I've been given this to do by the GENELAND tutorial to give population names to a dataset of populations of 60 individuals :
pop.mbrship1<-rep(c(1,2,3), each=60)
Nevertheless, my dataset comprises 10 populations of irregular sizes to which i would give the names of 1,2,3,4,5,6,7,8,9,10 and the distribution of my individuals (represented by one row each) would be :
1:24,25:39,40:58,59:79,80:103,104:126,127:147,148:171,172:191,192:214
I'd be tempted to use each population number as number of repeats which would make it
pop.mbrship1<-rep[c(1,2,3,4,5,6,7,8,9,10), each=c(24,15,19,21,24,23,21,24,20,23)]
Or try their distribution...
pop.mbrship1<-rep[c(1,2,3,4,5,6,7,8,9,10),
c(1:24,25:39,40:58,59:79,80:103,104:126,127:147,148:171,172:191,192:214)]
In both case, R gives me Error: unexpected '>' in ">"
I'm sure i'm really close to having it work but i've spent a shameful amount of time on this and i'd defenetly need a hand. Thanks a lot!
I'm looking at the geneland tutorial and I see that they have > at the beginning of the lines that you're copying/editing.
You are copying everything including the console pointer > all you need to copy/paste is :
# replicates each element 60 times
pop.mbrship1 <- rep(c(1,2,3),each=60)
# replicates each element, respectively
pop.mbrship2 <- rep(c(1,2,3),times=c(60,40,30))
Your answer is what Henrik said above, without a preceding>.
pop.mbrship1 <- rep(c(1,2,3,4,5,6,7,8,9,10), c(24,15,19,21,24,23,21,24,20,23))
# same as
pop.mbrship1 <- rep(c(...),times=c(...))
Related
I'm trying to analyse some tennis data and I'm hitting a problem with the code:
library(rpart)
library(rpart.plot)
library(ggplot2)
library(wesanderson)
train=read.csv("/ags_test.csv",header=T, na.strings=c("","NA"))
Please note this is a complete set, not one I've cobbled together through the code. All the gaps have NA values in them.
control=rpart.control(cp=0.007)
train$res=as.factor(train$res)
tree=rpart(res~Tournament+Surface+Round+J1Rank+J2Rank+J1Pts+J2Pts+DRank+DPts,data=train,control=control,parms=list(split="gini"))
All good until the last line when it kicks out:
Error in cbind(yval2, yprob, nodeprob) :
number of rows of matrices must match (see arg 2)
The data isn't a massive set but comprises of 17 columns and 50 lines.
Any ideas would be much appreciated.
Turns out that the problem the data is too certain, i.e. the pros are all in the same columns and the cons in a similar structure.
Therefore, there's little to run the decision tree against.
this sounds pretty basic but every time I try to make a histogram, my code is saying x needs to be numeric. I've been looking everywhere but can't find one relating to my problem. I have data with 240 obs with 5 variables.
Nipper length
Number of Whiskers
Crab Carapace
Sex
Estuary location
There is 3 locations and i'm trying to make a histogram with nipper length
I've tried making new factors and levels, with the 80 obs in each location but its not working
Crabs.data <-read.table(pipe("pbpaste"),header = FALSE)##Mac
names(Crabs.data)<-c("Crab Identification","Estuary Location","Sex","Crab Carapace","Length of Nipper","Number of Whiskers")
Crabs.data<-Crabs.data[,-1]
attach(Crabs.data)
hist(`Length of Nipper`~`Estuary Location`)
Error in hist.default(Length of Nipper ~ Estuary Location) :
'x' must be numeric
Instead of correct result
hist() doesn't seem to like taking more than one variable.
I think you'd have the best luck subsetting the data, that is, making a vector of nipper lengths for all crabs in a given estuary.
crabs.data<-read.table("whatever you're calling it")
names<-(as you have it)
Estuary1<-as.vector(unlist(subset(crabs.data, `Estuary Loc`=="Location", select = `Length of Nipper`)))
hist(Estuary1)
Repeat the last two lines for your other two estuaries. You may not need the unlist() command, depending on your table. I've tended to need it for Excel files, but I don't know what format your table is in (that would've been helpful).
