Displaying 20 results from an estimated 9000 matches similar to: "remove component from list or data frame"
2007 Jul 18
1
Neuman-Keuls
hello,
I have programmed this function to calculate the Neuman-Keuls test but I have a problem the function return an empty list and I don't know why.
summary(fm1)
E <- sqrt((summary(fm1)[[1]]["Residuals","Mean Sq"])/length(LR))
lst <- list()
lst1 <- list()
lst2 <- list()
NK <- function (x) {
if (length(x) == 2) {
Tstudent <- t.test(subset(exple,
2013 Aug 25
1
Capturing the whole output using R
Hi,
May be this helps:
#Creating some dummy data.?
set.seed(24)
lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE)))
set.seed(49)
lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE)))
??Find_Max_CCF()
#No vignettes or demos or help files found with alias or concept or
#title matching ?Find_Max_CCF? using regular expression matching.
Found a function with the same
2008 Oct 01
3
lapply where each list object has multiple parts
Hi. I have a list where each object in the list has multiple parts. I'd
like to take the mean of just one part of each object. Is it possible to do
this with lapply? If not, can you recommend another function? Thanks.
eric
> x1 <- c(0,1,2,3)
> x2 <- c(7,8)
> x3 <- c(2,6,6,8)
> x4 <- c(4,8)
>
> Lst1 <- list(label1 = x1,label2 = x2)
> Lst2 <-
2013 Apr 13
2
Comparison of Date format
Hi,
?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.?
DataA<- read.table(text="
ID,Status,Date1,Date2 ??? ??? ??????
1,A,3-Feb-01,15-May-01 ??? ???
1,B,15-May-01,16-May-01 ??? ???
1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ???
1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ???
1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???
2006 Jul 26
1
.Call question
Writing R Ext says to treat R objects that are arguments to .Call as
read only (i.e. don't modify).
I have a long list of lists that and I want to avoid the overhead of a
copy in my C code. I would just like to modify some of the elements
of list by replacing them with elements of exactly the same size/type.
below is an example of the essence of the problem. This seems to work.
Is this
2013 Jun 08
0
data
Hi,
Try this:
final3New<-read.table(file="real_data_cecilia.txt",sep="\t")
dim(final3New)
#[1] 5369??? 5
#Inside the split within split, dummy==1 for the first row.? For lists that have many rows, I selected the row with dummy==0 (from the rest) using the #condition that the absolute difference between the dimensions of those rows and the first row dimension was minimum
2013 Jun 07
4
matched samples, dataframe, panel data
I R-helpers
#I have a data panel of thousands of firms, by year and industry and
#one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not
#and another variable the represents the firm dimension (total assets in thousand of euros)
#I need to create two separated samples with the same number os firms where
#one firm in the first have a corresponding
2002 Nov 28
4
Mime-Version: 1.0
I am using expression() to incorporate text into graphics. To create a
superscript, I use the '^' character. Can someone please tell me the
character to use to create a subscript?
Thank you
-----------------------------------------------------
Christine Donnelly
Statistical Consulting Unit
The Graduate School
John Dedman Mathematical Sciences Building (Bldg 27)
Australian National
2004 Jul 06
2
lme: extract variance estimate
For a Monte Carlo study I need to extract from an lme model
the estimated standard deviation of a random effect
and store it in a vector. If I do a print() or summary()
on the model, the number I need is displayed in the Console
[it's the 0.1590195 in the output below]
>print(fit)
>Linear mixed-effects model fit by maximum likelihood
> Data: datag2
> Log-likelihood:
2013 Jan 11
3
Access comonents in lists of lists
Dear R users,I have a list of equally structured lists, how can I access e.g.
all 2nd compontents in those sub-lists?An example:lst <-
list(rep(list(1:3),3), rep(list(4:6),3))> lst[[1]][[1]][[1]][1] 1 2
3[[1]][[2]][1] 1 2 3[[1]][[3]][1] 1 2 3[[2]][[2]][[1]][1] 4 5 6[[2]][[2]][1]
4 5 6[[2]][[3]][1] 4 5 6What I want to get are all second sub-lists, in this
case:[[1]][[2]][1] 1 2
2013 Apr 10
6
means in tables
Hi.
