I am doing some R code practice and is wanting to generate some specific sequence.
I want to use the function seq() to generate sequence, which I have tried the following code, but it seems like character is not available in seq()
n <- c(1 : 100)
seq(from = 1, to = n )
I expect the sequence to be 1 1 2 1 2 3 1 2 3 4.... However, this code isn't able to work
In the following way:
sequence(1:100)
sequence works by creating a vector from 1 until the number passed to sequence() Since this function is vectorized, you can insert multiple numbers at once with c(1:100) at get the necessary output.
Related
I am trying to run a summation on each row of dataframe. Let's say I want to take the sum of 100n^2, from n=1 to n=4.
> df <- data.frame(n = seq(1:4),a = rep(100))
> df
n a
1 1 100
2 2 100
3 3 100
4 4 100
Simpler example:
Let's make fun1 our example summation function. I can pull 100 out because I can just multiply it in later.
fun <- function(x) {
i <- seq(1,x,1)
sum(i^2) }
I want to then apply this function to each row to the dataframe, where df$n provides the upper bound of the summation.
The desired outcome would be as follows, in df$b:
> df
n a b
1 1 100 1
2 2 100 5
3 3 100 14
4 4 100 30
To achieve these results I've tried the apply function
apply(df$n,1,phi)
and also with df converted into a matrix
mat <- as.matrix(df)
apply(mat[1,],1,phi)
Both return an error:
Error in seq.default(1, x, 1) : 'to' must be of length 1
I understand this error, in that I understand why seq requires a 'to' value of length 1. I don't know how to go forward.
I have also tried the same while reading the dataframe as a matrix.
Maybe less simple example:
In my case I only need to multiply the results above, df$b, by 100 (or df$a) to get my final answer for each row. In other cases, though, the second value might be more entrenched, for example a^i. How would I call on both variables, a and n?
Underlying question:
My underlying goal is to apply a summation to each row of a dataframe (or a matrix). The above questions stem from my attempt to do so using seq(), as I saw advised in an answer on this site. I will gladly accept an answer that obviates the above questions with a different way to run a summation.
If we are applying seq it doesn't take a vector for from and to. So we can loop and do it
df$b <- sapply(df$n, fun)
df$b
#[1] 1 5 14 30
Or we can Vectorize
Vectorize(fun)(df$n)
#[1] 1 5 14 30
this is probably a simple one, but I somehow got stuck...
I need to many loops to get the result of every sample in my support like the usual stacked loops:
for (a in 1:N1){
for (b in 1:N2){
for (c in 1:N3){
...
}
}
}
but the number of the for loops needed in this messy system depends on another random variable, let's say,
for(f in 1:N.for)
so how can I write a for loop to do deal with this? Or are there more elegant ways to do this?
note that the difference is that the nested for loops above (the variables a,b,c,...) do matter in my calculations, but the variable f of the for loop that controls for the number of for loops needed does not go into any of my calculations for my real purpose - all it does is count/ensure the number of for loops needed is correct.
Did I make it clear?
So what I am actually trying to do is generate all the possible combinations of a number of peoples preferences towards others.
Let's say I have 6 people (the simplest case for my purpose): Abi, Bob, Cath, Dan, Eva, Fay.
Abi and Bob have preference lists of C D E F ( 4!=24 possible permutations for each of them);
Cath and Dan have preference lists of A B and E F, respectively (2! * 2! = 4 possible permutations for each of them);
Eva and Fay have preference lists of A B C D (4!=24 possible permutations for each of them);
So all together there should be 24*24*4*4*24*24 possible permutations of preferences when taking all six them together.
I am just wondering what is a clear, easy and systematic way to generate them all at once?
I'd want them in the format such as
c.prefs <- as.matrix(data.frame(Abi = c("Eva", "Fay", "Dan", "Cath"),Bob = c("Dan", "Eva", "Fay", "Cath"))
but any clear format is fine...
Thank you so much!!
I'll assume you have a list of each loop variable and its maximum value, ordered from the outermost to innermost variable.
loops <- list(a=2, b=3, c=2)
You could create a data frame with all the loop variable values in the correct order with:
(indices <- rev(do.call(expand.grid, lapply(rev(loops), seq_len))))
# a b c
# 1 1 1 1
# 2 1 1 2
# 3 1 2 1
# 4 1 2 2
# 5 1 3 1
# 6 1 3 2
# 7 2 1 1
# 8 2 1 2
# 9 2 2 1
# 10 2 2 2
# 11 2 3 1
# 12 2 3 2
If the code run at the innermost point of the nested loop doesn't depend on the previous iterations, you could use something like apply to process each iteration independently. Otherwise you could loop through the rows of the data frame with a single loop:
for (i in seq_len(nrow(indices))) {
# You can get "a" with indices$a[i], "b" with indices$b[i], etc.
}
For the way of doing the calculation, an option is to use the Reduce function or some other higher-order function.
