similar to: Data frame reordering to time series

Displaying 20 results from an estimated 500 matches similar to: "Data frame reordering to time series"

2013 Mar 25
2
Faster way of summing values up based on expand.grid
Hello! # I have 3 vectors of values: values1<-rnorm(10) values2<-rnorm(10) values3<-rnorm(10) # In real life, all 3 vectors have a length of 25 # I create all possible combinations of 4 based on 10 elements: mycombos<-expand.grid(1:10,1:10,1:10,1:10) dim(mycombos) # Removing rows that contain pairs of identical values in any 2 of these columns: mycombos<-mycombos[!(mycombos$Var1
2010 Dec 18
3
dotchart for matrix data
Readers, I am trying to use the function dotchart. The data is: > testdot category values1 values2 values3 values4 1 a 10 27 56 709 2 b 4 46 47 208 3 c 5 17 18 109 4 d 6 50 49 308 The following error occurs > dotchart(testdot,groups=testdot[,2]) Error in dotchart(testdot, labels =
2006 Oct 01
3
aggregate function with 'NA'
Dear r-help reader, I have some problems with the aggregate function. My datframe looks like >frame Day Time V1 V2 1 M 0 3 NA 2 M 0 4 NA 3 M 0 5 2 4 M 1 NA 4 5 M 1 10 6 6 T 0 4 45 7 T 1 4 3 8 T 1 3 2 9 T 1 6 1 I used the aggegate function to obtain the mean in V1 and V2 over the grouping variable Time and Day
2013 Jul 17
0
usar partial=TRUE en rake
Estimados usuarios de R: Estoy usando un programa en que utilizo la función rake: *rake(ps,list(~A,~B,~C),list(pop.A, pop.B, pop.C),control = list(maxit=1000, epsilon = 1, verbose=TRUE))* ** *Obtengo como resultado:* ** , , C = 1 B A 1 2 3 4 5 A1 0.000000 0.000000 0.000000 0.000000 0.000000 A2 3.000000 3.000000 0.000000
2009 Jun 16
4
confusion on levels() function, and how to assign a wanted order to factor levels, intentionally?
Dear R-helpers, I want to make a series of boxplots on several numeric univariates with two group variables (species and population, population nested in species, and with population as the X-axis). In order to get a proper order of the individual populations in X-axis, I need to assign a wanted order to the factor (population). I used the levels() function to do this assignment, but it seemed
2007 Nov 16
0
Odp: R: ave and quantile
Hi Patrick Hausmann <c18g at uni-bremen.de> napsal dne 15.11.2007 18:59:06: > Hello Petr, > > one question solved, the next is standing in front of me... If you > have a minute to look .... great! > > > x > V1 V2 F1 > 1 A 2 0.1552277 > 2 A 3 0.1552277 > 3 A 4 0.1552277 > 4 B 3 0.8447723 > 5 B 2 0.8447723 > 6 C 6 0.2500000
2013 Jan 05
5
Need help on dataframe
Dear R users, I came up to a problem by taking means (or other summary statistics) of a big dataframe. Suppose we do have a dataframe: ID V1 V2 V3 V4 ........................ V71 1 6 5 3 2 ........................ 3 2 3 2 2 1 ........................ 1 3 6 5 3 2 ........................ 3 4 12 15 3 2 ........................ 100
2011 Jan 27
3
how to divide each element of a matrix by a specific value per column
Hi, I'd like to divide each element of a matrix by a specific value per column. These specific values are stored in a list. For example: > x <- c(1,2,3,4,5) > y <- matrix(c(1:30), nrow = 6) Now I want to divide each element in y[,1] by x[1], y[,2] by x[2] etc. I have tried this > my_function <- function(data, ind) data/ind > apply(y, 2, my_function, x) [,1] [,2]
2008 Oct 31
4
[ifelse] how to maintain a value from original matrix without probs?
