similar to: How to fix indeces in a loop

Displaying 20 results from an estimated 800 matches similar to: "How to fix indeces in a loop"

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]<-
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]]
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
2024 Jun 13
1
Create a numeric series in an efficient way
I apologize, I solved the problem, sorry for that. f. Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO < francesca.pancotto at unimore.it> ha scritto: > Dear Contributors > I am trying to create a numeric series with repeated numbers, not > difficult task, but I do not seem to find an efficient way. > > This is my solution > > blocB <- c(rep(x = 1,
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
2024 Jun 13
1
Create a numeric series in an efficient way
Maybe this was your solution? blocC <- c(rep(x=c(1:13), times=84)) blocC <- arrange(.data = data.frame(blocC), blocC) The second line sorts, but that may not be needed depending on application. The object class is also different in the sorted solution. Tim -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Francesca PANCOTTO via R-help Sent:
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
2024 Jun 13
2
Create a numeric series in an efficient way
Dear Contributors I am trying to create a numeric series with repeated numbers, not difficult task, but I do not seem to find an efficient way. This is my solution blocB <- c(rep(x = 1, times = 84), rep(x = 2, times = 84), rep(x = 3, times = 84), rep(x = 4, times = 84), rep(x = 5, times = 84), rep(x = 6, times = 84), rep(x = 7, times = 84), rep(x = 8, times = 84), rep(x = 9, times = 84),
2008 Jul 29
1
tensor product of equi-spaced B-splines in the unit square
Dear all, I need to compute tensor product of B-spline defined over equi-spaced break-points. I wrote my own program (it works in a 2-dimensional setting) library(splines) # set the break-points Knots = seq(-1,1,length=10) # number of splines M = (length(Knots)-4)^2 # short cut to splineDesign function bspline = function(x) splineDesign(Knots,x,outer.ok = T) # bivariate tensor product of
2010 May 17
2
best polynomial approximation
Dear R-users, I learned today that there exists an interesting topic in numerical analysis names "best polynomial approximation" (BSA). Given a function f the BSA of degree k, say pk, is the polynomial such that pk=arginf sup(|f-pk|) Although given some regularity condition of f, pk is unique, pk IS NOT calculated with least square. A quick google tour show a rich field of research
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
2008 Aug 08
3
Multivariate regression with constraints
Hi all, I am running a bivariate regression with the following: p1=c(184,155,676,67,922,22,76,24,39) p2=c(1845,1483,2287,367,1693,488,435,1782,745) I1=c(1530,1505,2505,204,2285,269,1271,298,2023) I2=c(8238,6247,6150,2748,4361,5549,2657,3533,5415) R1=I1-p1 R2=I2-p2 x1=cbind(p1,R1) y1=cbind(p2,R2) fit1=lm(y1~-1+x1) summary(fit1) Response 2: Coefficients: Estimate Std. Error t value
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
2013 Jan 29
0
On the calulation of crossed differences
Dear Contributors, I am back asking for help concerning the same type of dataset I was asking before, in a previous help request. I needed to sum data over subsample of three time series each of them made of 100 observations. The solution proposed were various, among which: db<-p dim( db ) <- c(25,4,3) db2 <- apply(db, c(2,3), sum) db3 <- t(apply(db2, 1, function(poff)
2013 Aug 22
1
corrgram (package corrgram): how to plot multiple correlograms in the same page?
Hello, I am trying to plot a few correlograms on the same figure, with the function corrgram() from the package corrgram. However, the function does not seem to use the base graphic system, as setting out the multiple figure layout with, e.g., par(mfrow=c(2, 2,)) does not work. Does anybody know a workaround for this? Many thanks in advance for any advice best giuseppe -- Giuseppe Pagnoni,
2024 Sep 16
2
(no subject)
Dear Contributors, I hope someone has found a similar issue. I have this data set, cp1 cp2 role groupid 1 10 13 4 5 2 5 10 3 1 3 7 7 4 6 4 10 4 2 7 5 5 8 3 2 6 8 7 4 4 7 8 8 4 7 8 10 15 3 3 9 15 10 2 2 10 5 5 2 4 11 20 20 2 5 12 9 11 3 6 13 10 13 4 3 14 12 6 4 2 15 7 4 4 1 16 10 0 3 7 17 20 15 3 8 18 10 7 3 4 19 8 13 3 5 20 10 9 2 6 I need to to average of groups, using the values of column
2011 Sep 28
1
Wilcox test and data collection
Dear Contributors I have a problem with the collection of data from the results of a test. I need to perform a comparative test over groups of data , recall the value of the pvalue and create a table. My problem is in the way to replicate the analysis over and over again over subsets of data according to a condition. I have this database, called y: gg t1 t2 d 40 1 1
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 <-
2008 May 29
2
Troubles plotting lrm output in Design Library
Dear R-helpers, I'm having a problem in using plot.design in Design Library. Tho following example code produce the error: > n <- 1000 # define sample size > set.seed(17) # so can reproduce the results > age <- rnorm(n, 50, 10) > blood.pressure <- rnorm(n, 120, 15) > cholesterol <- rnorm(n, 200, 25) > sex <-