similar to: is.null(mylist[1]) and is.null(mylist$a) returns different values

Hi, Try > is.null(mylist[[1]]) [1] TRUE Notice the double square brackets. From: ?`[` "The most important distinction between [, [[ and $ is that the [ can select more than one element whereas the other two select a single element." On Thu, Jun 15, 2017 at 11:33 AM, ce <zadig_1 at excite.com> wrote: > Hi > > I have a list : > > mylist <- list( a = NULL, b

I find that the str function is more helpful for understanding the difference between a null list and a list containing a null list than the implicit print function call that the interpreter invokes when you enter an expression at the console. str( mylist[1] ) -- Sent from my phone. Please excuse my brevity. On June 15, 2017 8:39:47 AM PDT, Huzefa Khalil <huzefa.khalil at umich.edu>

Thank you all , very informative, never thought of doing a str( mylist[1] ) -----Original Message----- From: "Jeff Newmiller" [jdnewmil at dcn.davis.ca.us] Date: 06/15/2017 11:56 AM To: r-help at r-project.org, "Huzefa Khalil" <huzefa.khalil at umich.edu>, "ce" <zadig_1 at excite.com> Subject: Re: [R] is.null(mylist[1]) and is.null(mylist$a) returns

or in base R : ?xtabs ?? as in: xtabs(~previous_location + current_location,data=x) (You can convert the 0s to NA's if you like) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, May 8, 2018 at 9:21 AM, Huzefa

Dear Bert and Huzefa, Apologies for the late reply, my account got hacked and I have just managed to recover it. Thank you very much for your replies and the solutions. Both work well. I was wondering if there was any way to ensure (force) that all possible combinations show up in the output. The full dataset has 25 cities but of course people have not moved from Boston to all the other 24

Hi James, If you want to specify the y-values, you need to use stat="identity" as below: ggplot(probability, aes(x=Fertilizer, y=prob)) + geom_bar(stat="identity", aes(fill=Treatment)) best, huzefa On Tue, Apr 12, 2016 at 1:02 PM, James Henson <jfhenson1 at gmail.com> wrote: > Dear R Community, > > Below is a problem with a simple ggplot2 graph. The code

Dear R Community, Below is a problem with a simple ggplot2 graph. The code returns the error message below. Error: stat_count() must not be used with a y aesthetic. My code is below and the data is attached as a ?text? file. # Graph of the probabilities library(digest) library(DT) datatable(probability) str(probability) probability$Fertilizer <- as.factor(probability$Fertilizer)

It looks like you made a copy/paste error below. Your ata.frame should be data.frame. Kevin On 05/04/2018 08:18 AM, Bill Poling wrote: > Good morning. > > Novice usR. Here. > > I am following this string, among many, learning as I go. > > Quick question please? > > I thought that perhaps ata.frame was part of the zoo pkg, b/c when I > searched it came up in

Good morning. Novice usR. Here. I am following this string, among many, learning as I go. Quick question please? I thought that perhaps ata.frame was part of the zoo pkg, b/c when I searched it came up in help? However, evidently not or I am not using it properly. Please advise, thank you. x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) x2 <- do.call(rbind,

xtabs does this automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 > lvls <- sort(union(x,y)) > x <- factor(x, levels = lvls) > y <-

Instead of Tsp = c(2016, 2018, 12) try Tsp = c(2016, 2018.25, 12) Hence, you can specify the object as structure(c(5973156.76, 5159011.2, 6695766.64, 6365359, 6495218.53, 7226302.39, 6835272.7, 7383501.57, 6962748.19, 7623278.72, 7274994.33, 7919421.8, 7360740.81, 7436693.35, 8545765.55, 7337269.76, 8180585.44, 8376635.05, 7758261.24, 10374641.22, 8000314.11, 9114958.9, 9805149.15,

Hi. I am trying to follow this forecasting tutorial at: https://www.r-bloggers.com/basic-forecasting/ Using my own data, I cannot get past the first step, lots of laughs. dat3 <- structure(c(5973156.76, 5159011.20, 6695766.64, 6365359.00, 6495218.53, 7226302.39, 6835272.70, 7383501.57, 6962748.19, 7623278.72, 7274994.33 ,7919421.80, 7360740.81, 7436693.35,

I can define a list containing NULL elements: > myList <- list("aaa",NULL,TRUE) > names(myList) <- c("first","second","third") > myList $first [1] "aaa" $second NULL $third [1] TRUE > length(myList) [1] 3 However, if I assign NULL to any of the list element then such element is deleted from the list: > myList$second <-

Dear Miluji, If I understand correctly, this should get you what you need. temp1 <- structure(list(id = 101:115, current_location = structure(c(2L, 8L, 8L, 3L, 6L, 5L, 1L, 2L, 7L, 4L, 2L, 8L, 8L, 3L, 6L), .Label = c("Austin", "Boston", "Cambridge", "Durham", "Houston", "Lynn", "New Orleans", "New York"), class =

Dear William and Ben, Thank you for your replies and elegant solutions. I am having trouble with the fact that two of the previous locations do not appear in current locations (that is no one moved to OKC and Dallas from other cities), so these two cities are not being included in the output. I have provided a better sample of the data and the ideal output (wide form - a 10x10 bilateral matrix)

I have data on current and previous location of individuals. I would like to have a matrix with bilateral movement between locations. I would like the final output to look like the second table below. I have tried using crosstab() from the ecodist but I do not have another variable to measure the flow. Ultimately I would like to compute the probability of movement between cities (movement to

Hi Martin, The corrected function would be RANDU <- function(num) { return ((65539*num)%%(2^31)) } You forgot the brackets for the return function. Hence, what was returned was always (65539 * num) On Mon, Aug 14, 2017 at 12:49 PM, Martin M?ller Skarbiniks Pedersen <traxplayer at gmail.com> wrote: > Dear all, > > I am trying to learn functions in R and 3D plotting so I

Dear all, I am trying to learn functions in R and 3D plotting so I decided to try to plot the famous bad PRNG Randu from IBM(1). However something is not correct in the function I have created. First I define the function RANDU like this: > RANDU <- function(num) { return (65539*num)%%(2^31) } and test that it works for a seed of 1: > RANDU(1) [1] 65539 but if I want the next

I can't install rugarch package because installation of nloptr package fails . I use opensuse 12.3 # uname -a Linux candide 3.7.10-1.11-desktop #1 SMP PREEMPT Thu May 16 20:27:27 UTC 2013 (adf31bb) x86_64 x86_64 x86_64 GNU/Linux my gcc version is 4.8.1 I compiled and installed R 3.01 . then I tried to install rugarch package but it fails because it can't install depended package nloptr.

Hi Kevin, There is probably a better way, but it can be done in two steps like this temp <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) temp <- lapply(names(temp), function(n, temp) { temp[[n]]$type <- n return(temp[[n]]) }, temp = temp) do.call(rbind, temp) On Wed, May 2, 2018 at 1:11 PM, Kevin E. Thorpe <kevin.thorpe at utoronto.ca> wrote: > I suspect