1 14.33992 14.81328
8 8 120hr trt expREP1 techREP3 14.36256 15.34493
9 8 120hr control expREP2 techREP3 14.01419 15.14270
10 8 120hr control expREP2 techREP2 14.09803 15.10079
11 8 120hr trt expREP1 techREP2 14.31522 15.39152
12 8 120hr trt expREP1 techREP1 14.45288 14.65430
13 8 24hr control expREP2 techREP1 13.95919 14.71188
14 8 24hr trt expREP2 techREP3 14.40640 14.36332
15 8 24hr trt expREP1 techREP2 14.04038 14.42856
16 8 24hr control expREP3 techREP3 14.22859 15.08463
17 8 24hr trt expREP1 techREP3 14.12598 14.53840
18 8 24hr trt expREP3 techREP1 13.59257 14.69984
19 8 24hr trt expREP3 techREP2 13.58308 14.85730
2010 120hr control expREP3 techREP1 13.02808 14.07448
On 3/7/07, ALAN SMITH [EMAIL PROTECTED] wrote:
Hello R users,
Problem...I do not understand how to use aggregate,by, or the
appropriate apply to perform a function on data with more than one
factor on unbalanced data...
I have a data frame in the long format that does not contain balanced
data. The ID is a unique identifier corresponding to the experimental
unit that will later be examined by ANOVA, T-tests etc. Y is the data
generated from the experiment. The factors represent the differences
between each sample or run measured.
str(mydata) ### sample of table at bottom of email ###
'data.frame': 129982 obs. of 6 variables:
$ ID: num 7 7 7 7 7 7 8 8 8 8 ...
$ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ...
$ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ...
$ expREP : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2
2 ...
$ techREP : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3
1 3 3 2 ...
$ Y : num 14.4 14.1 14.2 13.8 14.1 ...
Could someone please help with doing something like the following
1. I would like to find the median for each unique combination of
factors using the data in the long format (like finding the median
of a single column of data).
2. Create a new column where the median is repeated for the number of
rows of the unique factor combination
3. I would like to learn the most efficient way to do this because I
want to avoid recreating the table from scratch with many commands
like the series below. I will have to perform this operation on many
different data sets some, with many more factors then this example.
### help me learn to use an apply or other command that will do the
following #
m0-mydata$cpdID[mydata$time==24hr mydata$treatment==control
mydata$expREP==expREP1 mydata$techREP==techREP1]
m1-mydata$Y[mydata$time==24hr mydata$treatment==control
mydata$expREP==expREP1 mydata$techREP==techREP1]
m2-median(m1)
m3-cbind(ID=m0,time=rep(24hr,length(m1)),
treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)),
techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1)))
# I would like to avoid writing the above hundreds of times ##
I am able to reshape into wide format and then find the column
medians. However restacking the data and regenerating the factors
becomes very very messy on data sets with 150 columns. I am able to
preform this analysis is SAS easily using BY, but I would like to know
how to do it in R.
I have tried these commands in a number of different variations with
no luck and similar error messages
test1-aggregate(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(X[[1]], ...) : need numeric data ### Y in
numeric
test1-by(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(data[x, ], ...) : need numeric data
Thanks
Alan
winXP R 2.4.1
#Example data frame##
mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8,
8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 19, 19, 19,
Re: [R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work
Hi you can use aggregate to create table of medians with(mydata, aggregate(Y, list(time, tratment, expRep,), median) repeats of unique factors either by rle or aggregate with length function Then you can do replication by norep - rep(your.median, each = your replicates) Regards Petr On 7 Mar 2007 at 17:25, ALAN SMITH wrote: Date sent: Wed, 7 Mar 2007 17:25:54 -0600 From: ALAN SMITH [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject:[R] how to apply functions to unbalanced data in long format by factors..cant get by or aggregate to work Hello R users, Problem...I do not understand how to use aggregate,by, or the appropriate apply to perform a function on data with more than one factor on unbalanced data... I have a data frame in the long format that does not contain balanced data. The ID is a unique identifier corresponding to the experimental unit that will later be examined by ANOVA, T-tests etc. Y is the data generated from the experiment. The factors represent the differences between each sample or run measured. str(mydata) ### sample of table at bottom of email ### 'data.frame': 129982 obs. of 6 variables: $ ID: num 7 7 7 7 7 7 8 8 8 8 ... $ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ... $ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ... $ expREP : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2 2 ... $ techREP : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3 1 3 3 2 ... $ Y : num 14.4 14.1 14.2 13.8 14.1 ... Could someone please help with doing something like the following 1. I would like to find the median for each unique combination of factors using the data in the long format (like finding the median of a single column of data). 2. Create a new column where the median is repeated for the number of rows of the unique factor combination 3. I would like to learn the most efficient way to do this because I want to avoid recreating the table from scratch with many commands like the series below. I will have to perform this operation on many different data sets some, with many more factors then this example. ### help me learn to use an apply or other command that will do the following # m0-mydata$cpdID[mydata$time==24hr mydata$treatment==control mydata$expREP==expREP1 mydata$techREP==techREP1] m1-mydata$Y[mydata$time==24hr mydata$treatment==control mydata$expREP==expREP1 mydata$techREP==techREP1] m2-median(m1) m3-cbind(ID=m0,time=rep(24hr,length(m1)), treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)), techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1))) # I would like to avoid writing the above hundreds of times ## I am able to reshape into wide format and then find the column medians. However restacking the data and regenerating the factors becomes very very messy on data sets with 150 columns. I am able to preform this analysis is SAS easily using BY, but I would like to know how to do it in R. I have tried these commands in a number of different variations with no luck and similar error messages test1-aggregate(mydata[,-1], list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP) ,median, na.rm=T) Error in median.default(X[[1]], ...) : need numeric data ### Y in numeric test1-by(mydata[,-1], list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP) ,median, na.rm=T) Error in median.default(data[x, ], ...) : need numeric data Thanks Alan winXP R 2.4.1 #Example data frame## mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 23, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47), time = structure(as.integer(c(1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2,
