I'd like to produce a vector with the following repeating pattern:
1 1 2 1 2 3 1 2 3 4 ...
that ranges from one to some arbitrary stopping point.
I can hack it together using an sapply followed by an unlist, as in the following, but it sure feels like there should be a base call that is more direct than this.
repeating_function <- function(stop_point) {
res_list <- sapply(1:stop_point, FUN=function(x) {1:x}, simplify=TRUE)
res <- unlist(res_list)
return(res)
}
Which produces:
repeating_function(5)
[1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
An easier option would be
sequence(sequence(5))
#[1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
Wrapping in a function
repeating_function(val) {
sequence(sequence(val))
}
Related
There is a wide data set, a simple example is
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
I'm looking for a way to assign the values in "ax" to column "bx" and "cx". Here, imagine we have thousands of columns we intend to replace with "ax", so I want this to be done in an automated approach using R. The expected output look like
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(1,2,2,3,4,4),
"cx"=c(1,2,2,3,4,4))
I've thought of, and tried using mutate_at and ends_with, but this has not work for me. For example, I tried
df %>%
mutate_at(vars(ends_with("x")), labels = "ax")
and this prints an error. Not sure what's wrong or what's to be added to get this working, so I would like to request your help on this. Thank you very much!
A simple way using base R would be :
change_cols <- grep('x$', names(df))
df[change_cols] <- df$ax
df
# id ax bx cx
#1 1 1 1 1
#2 2 2 2 2
#3 3 2 2 2
#4 4 3 3 3
#5 5 4 4 4
#6 6 4 4 4
I would suggest this tidyverse approach using across() to select the range of variables you want:
library(tidyverse)
#Data
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
#Mutate
df %>% mutate(across(c(bx:cx), ~ ax))
Output:
id ax bx cx
1 1 1 1 1
2 2 2 2 2
3 3 2 2 2
4 4 3 3 3
5 5 4 4 4
6 6 4 4 4
Another option with mutate_at()
df %>%
mutate_at(vars(matches("x$")), ~ax)
# id ax bx cx
# 1 1 1 1 1
# 2 2 2 2 2
# 3 3 2 2 2
# 4 4 3 3 3
# 5 5 4 4 4
# 6 6 4 4 4
I'd think this would be simple using the rev() and seq() functions, but am struggling to get the reverse order part correct.
I'm trying to get 5432101234543210 from 5:0.
Not too hard to set as a function...
try_it <- function(x) {
c(rev(x), x[2:length(x-1)], rev(x)[2:length(x-1)])
}
try_it(0:5)
# [1] 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0
Edit
Extend function to have variable repeats
try_it <- function(x, reps) {
c(rev(x), rep(c(x[2:length(x-1)], rev(x)[2:length(x-1)]), (reps - 1) / 2))
}
try_it(0:5, 5)
# [1] 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0
Note: I've not worked hard to generalise this extension, it will not return the correct length for an even number of repetitions. I'm sure you could modify to suit your requirements.
This is my desired output:
> head(df)
String numSubStrings
1 1 1
2 1 1
3 1;1;1;1 4
4 1;1;1;1 4
5 1;1;1 3
6 1 1
Hi, I have a data frame which has a "String" column as above. I would like to add a column "numSubStrings" which contains the number of substrings separated by ";" in "String".
I have tried
lapply(df, transform, numSubStrings=length(strsplit(df$Strings,";")[[1]]))
which gives me 1s in the numSubStrings instead.
Please advice.
Thanks.
It sounds like you're looking for count.fields. Usage would be something like:
> count.fields(textConnection(mydf$String), sep = ";")
[1] 1 1 4 4 3 1
You may need to wrap the mydf$String in as.character, depending on how the data were read in or created.
Or, you can try lengths:
> lengths(strsplit(mydf$String, ";", TRUE))
[1] 1 1 4 4 3 1
We can use gsub to remove all the characters except the ; and count the ; with nchar
df$numSubStrings <- nchar(gsub('[^;]+', '', df$String))+1
df$numSubStrings
#[1] 1 1 4 4 3 1
Or another option is stri_count from library(stringi) to count the ; characters and add 1.
library(stringi)
stri_count_fixed(df$String, ';')+1
#[1] 1 1 4 4 3 1
You may use str_count from stringr package.
x <- " String
1 1
2 1
3 1;1;1;1
4 1;1;1;1
5 1;1;1
6 1 "
df <- read.table(text=x, header=T)
df$numSubStrings <- str_count(df$String, "[^;]+")
df
# String numSubStrings
# 1 1 1
# 2 1 1
# 3 1;1;1;1 4
# 4 1;1;1;1 4
# 5 1;1;1 3
# 6 1 1
Basically I want to generate a sequence, say:
n is 2, the sequence will be 112
n is 3, sequence is 112123
n is 5, sequence is 112123123412345
I did come up with a solution
n=5
seq=1
for (i in 2:n){
seq=c(seq,rep(1:n,len=i))
}
I am wondering if there is a way can do it without for loop?
Use sequence:
> sequence(1:5)
[1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
Here is one possibility:
n<-5
unlist(lapply(1:n,function(x) 1:x))
## [1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
It'd do something like:
do.call('c', sapply(1:5, seq, from = 1))
# [1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
I misread the question as "how to generate that annoying puzzler sequence," which goes
1,11,21,1112,3112,... :-). So I figured I might as well write a solution to that.
puzseq<-function(seqlen) {
theseq<- list(1)
for( j in 2:seqlen) {
thetab<-table(theseq[[j-1]])
theseq[[j]]<-unlist( sapply( 1:length(thetab), function(k) c(thetab[k], as.numeric(names(thetab)[k])) ))
}
return(theseq)
}
Basically I want to generate a sequence, say:
n is 2, the sequence will be 112
n is 3, sequence is 112123
n is 5, sequence is 112123123412345
I did come up with a solution
n=5
seq=1
for (i in 2:n){
seq=c(seq,rep(1:n,len=i))
}
I am wondering if there is a way can do it without for loop?
Use sequence:
> sequence(1:5)
[1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
Here is one possibility:
n<-5
unlist(lapply(1:n,function(x) 1:x))
## [1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
It'd do something like:
do.call('c', sapply(1:5, seq, from = 1))
# [1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
I misread the question as "how to generate that annoying puzzler sequence," which goes
1,11,21,1112,3112,... :-). So I figured I might as well write a solution to that.
puzseq<-function(seqlen) {
theseq<- list(1)
for( j in 2:seqlen) {
thetab<-table(theseq[[j-1]])
theseq[[j]]<-unlist( sapply( 1:length(thetab), function(k) c(thetab[k], as.numeric(names(thetab)[k])) ))
}
return(theseq)
}