Since your data is not inherently ordered (an individual is part of a set, its preferences are part of the set) I would keep indivudals in a factor and have eg preferences in lists named with the individuals. If you have large data you can store it in an environment.
The first code is just how to make it reproducible. the problem domain was akin for graph oriented naming. You just need to change in the first line and in runif to change the behavior.
#people
verts <- factor(c(LETTERS[1:10]))
#relations, disallow preferring yourself
edges<-lapply(seq_along(verts), function(ind) {
levels(verts)[-ind]
})
names(edges) <- levels(verts)
#directions
#say you have these stored in a list or something
pool <- levels(verts)
directions<-lapply(pool, function(vert) {
relations <- pool[unique(round(runif(5, 1, 10)))]
relations[!(vert %in% relations)]
})
names(directions) = pool
num_prefs <- (lapply(directions, length))
names(num_prefs) <- names(directions)
#First take factorial of each persons preferences,
#then reduce that with multiplication
combinations <-
Reduce(`*`,
sapply(num_prefs, factorial)
)
I hope this answers your question!
Let's say I have a vector a = (1,3,4).
I want to create new vector with integer numbers in range [1,length(a)]. But the i-th number should appear a[i] times.
For the vector a I want to get:
(1,2,2,2,3,3,3,3)
Would you explain me how to implement this operation without several messy concatenations?
You can try rep
rep(seq_along(a), a)
#[1] 1 2 2 2 3 3 3 3
data
a <- c(1,3,4)
I've been working on this problem and can't seem to figure out the proper solution. Ultimately I'm going to use dplyr in order to group by and apply a function to a column. I turned the column into a vector. Here is a snippet:
vec1 <- append(append(append(rep(1,3),rep(2,6)), rep(3,5)),rep(4,2))
if the number repeats more than 3 times, I want to change the following number to 1. So in the vector above, the number 2 occurs 6 times and the number 3 occurs 5 times. That means I want to replace the number 3 and 4 with 1. Ultimately in this snippet, the answer I'm looking for is:
c(1,1,1,2,2,2,2,2,2,1,1,1,1,1,1,1)
What I have below worked for cases when only one number was repeated more than 3 times, but not multiple. In addition, if I'm doing this inefficiently I'd like to learn how to better script it.
stack <- table(vec1)
stack1 <- list(as.numeric(rownames(data.frame(stack[stack>3]))) + 1)
replace(vec1,vec1 == stack1,1)
thanks in advance for any help
Try
inverse.rle(within.list(rle(vec1),
values[c(FALSE,(lengths >3)[-length(lengths)])] <- 1))
#[1] 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1
I'm using a for-loop to perform operations on specific subsets of my data. At the end of each iteration of the for loop, I have all the values that I need to fill a row of my dataframe.
So far I tried
df=NULL
for(...){
//stuff to calculate
newline=c(allthethingscalculated)
df=rbind(df,newline)
}
this results in the contents of the dataframe not being accessable using '$' , because the rows are then atomic vectors.
I also tried to append the values I get at the end of each iteration to an already existing vector and when the for loop ends create a dataframe from these vectors using but appending the values to the respective vector didn't work, the values weren't added.
x<-data.frame(a,b,c,d,...)
Any ideas on this?
Since my for loop iterates over IDs in my data, I realized I could do something like this:
uids=unique(data$id)
filler=c(1:length(uids))
df=data.frame(uids,filler,filler,filler,filler,filler,filler,filler,filler,filler)
for(i in uids){
...
df[i,]<-newline
}
I used filler to create a dataframe with the correct number of columns and rows so I don't get an error like 'replacement has length of 9, replacement has length of 1'
Is there a better way to do this? Using this approach I still have the values of filler in the respective row that I'd need to remove?
This should work, can your show us you data ?
R) x=data.frame(a=rep(1,3),b=rep(2,3),c=rep(3,3))
R) d=c(4,4,4)
R) rbind(x,d)
a b c
1 1 2 3
2 1 2 3
3 1 2 3
4 4 4 4
R) cbind(x,d)
a b c d
1 1 2 3 4
2 1 2 3 4
3 1 2 3 4