I'm trying to create a calculator that multiplies permutation groups written in cyclic form (the process of which is described in this post, for anyone unfamiliar: https://math.stackexchange.com/questions/31763/multiplication-in-permutation-groups-written-in-cyclic-notation). Although I know this would be easier to do with Python or something else, I wanted to practice writing code in R since it is relatively new to me.
My gameplan for this is take an input, such as "(1 2 3)(2 4 1)" and split it into two separate lists or vectors. However, I am having trouble starting this because from my understanding of character functions (which I researched here: https://www.statmethods.net/management/functions.html) I will ultimately have to use the function grep() to find the points where ")(" occur in my string to split from there. However, grep only takes vectors for its argument, so I am trying to coerce my string into a vector. In researching this problem, I have mostly seen people suggest to use as.integer(unlist(str_split())), however, this doesn't work for me as when I split, not everything is an integer and the values become NA, as seen in this example.
library(tidyverse)
x <- "(1 2 3)(2 4 1)"
x <- as.integer(unlist(str_split(x," ")))'
x
Is there an alternative way to turn a string into a vector when there are not just integers involved? I also realize that the means by which I am trying to split up the two permutations is very roundabout, but that is because of the character functions that I researched this seems like the only way. If there are other functions that would make this easier, please let me know.
Thank you!
Comments in the code.
x <- "(1 2 3)(2 4 1)"
out1 <- strsplit(x, split = ")(", fixed = TRUE)[[1]] # split on close and open bracket
out2 <- gsub("[\\(|\\)]", replacement = "", out1) # remove brackets
out3 <- strsplit(out2, " ") # tease out numbers between spaces
lapply(out3, as.integer)
[[1]]
[1] 1 2 3
[[2]]
[1] 2 4 1
There aren't really any scalars on R. Single values like 1, TRUE, and "a" are all 1-element vectors. grep(pattern, x) will work fine on your original string. As a starting point for getting towards your desired goal, I would suggest splitting the groups using:
> str_extract_all(x, "\\([0-9 ]+\\)")
[[1]]
[1] "(1 2 3)" "(2 4 1)"
If we need to split the strings with the brackets
strsplit(x, "(?<=\\))(?=\\()", perl = TRUE)[[1]]
#[1] "(1 2 3)" "(2 4 1)"
Or we can use convenient wrapper from qdapRegex
library(qdapRegex)
ex_round(x, include.marker = TRUE)[[1]]
#[1] "(1 2 3)" "(2 4 1)"
alternative: using library(magrittr)
x <- "(1 2 3)(2 4 1)"
x %>%
gsub("^\\(","c(",.) %>% gsub("\\)\\(","),c(",.) %>% gsub("(?=\\s\\d)",", ",.,perl=T) %>%
paste0("list(",.,")") %>% {eval(parse(text=.))}
result:
# [[1]]
# [1] 1 2 3
#
# [[2]]
# [1] 2 4 1
You could use chartr with read.table :
read.table(text= chartr("()"," \n",x))
# V1 V2 V3
# 1 1 2 3
# 2 2 4 1
Related
Suppose I have a long vector with characters which is more or less like this:
vec <- c("32, 25", "5", "15, 24")
I want to apply a function which give me the number of strings for any element separated by a comma and returns me a vector with any individual length. Using lapply and my toy vector, this is my approach:
lapply(vec, function(x) {
a <- strsplit(x, ",")
y <- length(a[[1:length(a)]])
unlist(y[1:length(y)])
})
[[1]]
[1] 2
[[2]]
[1] 1
[[3]]
[1] 2
This almost gives me what I want since first element has 2 strings, second element 1 string and third element 2 strings. The problem is I can't achieve that my function returns me a vector of the form c(2,1,2). I'm using this function to create a new variable on some data.frame which I'm working with.
Any idea will be much appreciated.
You could do:
stringr::str_count(vec, ",") + 1
#> [1] 2 1 2
Or, in base R:
nchar(gsub("[^,]", "", vec)) + 1
#> [1] 2 1 2
I have a list in R and I wanted to know how to get the element in it with the most characters.