Apparently if I try this:
# first grab the package
install.packages("stringi")
library(stringi)
# and then try to generate some serious dummy data
my_try <- as.vector(sample(1111111111:99999999999,3000000,replace=T))
R will say NOPE, sorry:
Error: cannot allocate vector of size 736.8 Gb
Should I buy more RAM*?
*this is a joke, but I seriously appreciate any help!
EDIT:
The desired output is a dataframe of 20 variables, and 3x10^6 rows. Some columns/variables should be strings, some integers. All in lengths ranging from 2 to 12.
The error isn't coming from sampling 3 million values, it's from trying to create a population of about 90 billion values 1111111111:99999999999 from which to sample. If you want to sample from that range, sample from the range 1:88888888889 and add 11111111110 using
sample(88888888889, 3000000,replace=TRUE) + 11111111110
There's no need for as.vector at the end, it's already a vector.
P.S. I believe in R-devel the range 1111111111:99999999999 will be stored much more efficiently (basically just the limits), but I don't know if sample() will be modified to work with it that way.
I'm not sure if my title is properly expressing what I'm asking. Once I'm done writing, it'll make sense. Firstly, I just started learning R, so I am a newbie. I've been reading through tutorial series and PDF's I've found online.
I'm working on a data set and I created a data frame of just the year 2001 and the DAM value Bon. Here's a picture.
What I want to do now is create a matrix with 3 columns: Coho Adults, Coho Jacks and the third column the ratio of Coho Jacks to Adults. This is what I'm having trouble with. The ratio between Coho Jacks to Adults.
If I do a line of code like this I get a normal output.
(cohoPassage <- matrix(fishPassage1995BON[c(5,6, 7)], ncol = 3))
The values are 259756, 6780 114934.
I'm figuring in order to get the ratio, I should divide column 5 and column 6's values. So basically 259756/6780 = 38.31
I've tried many things like:
(cohoPassage <- matrix(fishPassage1995BON[c(5,6, 5/6)], ncol = 3))
This just outputs the value of the fifth column instead of dividing for some reason
I've tried this:
matrix(fishPassage1995BON[c(5,6)],fishPassage1995BON[,5]/fishPassage1995BON[,6], ncol = 3)
Which gives me an incorrect output
I decided to break down the problem and divide the fifth and sixth columns separately and it gave the correct ratio.
If I create a matrix like this
matrix(fishPassage1995BON[,5]/fishPassage1995BON[,6])
It outputs the correct ratio of 38.31209. But when I try to combine everything, I just keep getting errors.
What can I do? Any help would be appreciated. Thank you.
I need help on 4b please
‘Warpbreaks’ is a built-in dataset in R. Load it using the function data(warpbreaks). It consists of the number of warp breaks per loom, where a loom corresponds to a fixed length of yarn. It has three variables namely, breaks, wool, and tension.
b. For the ‘AM.warpbreaks’ dataset, compute for the mean and the standard deviation of the breaks variable for those observations with breaks value not exceeding 30.
data(warpbreaks)
warpbreaks <- data.frame(warpbreaks)
AM.warpbreaks <- subset(warpbreaks, wool=="A" & tension=="M")
mean(AM.warpbreaks<=30)
sd(AM.warpbreaks<=30)
This is what I understood this problem and typed the code as in the last two lines. However, I wasn't able to run the last two lines while the first 3 lines ran successfully. Can anybody tell me what is the error here?
Thanks! :)
Another way to go about it:
This way you aren't generating a bunch of datasets and then working on remembering which is which. This is more a personal thing though.
data(warpbreaks)
mean(AM.warpbreaks[which(AM.warpbreaks$breaks<=30),"breaks"])
sd(AM.warpbreaks[which(AM.warpbreaks$breaks<=30),"breaks"])
There are two problems with your code. The first is that you are comparing to 30, but you're looking at the entire data frame, rather than just the "breaks" column.
AM.warpbreaks$breaks <= 30
is an expression that refers to the breaks being less than thirty.
But mean(AM.warpbreaks$breaks <= 30) will not give the answer you want either, because R will evaluate the inner expression as a vector of boolean TRUE/FALSE values indicating whether that break is less than 30.
Generally, you just want to take another subset for an analysis like this.
AM.lt.30 <- subset(AM.warpbreaks, breaks <= 30)
mean(AM.lt.30$breaks)
sd(AM.lt.30$breaks)