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).
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
This question already has answers here:
Convert comma separated string to integer in R
(3 answers)
Closed 1 year ago.
I am using a function where Timepoints need to be defined as
Timepoints = c(x,y,z)
Now i have a chr list
List
$ chr: "1,2,3,4,5,6,7"
with the timepoints i need to use, already seperated by commas.
I want to use this list in the function and lose the quotation marks, so the function can read my timepoints as
Timepoints= c(1,2,3,4,5,6,7)
I tried using noquote(List), but this is not accepted.
Am is missing something ? printing the list with noquote() results in the desired line of characters 1,2,3,4,5,6,7
1) Base R - scan Assuming that you have a list containing a single character string as shown in L below use scan as shown.
L <- list("1,2,3,4,5,6")
scan(text = L[[1]], sep = ",", quiet = TRUE)
## [1] 1 2 3 4 5 6
2) gsubfn::strapply Another possibility is to use strapply to match each string of digits, convert them to numeric and return it as a vector. (We assume that the numbers have no signs or decimal points but that could readily be added if needed.)
library(gsubfn)
strapply(L[[1]], "\\d+", as.numeric, simplify = unlist)
[1] 1 2 3 4 5 6
Added
In a comment the poster indicated an interest in having a list of character strings as input. The output was not specified but if we assume we want a list of numeric vectors then
L2 <- list(A = "1,2,3,4,5,6", B = "1,2")
Scan <- function(x) scan(text = x, sep = ",", quiet = TRUE)
lapply(L2, Scan)
## $A
## [1] 1 2 3 4 5 6
##
## $B
## [1] 1 2
library(gsubfn)
strapply(L2, "\\d", as.numeric)
## $A
## [1] 1 2 3 4 5 6
##
## $B
## [1] 1 2
Here is an option with strsplit.
as.integer(unlist(strsplit(L[[1]], ",")))
#[1] 1 2 3 4 5 6
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
I have a list:
list(c(1,2,3,4), c(3,2,6,8),c(6,4,3))
How would I be able to filter out for the list that contains 2 and 3 in each of the vectors? (They do not necessary have to be in descending/ ascending order)
Thank you!
Use Filter like this:
L <- list(c(1,2,3,4), c(3,2,6,8), c(6,4,3))
Filter(function(x) all(2:3 %in% x), L)
giving:
[[1]]
[1] 1 2 3 4
[[2]]
[1] 3 2 6 8
The above uses no packages but if we were to use fn from the gsubfn package it could be shortened to the following. The formula is regarded as the specification of a function whose body is the right hand side and whose arguments are the free variables in the body, in this case just x.
library(gsubfn)
fn$Filter(~ all(2:3 %in% x), L)
If we are using tidyverse, one option is keep from purrr
library(purrr)
keep(lst, ~all(2:3 %in% .x))
#[[1]]
#[1] 1 2 3 4
#[[2]]
#[1] 3 2 6 8
I would like to find the location of a character in a string.
Say: string = "the2quickbrownfoxeswere2tired"
I would like the function to return 4 and 24 -- the character location of the 2s in string.
You can use gregexpr
gregexpr(pattern ='2',"the2quickbrownfoxeswere2tired")
[[1]]
[1] 4 24
attr(,"match.length")
[1] 1 1
attr(,"useBytes")
[1] TRUE
or perhaps str_locate_all from package stringr which is a wrapper for gregexpr stringi::stri_locate_all (as of stringr version 1.0)
library(stringr)
str_locate_all(pattern ='2', "the2quickbrownfoxeswere2tired")
[[1]]
start end
[1,] 4 4
[2,] 24 24
note that you could simply use stringi
library(stringi)
stri_locate_all(pattern = '2', "the2quickbrownfoxeswere2tired", fixed = TRUE)
Another option in base R would be something like
lapply(strsplit(x, ''), function(x) which(x == '2'))
should work (given a character vector x)
Here's another straightforward alternative.
> which(strsplit(string, "")[[1]]=="2")
[1] 4 24
You can make the output just 4 and 24 using unlist:
unlist(gregexpr(pattern ='2',"the2quickbrownfoxeswere2tired"))
[1] 4 24
find the position of the nth occurrence of str2 in str1(same order of parameters as Oracle SQL INSTR), returns 0 if not found
instr <- function(str1,str2,startpos=1,n=1){
aa=unlist(strsplit(substring(str1,startpos),str2))
if(length(aa) < n+1 ) return(0);
return(sum(nchar(aa[1:n])) + startpos+(n-1)*nchar(str2) )
}
instr('xxabcdefabdddfabx','ab')
[1] 3
instr('xxabcdefabdddfabx','ab',1,3)
[1] 15
instr('xxabcdefabdddfabx','xx',2,1)
[1] 0
To only find the first locations, use lapply() with min():
my_string <- c("test1", "test1test1", "test1test1test1")
unlist(lapply(gregexpr(pattern = '1', my_string), min))
#> [1] 5 5 5
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(min) %>%
unlist()
#> [1] 5 5 5
To only find the last locations, use lapply() with max():
unlist(lapply(gregexpr(pattern = '1', my_string), max))
#> [1] 5 10 15
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(max) %>%
unlist()
#> [1] 5 10 15
You could use grep as well:
grep('2', strsplit(string, '')[[1]])
#4 24