x) > 1) x)
> l2 <- do.call(rbind, l2)
> l2
>
> # Create a new variable
> l3 <- lapply(sl, function(x) cbind(x, NEW_VARIABLE=seq_len(nrow(x)))**)
> l3 <- do.call(rbind, l3)
> l3
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 25-07-2012 01:59, AC
Hi,
I am trying to reshape data from a long to wide format but have a specific
task that I cannot get to output properly.
# SAMPLE DATA;
id <- c(1,2,2,3,3,3)
time <-c(0,0,5, 0, 2, 10)
x <- rnorm(length(id))
long <- data.frame(id,time,x)
# To reshape, I would like to exclude 'id' values that have
Hi,
Is there an easy way to remove dataframe rows without duplicated values of
a specified column ('id')? e.g.,
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4), value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
2 1 6 4
3 1 7 3
4 2 4 3
5
Thats a good start but I am looking for more. Specifically (from previous
post):
4x4 matrix (based on the maximum number of factor levels in 'o') for each
'id' as a list.
Each matrix would have v in the diagonal and r*s*s in the off-diagonal
(e.g.,
.5*2.0814*.1095 = 0.1139), where
r is constant (
Hi,
I have this sample data below and would like to create a list of matricies.
setseed(1254)
id <- c(1,1,1,1 ,2,2,2)
o <- as.factor(c(1:4, 1, 3, 4))
r <- rep(.5, 7)
v <- rnorm(7)
s <- rnorm(7)
dat <-data.frame(id, o, r, v, s)
dat
#> dat
# id o r v s
# 1 1 0.5 0.7024631 2
x, 1))
> id value
> 1 1 5
> 2 2 4
>
> The formula version of aggregate() requires R-2.11.0 +
>
> Dennis
>
> On Sun, Sep 25, 2011 at 1:22 PM, AC Del Re wrote:
> > Hi,
> >
> > I am trying to select the first row of a variable with data in
> l
Hi,
I am trying to select the first row of a variable with data in long-format,
e.g.,
# sample data
id <- c(1,1,1,2,2)
value <- c(5,6,7,4,5)
dat <- data.frame(id, value)
dat
How can I select/subset the first 'value' for each unique 'id'?
Thanks,
AC
[[alternative HTML version deleted]
Hi,
I am wanting to simulate data where a percentage of the data has
multiple duplicated id variables (with unique values of another factor
variable for the dupicated id variables). Im having trouble figuring
out an efficent way to do so.
For example, consider this mock output [Note: Although the
Wonderful, David! Thank you much for your help on this.
AC
On Fri, Jul 9, 2010 at 6:49 PM, David Winsemius wrote:
>
> On Jul 9, 2010, at 7:19 PM, Erik Iverson wrote:
>
> AC Del Re wrote:
>>
>>> Hi David,
>>> I don't have the function in order yet
us wrote:
>
> On Jul 9, 2010, at 3:51 PM, AC Del Re wrote:
>
> Hi All,
>>
>> I am interested in printing column names without quotes and am struggling
>> to
>> do it properly. The tough part is that I am interested in using these
>> column
>> nam
Hi All,
I am interested in printing column names without quotes and am struggling to
do it properly. The tough part is that I am interested in using these column
names for a function within a function (e.g., lm() within a wrapper
function). Therefore, cat() doesnt seem appropriate and print() is n
Hi All,
>
> Attempt #2. any help is much appreciated!:
> I am reading through section 2.6 (Mathematics) of the "Writing R
> Extensions" manuscript and am wondering where I can find more
> examples/documentation on the \deqn{ } function. I would like to learn how
> to display equations using th
Hi All,
I am reading through section 2.6 (Mathematics) of the "Writing R
Extensions" manuscript and am wondering where I can find more
examples/documentation on the \deqn{ } function. I would like to learn how
to display equations using this function but am not sure how to go about
doing it. The
;- facts(testdata, "b", "newb")
> > str(test2)
> 'data.frame': 3 obs. of 2 variables:
> $ a : num 1 2 3
> $ newb: Factor w/ 2 levels "1","2": 1 1 2
>
> Sarah
>
On Wed, Mar 17, 2010 at 2:23 PM, Henrique Dallazuanna wrote
Hi All,
Im interested in creating a function that will convert a variable within a
data.frame to a factor while retaining the original name (yes, I know that I
can just: var <-factor(var) but I need it as a function for other
purposes). e.g.:
# this was an attempt but fails.
facts <- function(
=st[i]],
n.2 = n.2[id==st[i]], cor)[2])
out$n.1[i] <- round(mean(n.1[id==st[i]]),0)
out$n.2[i] <- round(mean(n.2[id==st[i]]),0)
}
return(out)
}
AC
On Thu, Mar 4, 2010 at 9:35 AM, AC Del Re wrote:
> Hi All,
>
> I am using a specialized aggre
Hi All,
I am using a specialized aggregation function to reduce a dataset with
multiple rows per id down to 1 row per id. My function work perfect when
there are >1 id but alters the 'var.g' in undesirable ways when this
condition is not met, Therefore, I have been trying ifthen() statements to
ke
Hi All,
Is there an easy way to reduce a data.frame to 1 'id' per row while keeping
information from the other rows of that same variable, if applicable? e.g.:
# data
multi[1:15,]
id r n wi wi.tau z k alliance a.rater eml
treatment outcome o.rater german
1 100 0.2
(.data), 1),]))
> id mod1 r
> 1 1 2 0.295
> 4 4 1 0.640
> 5 5 2 0.490
> 6 6 3 -0.040
> 7 7 1 0.490
> 8 8 2 0.330
> 9 9 3 0.580
> 10 10 1 0.180
> 11 11 2 0.600
> 12 12 3 0.210
>>
>>
>
>
gt;> id mod1 r
>> 1 1 1 0.98
>> 2 4 1 0.64
>> 3 7 1 0.49
>> 4 10 1 0.18
>> 6 5 2 0.49
>> 7 8 2 0.33
>> 8 11 2 0.60
>> 9 6 3 -0.04
>> 10 9 3 0.58
>> 11 12 3 0.21
>>
>
1 1 0.980
> 2 4 1 0.640
> 3 7 1 0.490
> 4 10 1 0.180
> 5 1 2 0.295
> 6 5 2 0.490
> 7 8 2 0.330
> 8 11 2 0.600
> 9 6 3 -0.040
> 10 9 3 0.580
> 11 12 3 0.210
>>
>
>
> On Sat, Feb 20, 2010 at 6:54 PM,
Hi All,
I am interested in aggregating a data frame based on 2
categories--mean effect size (r) for each 'id's' 'mod1'. The
'with' function works well when aggregating on one category (e.g.,
based on 'id' below) but doesnt work if I try 2 categories. How can
this be accomplished?
# sample data
i
Hi All,
I have a function that is used with data frames having multiple id's
per row and it aggregates the data down to 1 id per row. It also
randomly selects one of the within-id values of a variable (mod),
which often differ within-id. Assume this data frame (below) is much
larger and I want to
Perfect!
Thank you, Dimitris.
AC
On Mon, Feb 15, 2010 at 1:45 PM, Dimitris Rizopoulos
wrote:
> try this:
>
> MRfit <- function (...) {
> models <- list(...)
>> do.call(anova, models)
>> }
>>
>>
>> I hope it helps.
>>
>>
Hi All,
I am interested in creating a function that will take x number of lm
objects and automate the comparison of each model (using anova). Here
is a simple example (the actual function will involve more than what
Im presenting but is irrelevant for the example):
# sample data:
id<-rep(1:20)
Hi All,
I have recently created an Rcmdr plugin package and it passed all the
checks and was uploaded to CRAN. I then downloaded it from CRAN and
tried running it from my local R program and received this error:
...
Error in f(libname, pkgname) : could not find function "getRcmdr"
Error in librar
3 0.21 wai
> > > > > 6 3 0.31 wai
> > > > >
> > > > > # I would like to reduce the entire data.frame like this:
> > >
> > > E.g. aggregate
> > >
> > > a
egards
> Petr
>
> > >
> > > id es mod1 mod2
> > >
> > > 1 .30 2wai
> > > 2 .15 4other
> > > 3 .20 1 itas
> > >
> > > # If possible, I would also like the option of this (collapsing on
> 2 1 2 0.30
> 3 2 4 0.15
>
> ddply() is a little less painless and sorts the output for you
> automatically.
>
> HTH,
> Dennis
>
> On Wed, Jan 27, 2010 at 7:34 PM, AC Del Re wrote:
>
>> Hi All,
>>
>> I'm cond
wai
2 .15 4other
3 .20 1 itas
# If possible, I would also like the option of this (collapsing on id and
mod2):
id es mod1 mod2
1 .30 2wai
2 0.1 4 other
2 0.2 4calpas
3 0.1 1 itas
3 0.251 wai
Any
psing on id and
> mod2):
>
> id es mod1 mod2
> 1 .30 2wai
> 2 0.1 4 other
> 2 0.2 4calpas
> 3 0.1 1 itas
> 3 0.251 wai
>
> Any help is much appreciated!
>
> AC Del Re
>
[[alternative
Thanks for your help on this Hadley and David!
Dennis Murphy also had a good solution (changing list(data.out2[-1])...etc
to names(data.out2[-1]}...):
> data.out3 <- reshape(data.out2, direction = 'long', varying =
names(data.out2[-1]),
+ idvar = 'id')
> data.out4 <- split(data.out3, da
Hi All,
I am wanting to convert a data.frame from a wide format to a long format
(with >1 variable) and am having difficulties. Any help is appreciated!
#current wide format
> head(data.out2)
id rater.1 n.1 rater.2 n.2 rater.3 n.3 rater.4 n.4
11 11 0.118 79NA NA
>
> Wonderful, Phil! Your suggestion produced the exact format I desired.
> Thank you kindly,
>
AC Del Re
>
>
> On Sun, Oct 4, 2009 at 2:03 PM, Phil Spector wrote:
>
>> AC -
>> The easiest way I can think of is to create a time variable,
>> so res
Dear R Community,
I am attempting to transpose a dataset from rows to columns but am stuck. I
have tried using reshape() with little luck, possibly due to the categorical
nature of the data. For example:
id<-c(1,2,2,3,3,3)
author<-c("j","k","k","l","l","l")
tmt<-c("cbt","act","dbt","act","act","cb
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