I have 2 tables, with same dimensions (8000 x 5). Something like:
tab1:
V1 V2 V3 V4 V5
14.23 1.71 2.43 15.6 127
13.20 1.78 2.14 11.2 100
13.16 2.36 2.67 18.6 101
14.37 1.95 2.50 16.8 113
13.24 2.59 2.87 21.0 118
tab2:
V1 V2 V3 V4 V5
1.23 1.1 2.3 1.6 17
1.20 1.8 2.4 1.2 10
1.16 2.6 2.7 1.6 11
1.37 1.5 2.0 1.8 13
1.24 2.9 2.7 2.0 18
I need generate a table of averages, the
2006 Mar 07
4
POSIX time zone codes
The manual entry for as.POSIX says this about time zone codes...
Usage
as.POSIXct(x, tz = "")
tz
A timezone specification to be used for the conversion...
but it fails to mention what these "specifications" are. So far, I
have tried...
as.POSIX(x, tz="UTC") ... works, gives UTC times
as.POSIX(x, tz="UTC") ... works, gives EST times
as.POSIX(x,
2013 Sep 05
2
binary symmetric matrix combination
Hi,
May be this helps:
m1<- as.matrix(read.table(text="
y1 g24
y1 0 1
g24 1 0
",sep="",header=TRUE))
m2<-as.matrix(read.table(text="y1 c1 c2 l17
?y1 0 1 1 1
?c1 1 0 1 1
?c2 1 1 0 1
?l17 1 1 1 0",sep="",header=TRUE))
m3<- as.matrix(read.table(text="y1 h4??? s2???? s30
?y1 0 1 1 1
?h4 1 0 1 1
?s2 1 1 0 1
?s30 1 1 1
2013 Jun 04
0
choose the lines2
Hi,
May be this helps:
dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t")
dat.bru<- dat1[!is.na(dat1$evnmt_brutal),]
fun2<- function(dat){??
????? lst1<- split(dat,dat$patient_id)
??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,])
??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])
2013 Sep 20
3
search species with all absence in a presence-absence matrix
Dear list
I have a matrix composed of islandID as rows and speciesID as columns.
IslandID: Island A, B, C….O (15 islands in total)
SpeciesID: D0001, D0002, D0003….D0100 (100 species in total)
The cell of the matrix describes presence (1) or absence (0) of the species
in an island.
Now I would like to search the species with absence (0)
in all the islands (Island A to Island O.)
2011 Oct 14
3
Split a list
I have a list of dataframes i.e. each list element is a dataframe with three columns and differing number of rows. The third column takes on only two values. I wish to split the list into two sublists based on the value of the third column of the list element.
Second issue with lists as well. I would like to reduce each of the sublist based on the range of the second column, i.e. if the range of
2013 Apr 18
6
count each answer category in each column
Hey,
Is it possible that R can calculate each options under each column and
return a summary table?
Suppose I have a table like this:
Gender Age Rate
Female 0-10 Good
Male 0-10 Good
Female 11-20 Bad
Male 11-20 Bad
Male >20 N/A
I want to have a summary table including the information that how many
answers in each category, sth like this:
X
2013 Nov 08
2
making chains from pairs
Hello,
having a data frame like test with pairs of characters I would like to
create chains. For instance from the pairs A/B and B/I you get the vector A
B I. It is like jumping from one pair to the next related pair. So for my
example test you should get:
A B F G H I
C F I K
D L M N O P
> test
V1 V2
1 A B
2 A F
3 A G
4 A H
5 B F
6 B I
7 C F
8 C I
9 C K
10 D L
2009 Sep 29
1
Comparing vectors from lists
Hi guys,
I still did not solve my problem properly! I have to compare the values of two lists of 250 numbers as a result of using the ?by function!
List1 of 250
$ 0 : num [1:28] 22 11 31...
$ 1 : num [1:15] 12 14 9 ...
..
..
..
- attr(*, "dim")= int 250
- attr(*, "dimnames")=List of 1
List2 of 250
$ 0 : num [1:24] 20 12 22...
$ 1 : num [1:17] 11 12 19 ...
..
..
2005 Oct 11
2
non-zero sequence of numbers
Can anyone think of a way to create a pretty() sequence that excludes
zero? Or a way to remove the zero from a sequence after using pretty()?
Thanks,
- Jason
Jason Horn
Boston University Department of Biology
5 Cumington Street Boston, MA 02215
jhorn@bu.edu
office: 617 353 6987
cell: 401 588 2766
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