I think it's something using which() and nchar()? This is what I tried:
cnt <- sapply(unformatted_list, nchar)
unformatted_list[which.max(cnt)]
If you don't care about ties:
L <- c("Apple","Banana","Monkey","Drugs")
which.max(nchar(L))
[1] 2
If you care about ties:
which(nchar(L) %in% max(nchar(L)))
[1] 2 3
I am trying to paste together the rowname along with the data in the desired column. I wrote the following code but somehow couldnot find a way to do it correctly.
The desired output will be: "a,1,11" "b,2,22" "c,3,33"
x = data.frame(cbind(f1 = c(1,2,3), f2 = c(5,6,7), f3=c(11,22,33)), row.names= c('a','b','c'))
x
# f1 f2 f3
# a 1 5 11
# b 2 6 22
# c 3 7 33
do.call("paste", c(rownames(x), x[c('f1','f3')], sep=","))
# [1] "a,b,c,1,11" "a,b,c,2,22" "a,b,c,3,33"
Two main points:
Use apply instead of do.call(paste, .)
Use cbind instead of c in this case.
If you would rather use c, you would need to coerce the row names to a list or column first, eg: c(list(rownames(x)), x)
Try the following:
apply(cbind(rownames(x), x[c('f1','f3')]), 1, paste, collapse=",")
a b c
"a,1,11" "b,2,22" "c,3,33"
Your do.call instructs R to paste the list c(rownames(x), x[c('f1','f3')]) together. But take a look at your list.
> c(rownames(x), x[c('f1','f3')])
[[1]]
[1] "a"
[[2]]
[1] "b"
[[3]]
[1] "c"
$f1
[1] 1 2 3
$f3
[1] 11 22 33
The c command takes the elements of each argument and joins them together. This properly deconstructs x[c('f1','f3')] but also deconstructs rownames(x) in a way you don't want. Obeying the standard recycling rule, paste then takes an item from each list element and patches them together with sep=",".
You could fix this by encapsulating rownames(x) inside a list structure so that your list of arguments comes out properly:
do.call("paste", c(list(rownames(x)), x[c('f1','f3')], sep=","))
No need for do.call or apply:
paste(rownames(x),x[[1]],x[[3]] , sep=",")
[1] "a,1,11" "b,2,22" "c,3,33"
I have a list of strings which contain random characters such as:
list=list()
list[1] = "djud7+dg[a]hs667"
list[2] = "7fd*hac11(5)"
list[3] = "2tu,g7gka5"
I'd like to know which numbers are present at least once (unique()) in this list. The solution of my example is:
solution: c(7,667,11,5,2)
If someone has a method that does not consider 11 as "eleven" but as "one and one", it would also be useful. The solution in this condition would be:
solution: c(7,6,1,5,2)
(I found this post on a related subject: Extracting numbers from vectors of strings)
For the second answer, you can use gsub to remove everything from the string that's not a number, then split the string as follows:
unique(as.numeric(unlist(strsplit(gsub("[^0-9]", "", unlist(ll)), ""))))
# [1] 7 6 1 5 2
For the first answer, similarly using strsplit,
unique(na.omit(as.numeric(unlist(strsplit(unlist(ll), "[^0-9]+")))))
# [1] 7 667 11 5 2
PS: don't name your variable list (as there's an inbuilt function list). I've named your data as ll.
