similar to: help on data.frames/data.entry

Displaying 20 results from an estimated 6000 matches similar to: "help on data.frames/data.entry"

2012 Dec 13
2
Position available University of Oxford UK
*Postdoctoral Researcher - Bioinformatics/Statistics University of Oxford, UK *An exciting opportunity has arisen for a Postdoctoral Researcher in Bioinformatics/Statistics to join the Department of Oncology at the University of Oxford. The postholder will work under the supervision of Dr Francesca Buffa and Prof Adrian Harris, and will work closely with staff in the Molecular Oncology
2004 Jan 06
1
help on rmeta
Hello I'm trying to plot hazard risk values using the function metaplot with the specifications: > metaplot(HR,SE,W,labels=row.names(lc),xlab="Hazard Ratio",ylab="Covariates", logeffect=TRUE,logticks=FALSE,colors=meta.colors(box="black",lines="dark gray",zero="darkgray"),cex=1.5,cex.lab=1.5,font=3) However, in the plot the x axis
2007 Nov 25
1
Constructin a call of function including permutation of column names - how to escape parentheses?
Dear R-users, I would like to construct a list of arguments for a function in a format function (list(item1=c("A","B"), item2=c("B","C")), item3=...): The individual vectors in the list are permutations of colnames of a dataframe. The trouble is that I am not able to handle escaping of parentheses correctly. I was trying the following: library(gregmisc)
2011 Oct 03
2
Merge two data frames and find common values and non-matching values
Hi, I am trying to find a function to match two data frames of different lengths for one field only. So, for example, df1 is: Name Position location francesca A 75 cristina B 36 And df2 is: location Country 75 UK 56 Austria And I would like to match on "Location" and the output to be something like: Name Position Location Match francesca A 75 1 cristina B 36 0 I have tried with
2012 May 25
2
Collecting results of a test with array
Dear contributors I have tried this experiment: x<-c() for (i in 1:12){ x[i]<-list(cbind(x1[i],x2[i])) #this is a list of 12 couples of time series I am using to perform a test } # that compares them 2 by 2 # ################# #trace statistic test<-data.frame() cval<-array( , dim=c(2,3,12)) for (i in 2:12){ for (k in 1:2){ for (j in 1:3){ result[k,j,i]<-
2011 Jul 27
3
Reorganize(stack data) a dataframe inducing names
Dear Contributors, thanks for collaboration. I am trying to reorganize data frame, that looks like this: n1.Index Date PX_LAST n2.Index Date.1 PX_LAST.1 n3.Index Date.2 PX_LAST.2 1 NA 04/02/07 1.34 NA 04/02/07 1.36 NA 04/02/07 1.33 2 NA 04/09/07 1.34 NA 04/09/07
2012 Jan 20
3
Istalling Ruby 1.9.2
I just installed 1.9.2 on my 2nd machine, but after I installed I checked the version with ruby -v and it sent me 1.8.7 Here''s the terminal output...anyone know what''s wrong (I have RVM installed): Francescas-MacBook-Air:~ fkrihely$ rvm install 1.9.2 Fetching yaml-0.1.4.tar.gz to /Users/fkrihely/.rvm/archives Extracting yaml-0.1.4.tar.gz to /Users/fkrihely/.rvm/src Configuring
2017 Oct 19
0
Select part of character row name in a data frame
(Re-)read the discussion of indexing (both `[` and `[[`) and be sure to get clear on the difference between matrices and data frames in the Introduction to R document that comes with R. There are many ways to create numeric vectors, character vectors, and logical vectors that can then be used as indexes, including the straightforward way: df[ c( "Unique to strat ", "Unique
2012 May 18
3
How to fix indeces in a loop
Dear Contributors, I have an easy question for you which is puzzling me instead. I am running loops similar to the following: for (i in c(100,1000,10000)){ print((mean(i))) #var<-var(rnorm(i,0,1)) } This is what I obtain: [1] 100 [1] 1000 [1] 10000 In this case I ask the software to print out the result, but I would like to store it in an object. I have tried a second loop, because if I
2013 Jan 27
2
Loops
Dear Contributors, I am asking help on the way how to solve a problem related to loops for that I always get confused with. I would like to perform the following procedure in a compact way. Consider that p is a matrix composed of 100 rows and three columns. I need to calculate the sum over some rows of each column separately, as follows: fa1<-(colSums(p[1:25,])) fa2<-(colSums(p[26:50,]))
