I started with library(ggtern) ggtern(df,aes(GRAVEL,SAND,MUD))+geom_point() Adding fill=Root within the aes() function and shape=21 outside would colour in the points according to the value of some other variable, but it only makes most sense to colour the points if you have a separate variable in your data set which could determine the colour — your example … Read more
# Filter dat.filtered <- subset(dat, f2 > 0) # Convert to time-series library(xts); xts.sample <- xts(dat.filtered$f2, order.by = as.Date(dat.filtered$f1, “%d/%m/%Y”)) Sample data dat <- read.table(text = ” f1 f2 1 11/1/16 0 2 12/1/16 0 3 11/2/16 56.25 4 12/2/16 0 5 11/3/16 56.25 6 12/3/16 0 7 11/4/16 111 8 12/4/16 0 9 11/5/16 … Read more
Here’s a crack at what you want. I think your problem starts with how SDB is stored. To see what an object looks like use str. I don’t know what an xts object is but using str helps me determine how to pull out the pieces I want. str(SDB) > str(SDB) An ‘xts’ object from … Read more
You can still use aes in each stat_smooth including only the desired color. This will add legends to the plot. Then you can use scale_color_identity to match and rename the colors ggplot(mtcars, aes(x=wt, y = mpg, col=”green”)) + geom_point(col=”blue”) + stat_smooth(method=’loess’,linetype=”dashed”, aes(color = “red”), span=0.1) + stat_smooth(method=’loess’,linetype=”dashed”, aes(color = “orange”), span=0.25) + stat_smooth(method=’loess’,linetype=”dashed”, aes(color = … Read more
A bit untidy, but does the job: d <- data.frame(a=a[-(1:2)], diff=diff(a, 2)) d$br <- 0 for (i in 1:nrow(d)) { if (i==1 & d$diff[1]==2) { d$br[1] <- 1 } else if (i==2 & d$diff[2]==2 & d$br[1]!=1) { d$br[2] <- 1 } if (d$diff[i]==2 & !any(sum(d$br[c(i-1, i-2)])>0)) d$br[i] <- 1 } t(sapply(d$a[d$br==1], function(x) (x-2):x)) # [,1] … Read more
Try library(sp) S@coords[,’roadtype’][S@coords[,’jointcount’]!=1] <- NA S # SpatialPoints: # jointcount roadtype #[1,] 1 3 #[2,] 4 NA #[3,] 3 NA #[4,] 1 1 #[5,] 1 4 data jointcount = c(1,4,3,1,1) roadtype = c(3,2,5,1,4) S <- SpatialPoints(data.frame(jointcount,roadtype)) solved Replace Values in a column wiht NA if values from another column are not 1 [closed]
Assuming the “df” dataframe has the relevant variables under the names “channel” and “email”, then: To get the number of unique channel-email pairs: dim(unique(df[c(“channel”, “email”)]))[1] To get the sum of all channel-email observations: sum(table(df$channel, df$email)) To get the number of duplicates, simply subtract the former from the later: sum(table(df$channel, df$email)) – dim(unique(df[c(“channel”, “email”)]))[1] 0 solved … Read more
In R, a vector (a in your question) can be subset with a logical index vector (b in your question). If the corresponding members are TRUE in the logical vector, the elements (of a) are retained. a = c (3, 4, 7, 8) b = c(TRUE, TRUE, FALSE, FALSE) a[b] #[1] 3 4 a[b] will … Read more
#margins parameters mar_save <- par(“mar”) par(mar=c(2, 2.3, 0, 0)) #plot windows and initialisation xmin <- -1 xmax <- 11 ymin <- 0 ymax <- 10 plot(x = c(xmin,xmax), y = c(ymin,ymax), xlim = c(xmin,xmax), ylim = c(ymin,ymax), type = “n”, xaxs = “i”, xaxt = “n”, yaxs = “i”, yaxt = “n”) # axes labels … Read more
Try to look in the .GlobalEnv: srch = function(srchstr){ print(ls(pattern=srchstr, envir = .GlobalEnv)) } edit joran was faster… solved command ls with pattern in a function is not working in R
Well, I am not sure if Tail and Class are part of the same dataframe or are two seperate vectors. If they are two seperate vectors, maybe you can merge the two vectors in a dataframe df <- data.frame(Tail = as.character(Tail), Class = as.character(Class)) and then with dplyr you can try, library(dplyr) df %>% group_by(Tail) … Read more
Sui_Q1<- list() is written more than once. it cause the list() to override. The solution is only to put Sui_Q1<- list() and Sui_Q2<-list() before looping. solved R : assigning value in looping and else if
You can do something like this with dplyr set.seed(43) df<-data.frame(location=sample(LETTERS[1:3],20,replace=TRUE), site=sample(c(“bang”,”mys”,”hubl”,”dar”),20,replace=TRUE)) library(dplyr) df%>%group_by(location,site)%>%summarize(Count=n())%>%arrange(desc(Count))%>%slice(1)%>%ungroup()%>%select(location,site) 3 solved To get the most frequent levels in a data frame
You’re missing a closing bracket in the first line of the ggplot call.. p <- ggplot(cabbage_exp, aes(Date, Weight, fill = Cultivar)**)** + Give that a shot. solved Simple R Function, What Is Wrong? [closed]
The following code does what the question asks for. The function that does all the work is append_one. It Creates a vector Y repeating the prefix length(x) times. Gets the runs of vector y. Cleans the runs’ values to the empty string “” if the runs’ lengths are less than N. Reverses the run-length encoding. … Read more