I am a beginner in R who got this question:
One of the functions we will be using often is sample(). Read the help file for sample() using ?sample. Now take a random sample of size 1 from the numbers 13 to 24 and report back the weight of the mouse represented by that row. Make sure to type set.seed(1) to ensure that everybody gets the same answer.
I tried this:
set.seed(1)
i <- sample( 13:24, 1)
dat$Bodyweight[i]
And got the answer 25.34. But apparently, that's wrong. What am I doing wrong?!
you need to detect which col u want to choose from first
set.seed(1)
sample(data$Bodyweight[13:24] ,1 )
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 think I have a rather simple problem but I can't figure out the best approach. I have a vector with 30 different values. Now I need to divide the vector into 10 groups in such a way that the mean within group variance is as small as possible. the size of the groups is not important, it can anything between one and 21.
Example. Let's say I have vector of six values, that I have to split into three groups:
Myvector <- c(0.88,0.79,0.78,0.62,0.60,0.58)
Obviously the solution would be:
Group1 <-c(0.88)
Group2 <-c(0.79,0.78)
Group3 <-c(0.62,0.60,0.58)
Is there a function that gives the same outcome as the example and that I can use for my vector withe 30 values?
Many thanks in advance.
It sounds like you want to do k-means clustering. Something like this would work
kmeans(Myvector,3, algo="Lloyd")
Note that I changed the default algorithm to match your desired output. If you read the ?kmeans help page you will see that there are different algorithms to calculate the different clusters because it's not a trivial computational problem. They might necessarily guarantee optimality.
I'm working on a dataset that consists of ~10^6 values which clustered into a variable number of bins. In the course of my analysis, I am trying to randomize my clustering, but keeping bin size constant. As a toy example (in pseudocode), this would look something like this:
data <- list(c(1,5,6,3), c(2,4,7,8), c(9), c(10,11,15), c(12,13,14));
sizes <- lapply(data, length);
for (rand in 1:no.of.randomizations) {
rand.data <- partition.sample(seq(1,15), partitions=sizes, replace=F)
}
So, I am looking for a function like "partition.sample" that will take a vector (like seq(1,15)) and randomly sample from it, returning a list with the data partitioned into the right bin sizes given already by "sizes".
I've been trying to write one such function myself, since the task seems to be not so hard. However, the partitioning of a vector into given bin sizes looks like it would be a lot faster and more efficient if done "under the hood", meaning probably not in native R. So I wonder whether I have just missed the name of the appropriate function, or whether someone could please point me to a smart solution that is around :-)
Your help & time are very much appreciated! :-)
Best,
Lymond
UPDATE:
By "no.of.randomizations" I mean the actual number of times I run through the whole "randomization loop". This will, later on, obviously include more steps than just the actual sampling.
Moreover, I would in addition be interested in a trick to do the above feat for sampling without replacement.
Thanks in advance, your help is very much appreciated!
Revised: This should be fairly efficient. It's complexity should be primarily in the permutation step:
# A single step:
x <- sample( unlist(data))
list( one=x[1:4], two=x[5:8], three=x[9], four=x[10:12], five=x[13:16])
As mentioned above the "no.of.randomizations" may have been the number of repeated applications of this proces, in which case you may want to wrap replicate around that:
replic <- replicate(n=4, { x <- sample(unlist(data))
list( x[1:4], x[5:8], x[9], x[10:12], x[13:15]) } )
After some more thinking and googling, I have come up with a feasible solution. However, I am still not convinced that this is the fastest and most efficient way to go.
In principle, I can generate one long vector of a uniqe permutation of "data" and then split it into a list of vectors of lengths "sizes" by going via a factor argument supplied to split. For this, I need an additional ID scheme for my different groups of "data", which I happen to have in my case.
It becomes clearer when viewed as code:
data <- list(c(1,5,6,3), c(2,4,7,8), c(9), c(10,11,15), c(12,13,14));
sizes <- lapply(data, length);
So far, everything as above
names <- c("set1", "set2", "set3", "set4", "set5");
In my case, I am lucky enough to have "names" already provided from the data. Otherwise, I would have to obtain them as (e.g.)
names <- seq(1, length(data));
This "names" vector can then be expanded by "sizes" using rep:
cut.by <- rep(names, times = sizes);
[1] 1 1 1 1 2 2 2 2 3 4 4 4 5
[14] 5 5
This new vector "cut.by" can then by provided as argument to split()
rand.data <- split(sample(1:15, 15), cut.by)
$`1`
[1] 8 9 14 4
$`2`
[1] 10 2 15 13
$`3`
[1] 12
$`4`
[1] 11 3 5
$`5`
[1] 7 6 1
This does the job I was looking for alright. It samples from the background "1:15" and splits the result into vectors of lengths "sizes" through the vector "cut.by".
However, I am still not happy to have to go via an additional (possibly) long vector to indicate the split positions, such as "cut.by" in the code above. This definitely works, but for very long data vectors, it could become quite slow, I guess.
Thank you anyway for the answers and pointers provided! Your help is very much appreciated :-)