Here is yet another answer, this one using gregexpr to find the numbers, and regmatches to extract them:
l <- c("djud7+dg[a]hs667", "7fd*hac11(5)", "2tu,g7gka5")
temp1 <- gregexpr("[0-9]", l) # Individual digits
temp2 <- gregexpr("[0-9]+", l) # Numbers with any number of digits
as.numeric(unique(unlist(regmatches(l, temp1))))
# [1] 7 6 1 5 2
as.numeric(unique(unlist(regmatches(l, temp2))))
# [1] 7 667 11 5 2
A solution using stringi
# extract the numbers:
nums <- stri_extract_all_regex(list, "[0-9]+")
# Make vector and get unique numbers:
nums <- unlist(nums)
nums <- unique(nums)
And that's your first solution
For the second solution I would use substr:
nums_first <- sapply(nums, function(x) unique(substr(x,1,1)))
You could use ?strsplit (like suggested in #Arun's answer in Extracting numbers from vectors (of strings)):
l <- c("djud7+dg[a]hs667", "7fd*hac11(5)", "2tu,g7gka5")
## split string at non-digits
s <- strsplit(l, "[^[:digit:]]")
## convert strings to numeric ("" become NA)
solution <- as.numeric(unlist(s))
## remove NA and duplicates
solution <- unique(solution[!is.na(solution)])
# [1] 7 667 11 5 2
A stringr solution with str_match_all and piped operators. For the first solution:
library(stringr)
str_match_all(ll, "[0-9]+") %>% unlist %>% unique %>% as.numeric
Second solution:
str_match_all(ll, "[0-9]") %>% unlist %>% unique %>% as.numeric
(Note: I've also called the list ll)
Use strsplit using pattern as the inverse of numeric digits: 0-9
For the example you have provided, do this:
tmp <- sapply(list, function (k) strsplit(k, "[^0-9]"))
Then simply take a union of all `sets' in the list, like so:
tmp <- Reduce(union, tmp)
Then you only have to remove the empty string.
Check out the str_extract_numbers() function from the strex package.
pacman::p_load(strex)
list=list()
list[1] = "djud7+dg[a]hs667"
list[2] = "7fd*hac11(5)"
list[3] = "2tu,g7gka5"
charvec <- unlist(list)
print(charvec)
#> [1] "djud7+dg[a]hs667" "7fd*hac11(5)" "2tu,g7gka5"
str_extract_numbers(charvec)
#> [[1]]
#> [1] 7 667
#>
#> [[2]]
#> [1] 7 11 5
#>
#> [[3]]
#> [1] 2 7 5
unique(unlist(str_extract_numbers(charvec)))
#> [1] 7 667 11 5 2
Created on 2018-09-03 by the reprex package (v0.2.0).
I have a csv that looks like
Deamon,Host,1:2:4,aaa.03
Pixe,Paradigm,1:3:5,11.us
I need to read this into a dataframe for analysis but the 3rd column in my data is separated by : and need to be read like a set or list 1.e splitted by : so that it returns (1,2,4) . Is it possible to have a columns that has a class list in R . Or How best do you think i can approach this problem.
You can use strsplit to split a character vector into a list of components:
x <- c("1:2:4", "1:3:5")
strsplit(x, split=":")
[[1]]
[1] "1" "2" "4"
[[2]]
[1] "1" "3" "5"
As noted above, the answer will vary depending on if the number of separators in the columns are consistent or not. The answer is more straight forward if that number is consistent. Here's one answer to do that building off of Andrie's strsplit answer:
dat <- read.csv("yourData.csv", header=FALSE, stringsAsFactors = FALSE)
#If always going to be a consistent number of separators
dat <- cbind(dat, do.call("rbind", strsplit(dat[, 3], ":")))
V1 V2 V3 V4 1 2 3
1 Deamon Host 1:02:04 aaa.03 1 02 04
2 Pixe Paradigm 1:03:05 11.us 1 03 05
Note that the above is essentially how colsplit.character from package reshape is implemented and may be a better option for you as it forces you to give proper names.
If the number of separators is different, then using rbind.fill is an option from package plyr. rbind.fill expects data.frames which was a bit annoying, and I couldn't figure out how to get a one row data.frame without first converting to a matrix, so I imagine this can be made more efficient, but here's the basic idea:
library(plyr)
x <- c("1:2:4", "1:3:5:6:7")
rbind.fill(
lapply(
lapply(strsplit(x, ":"), matrix, nrow = 1)
, as.data.frame)
)
V1 V2 V3 V4 V5
1 1 2 4 <NA> <NA>
2 1 3 5 6 7
Which can then be cbinded as shown above.
Try using gsub to replace that character:
R> str <- "1:2:4"
R> str
[1] "1:2:4"
R> gsub(":", ",", str)
[1] "1,2,4"
Make sure the column is a string not a factor beforehand.