similar to: select most frequent value in set of variables

I have 2 mysql tables, Product and Color: Color ID ColorName 1 Red 2 Green 3 Yellow 4 Blue Products ID Color1 Color2 Color3 ProductName 1 ? ? ? Orco 2 ? ? ? Skeletor 3 ? ? ? He-Man I need to display the ColorName to

Hello there! I'd just like to say in advance, "Thank you," for any help and/or advice. My problem is as follows: I have a dataset that is made up of percentages. I've assigned my "error" percentages a value of '-100', my "non-existent" percentages a value of '0', and all my other percentages are normal values that range from the high

Hi everyone I am using barchart to make my graphs. Here is my code. barchart(percent_below ~ factor(Year)| factor(Season, levels=unique(Season)), data= .season_occurrence, origin = 0, layout = c(4, 1), scales=list(tick.number=ticknum,labels=NULL), ylim=c(0, ymax), group = factor(Year), xlab= "Year", auto.key = list(points = FALSE, rectangles =

Hi , I am new to R. I am trying to run a simple R script as shown below: aov.R ------ data1<-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste("subj", 1:12), c("Shape1.Color1", "Shape2.Color1", "Shape1.Color2",

Dear All, Consider the following trivial code snippet rm(list=ls()) name_vec <- c("color1", "color2") pdf("test_color.pdf") plot(seq(5), seq(5), main=paste(name_vec[1]," and ",name_vec[2], sep="")) dev.off() What I would like to achieve is rather simple to explain, but it is giving me a headache: how can I have two colors in main? Let us

If you are not using an anonymous function and say you had written the function out The below gives me the error > 'f(colordata2$color1)' is not a function, character or symbol' But then how is the anonymous function working? f <- function(col){ ifelse(col == 'blue', 1, 0) } responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )

Lapply is not a vectorized function. It is compact to read, but it would not be worth using for this calculation. However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a for loop to repeat this operation. colordata2 <- data.frame(id = c(1,2,3,4,5), color1 =

lapply(colordata2[ -1 ], f ) When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function. -- Sent from my phone. Please excuse my brevity. On April 7, 2016 7:31:18 AM PDT, Michael Artz <michaeleartz at

Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: >

Hola, Tienes varios ejemplos de cómo hacer esto en esta galería de gráficos, tanto de R como de Python. https://r-graph-gallery.com/heatmap.html Gracias, Carlos Ortega www.qualityexcellence.es El mar, 17 dic 2024 a las 14:38, Javier Marcuzzi (< javier.ruben.marcuzzi en gmail.com>) escribió: > Estimados > > Estoy pensando en buscar la forma para explicar algo, dentro de la forma

Hi, I used to get a really useful graph when I ran the following command using the MASS library: > cov.rob(cbind(dekeyser$AGE,dekeyser$GJTSCORE),cor=T) Besides the regular output, a graph appeared that had the classical correlation and the robust correlation, and two ellipses, one surrounding the data that would be used in the classical correlation and the other surrounding the data in the

Hello R usuer, The code given below superimposes a pie diagram on another plot containing some points. However, I would like to center the pie diagram on the xy location on the plot, but not on the center. is there any way to re-center pic diagram. Any suggestion or better alternative are highly appreciated. Thank you in advance for your help. Regards, Bibke library(visualFields) library(car)

Good afternoon, This is my first post, so please be gentle ;) Anyway, here's my question....I've recently put together a Linux Server for our company, replacing an old Novel server. I have the box set up with Samba, as a PDC on our network. So far, just about everything is working correctly. One issue I'm coming up against, though, is printing from DOS. I have Samba

Hi, I have a question about exporting interactive 3D plots. I use the following code to plot a contour of a trivariate normal distribution: library(mvtnorm) library(rgl) library(misc3d) n=25 x=seq(-3,3,length=n) X=cbind(rep(x,each=n**2),rep(rep(x,each=n),n),rep(x,n**2)) p=array(dmvnorm(X,sigma=diag(3)*0.5+0.5),c(n,n,n)) contour3d(p,x,x,x,level=mean(p)) lim=c(-3,3)

Dear Members of the R-Help, While using a R function - 'aggregate' that you developed, I become to have a question. In that function, > aggregate(x, by, FUN, ..., simplify = TRUE) I was wondering about what type of FUN I should write if I want to get "the most frequent value of the subgroup" as a summary statistics of the subgroups. I will appreciate if I can get

I am bootstrapping using a function that I have defined. The "Statistic" of the function is an array of 10 numbers. Therefore if I use 1000 replications, the "t" matrix will have 1000 rows each of which is a bootstrap replicate of this 10 number array (10 columns). Is there any easy way in R to determine which row appears the most frequently? Thanks, Lisa Pappas Huntsman

Hello! I have a continous distribution and would like to get mode (the most frequent value in distribution). I easily found mean, median and other basic thing but not mode function. Can anyone help? I know there my might be problems with multiple modes, but still I think that there should be a mode function in R. Please send mail to R-help list and me, so I can get response faster. Thank

Dear R-helpers, Suppose I have dataset like this below: data(HairEyeColor) dfHEC <- as.data.frame(as.table(HairEyeColor)) my.dfHEC <- data.frame(Hair=rep(dfHEC$Hair,dfHEC$Freq), Eye=rep(dfHEC$Eye,dfHEC$Freq), Sex=rep(dfHEC$Sex,dfHEC$Freq)) my.dfHEC my.dfHEC$Hair my.dfHEC$Eye my.dfHEC$Sex and I know all levels for Hair, Eye and Sex. In my case, I

Hi, Can someone set me straight as to how to write formulas in R to indicate: complete independence [A][B][C] joint independence [AB][C] conditional independence [AC][BC] nway interaction [AB][AC][BC] ? For example, if I have 4 factors: hair colour, eye colour, age, sex does > mosaicplot( frequency ~ hair + eye + age + sex) mean that the model fitted is of complete independence of

I have been tinkering with mosaicplot() and friends as a way of learning R. As part of this, I've written a pair.table() method for mosaic matrices, and would like to extend mosaicplot to work with loglin and logln (MASS) objects. I'm using R 1.4.0 on Win 98. I've been trying to figure out the formula interface, and think there's a bug, but not sure how to find it, yet alone fix