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Hi all, I tried to have plot of many vector in one plot and i have got a nice plot but i have problem with x-axis. I want to have month and year only(Jul.07 means July 2007) in x-axis without appearing other number behaind it. I would appercit any help. The R code: F<-c(7.49,6.91,6.78,6.99,7.44,7.42) M<-c(4.81,4.51,5.21,4.65,4.75,3.86) P<-c(7.49,15.03,15.19,15.32,15.42,15.45)

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",

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

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) )

https://bugs.freedesktop.org/show_bug.cgi?id=57484 Priority: medium Bug ID: 57484 Assignee: nouveau at lists.freedesktop.org Summary: NV43: fail resume after suspend QA Contact: xorg-team at lists.x.org Severity: normal Classification: Unclassified OS: Linux (All) Reporter: xoddark at gmail.com

Hi, I would like to select the most frequent value level in a set of three variables. Three different observators have judged hair color in study subjects. Mostly they judge the same color, sometimes there is a slight difference. I want to know what most of the observators have chosen (so at least 2) from the 3 observations. E.g. If two out of three observators decide the hair is black, then

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

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 =

Hi, In the example dataset below - how can I cahnge "gray20", to "blue" # data black <- rep(c("black","red"),10) gray <- rep(c("gray10","gray20"),10) black_gray <- data.frame(black,gray) # none of this desperate things works # replace(black_gray$gray, gray=="gray20","red") #

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

Would anyone be able to give me some ideas on how to go about changing the output of pairs()? What I want to do is : * plot each value as a point (instead of those hollow circles) * remove the gap between each scatter plot * make the frame around each scatter plot pale grey Thanks for you time, Hadley -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help

Dear R list, I am having some trouble drawing a bar-graph with two groups, both of which are stacked. A sample of my data (IUCN and national conservation status for marine and terrestrial species) look like this: Status IUCN Terrestrial IUCN Marine National CS Terrestrial National CS Marine NE 168 69 16 69 LC 41 6 148 6 NT 5 4 7 4 VU 4 2 7 2 EN 4 2 4 2 CR 1

Dear R list, It appears that my request from yesterday got a bit garbled. I have tried to rephrase and retype it - please let me know if you can help me - I am having some trouble drawing a bar-graph with two groups, both of which are stacked. I would like to plot the conservation status according to two classifications (i.e. my groups - IUCN status and national status), and for each of those

Hello, I have problem running the histogram function "hist". The area under the histogram is much lower than 1. Could anyone tell me what the problem is? Thanks, (The total number of observation is 992 (close to 1000), so the probability that 0<Y1<35 is approximately 0.277) miao rm(list=ls()) par(mfrow=c(1, 1)) Y <- cbind(matrix(35*0.5,1,277), matrix(35*1.5, 1, 146),

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

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: >

Hello, I'd like to add custom labels to my pair() plot. These labels include math expression but they aren't correctly displayed... Further, I want that the boxes for the text.panel (diagonal) have an other background color (grey80). Is that generally possible? If yes how do I have to set it? What I've so far is: panel.cor <- function(x, y, digits=2, prefix="",

I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp <-

Dear R users, First just want to say thank you to all for developing such a wonderful software and packages. I need to produce a wind rose plot. Tried with packages circular and plotrix and couldn't quite get what I want. Moved to package ggplot2 and it's going great. However stuck in how to move axis labels. I am using the wind rose from the help to learn how to do what I need (code

Hi R fans, As a newbie following the five-hour rule (after hitting my head against the wall for five hours, post to this list), I am appealing for some help understanding geom_rect() in ggplot2. What I want to do is very simple. I want to generate a plot of rectangles. Each one represents a business cycle. The x-values will be pairs representing the start and end of each cycle. The y-values