2017 Oct 19
2
Select part of character row name in a data frame
Thanks a lot, so simple so efficient! I will study more the grep command I did not know. Thanks! Francesca Pancotto > Il giorno 19 ott 2017, alle ore 12:12, Enrico Schumann <es at enricoschumann.net> ha scritto: > > df[grep("strat", row.names(df)), ] [[alternative HTML version deleted]]
2010 Jan 18
2
Rotating pca scores
Dear Folks I need to rotate PCA loadings and scores using R. I have run a pca using princomp and I have rotated PCA results with varimax. Using varimax R gives me back just rotated PC loadings without rotated PC scores. Does anybody know how I can obtain/calculate rotated PC scores with R? Your kindly help is appreciated in advance Francesca [[alternative HTML version deleted]]
2004 Sep 20
3
montecarlo simulation
Hy! I would like to know how run a montecarlo simulation with R. Thank you!!!! Francesca Matalucci __________________________________________________________________ Accesso Internet Gratis per utenti Excite! Attivalo subito! http://www.excite.it/hitech/accesso Il Mio Excite. Personalizza la tua Home page Excite come vuoi tu! http://www.excite.it AAA/Relazioni. Sfoglia gli annunci e trova la
2017 Oct 19
2
Select part of character row name in a data frame
Dear R contributors, I have a problem in selecting in an efficient way, rows of a data frame according to a condition, which is a part of a row name of the table. The data frame is made of 64 rows and 2 columns, but the row names are very long but I need to select them according to a small part of it and perform calculations on the subsets. This is the example: X Y "Unique to
2012 Apr 14
2
master thesis
Hi, For my master thesis I have 24 micro-plots on which I did measurements during 3 months. The measurements were: - Rainfall and runoff events throughout 3monts (runoff being dependant on the rainfall, a coefficient (%) has been made per rainfall event and per 3 months) - Soil texture (3 different textures were differentiated) - Slope (3 classes of slopes) - Stoniness (one time measurement)
2011 Nov 12
1
Simulation over data repeatedly for four loops
Dear Contributors, I am trying to perform a simulation over sample data, but I need to reproduce the same simulation over 4 groups of data. My ability with for loop is null, in particular related to dimensions as I always get, no matter what I try, "number of items to replace is not a multiple of replacement length" This is what I intend to do: replicate this operation for four
2008 Feb 25
1
Error accessing [homes] after 3.0.25b update (uppercase)
OS is smeserver, based on CentOS/RHEL 4. After update to samba 3.0.25b (included in RHEL U6) users who had "mapped" their home in Network places were unable to access the share. All other shares worked. On every attempt a line similar to the following was logged: Feb 14 10:11:49 nethservice smbd[6258]: '/home/e-smith/files/users/FRANCESCA/home' does not exist or permission
2017 Oct 19
0
Select part of character row name in a data frame
Quoting Francesca PANCOTTO <f.pancotto at unimore.it>: > Dear R contributors, > > I have a problem in selecting in an efficient way, rows of a data > frame according to a condition, > which is a part of a row name of the table. > > The data frame is made of 64 rows and 2 columns, but the row names > are very long but I need to select them according to a small
2005 May 27
3
Re: Shorewall development web site (Mike Noyes)
Hello, I leave for a couple days .. (Well months) and look at what has happened. :-) I would throw my support in behind Xoops .. to be honest .. If a portal is what we are trying to achieve here. I just happen to think that sometimes .. More work goes into web design etc than goes into actual Code. But thats because I am a lamer at web design :-) I am coming in here a bit late .. But tell
2011 Nov 11
8
Help
Dear Contributors I would like to perform this operation using a loop, instead of repeating the same operation many times. The numbers from 1 to 4 related to different groups that are in the database and for which I have the same data. x<-c(1,3,7) datiP1 <- datiP[datiP$city ==1,x]; datiP2 <- datiP[datiP$city ==2,x]; datiP3 <- datiP[datiP$city ==3,x] datiP4 <-