Dear all, I have a matrix with positive and negative values. >From this I would like to produce 2 matrices: 1st - retaining positives and putting NA in other positions 2nd - retaining negatives and putting NA in other positions and then apply rowMeans for both. I am trying to use the function ifelse in the exemplified form: ifelse(A>0,A,NA) but by putting A as a 2nd parameter it
2007 Jun 05
1
Gruff Stacked Bar order of data
I am making a stacked bar graph in Gruff. Could anyone help me by telling me how I can change the order in which the data is stacked. No matter what order I put the data in, I get the same result on the png picure. Any help would be much appreciated. Thanks. -- Posted via http://www.ruby-forum.com/. --~--~---------~--~----~------------~-------~--~----~ You received this message because
2004 Apr 07
4
Problems with rlm
Dear all, When calling rlm with the following data, I get an error. (R v.1.8.1, WinXP Pro 2002 with service pack 1.) > d <- na.omit(data.frame(CPRATIO, HEIGHTZ, FAMILYID)) > c <- tapply(d$CPRATIO, d$FAMILYID, mean) > h <- tapply(d$HEIGHTZ, d$FAMILYID, mean) > c 1 2 3 6 7 9 10 11 6.000000 2.500000 3.250000
2011 Oct 11
2
matrix multiplication
Dear all, Sorry to bother you with such a stupid question, but I just cannot find the solution to my problem. I'd like to use matrix multiplication for meanA and factorial 3. I use the command meanA%*%factorial 3. But everything I get is: Error in factorial3 %*% A : non-conformable arguments I know that the number of the columns of the first vector has to be the same number of rows of the
2018 Mar 14
3
the same function returning different values when called differently..
dear members, I have a function ygrpc which acts on the daily price increments of a stock. It returns the following values: ygrpc(PFC.NS,"h") [1] 2.149997 1.875000 0.750000 0.349991 2.100006 0.199997 4.000000 2.574996 0.500000 0.349999 1.500000 0.700001 [13] 0.500000 1.300003 0.449997 2.800003 2.724998 66.150002 0.550003 0.050003 1.224991 4.899994 1.375000
2013 Mar 18
2
Confirmatory factor analysis using the sem package. TLI CFI and RMSEA absent from model summary.
Hi R-help, I am using the sem package to run confirmatory factor analysis (cfa) on some questionnaire data collected from 307 participants. I have been running R-2.15.3 in Windows in conjunction with R studio. The model I am using was developed from exploratory factor analysis of a separate dataset (n=439); it includes 18 items that load onto 3 factors. I have used the sem package documentation
2007 May 04
1
Bug in qr.R ? (PR#9655)
Ladies and Gentlemen, using > A <- structure(c(1, 0, 0, 3, 2, 1, 4, 5, -3, -2, 1, 0), .Dim = as.integer(c(3,4))) I get > dim(A) [1] 3 4 > qr.R(qr(A),complete=TRUE) [,1] [,2] [,3] [,4] [1,] -1 -3.000000 -4.000000 2.0000000 [2,] 0 -2.236068 -3.130495 -0.8944272 [3,] 0 0.000000 -4.919350 -0.4472136 > qr.R(qr(A),complete=FALSE) [,1]
2019 Oct 23
2
Unexpected behavior when using macro to loop over vector
Hi all, I found an unexpected behavior when I was trying to use the macro defined in "R_ext/Itermacros.h" to loop over an atomic vector. Here is a minimum example: C++ code ``` #include "R_ext/Itermacros.h" #define GET_REGION_BUFSIZE 2 //Redefine the macro since C++ is not happy with the implicit type conversion #define ITERATE_BY_REGION_PARTIAL(sx, px, idx, nb, etype,
2018 Mar 14
0
Fwd: the same function returning different values when called differently..
Hi Akshay, (Please include r-help when replying) You have learned that PFC.NS and snl[[159]] are not identical. Now you have to figure out why they differ. This could also point to a bug or a logic error in your program. Figuring out how two objects differ can be a bit tricky, but with experience it becomes easier. (Some others may even have some suggestions for good ways to do it.) Basically
2019 Oct 25
2
Unexpected behavior when using macro to loop over vector
On 10/25/19 11:01 AM, Tomas Kalibera wrote: > On 10/23/19 6:45 AM, Wang Jiefei wrote: >> Hi all, >> >> I found an unexpected behavior when I was trying to use the macro >> defined >> in "R_ext/Itermacros.h"? to loop over an atomic vector. Here is a >> minimum >> example: >> >> C++ code >> ``` >> #include
2011 Jul 22
2
averaging rows based on string¿?
Hi Folks, Ran into something I'd really like to do in R simply/elegantly, but my R - coding skills seem surpassed. This is the thing. Imagine the following data: labs<-c("abcdef","abcgg","tgthefdk","tgtijuel","tgtnjmoi","gbnt","dlift") dat<-c(0.5,0.25,1,2,16,0.250,4) dframe<-data.frame(labs,dat) I would like to
2013 Mar 29
1
Dataframe manipulation
Hi Adam, I hope this is what you wanted: dat1<- read.csv("example.csv",sep="\t",stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 102 obs. of? 5 variables: # $ species? : chr? "B. barbastrellus" "E. nilssonii" "H. savii" "M. alcathoe" ... # $ period?? : chr? "dusk" "dusk" "dusk"