I have a table with the cuts in intervals like:
bin targets casos prop phyp logit
(-2,-1] 193 6144 0.0314 0 -3.4286244
(-1,3] 128 431 0.2970 1 -0.8617025
(3,11] 137 245 0.5592 1 0.2378497
I want to get the original cuts. I tried with:
a<-strsplit(as.character(pl$table[,'bin']), ' ')
And then I tried to split each row with:
lapply(a, function(x) strsplit(x, ",")[1] )
But I don't get the expected result, which is:
(-1,3,11)
Is there a better way to achieve this? What else do I need to do to get to the result?
Thanks.
If your data is consistently in this format, you could use gsub().
df <- data.frame(bin = c('(-2,-1]','(1,3]','(3,11]'),
targets = c(193, 128, 137),
casos = c(6144, 431, 245),
prop = c(0.0314, 0.297, 0.5592),
phyp = c(0,1,1),
logit = c(-3.4286244,-0.8617025, 0.2378497), stringsAsFactors = F)
a <- strsplit(df$bin, ',')
sapply(a, function(x) gsub("]", "", x))[2,]
sapply(a, function(x) gsub("\\(", "", x))[1,]
Which gives you
[1] "-1" "3" "11"
[1] "-2" "1" "3"
In your example, there are more bounds than you say you are hoping to retrieve. This will give you all bounds:
d <- read.table(text=' bin targets casos prop phyp logit
"(-2,-1]" 193 6144 0.0314 0 -3.4286244
"(1,3]" 128 431 0.2970 1 -0.8617025
"(3,11]" 137 245 0.5592 1 0.2378497', header=T)
strings <- as.character(levels(d$bin))
strings <- substr(strings, 2, nchar(strings)-1)
unique(unlist(strsplit(strings, ",")))
# [1] "-2" "-1" "1" "3" "11"
If you only wanted the upper bounds, this will work:
strings <- as.character(levels(d$bin))
strings <- sapply(strsplit(strings, ","), function(l){ l[2] })
strings <- substr(strings, 1, nchar(strings)-1)
unique(strings)
# [1] "-1" "3" "11"
Another way would be:
a<-strsplit(as.character(pl$table[,'bin']), ' ')
lapply(a, function(x) unlist(strsplit(x, ",|]"))[2])
Related
I want to create a function such that, given a vector of n numbers, create all possible combinations from them (the same expand.grid() does, really).
For example, given the vector [1, 2] I want to create a function such that it outputs all possible combinations, i.e., "11", "12", "21", "21".
This is what I've come up with:
V <- trunc(runif(3, 0, 9))
FE2 <- function(V){
C <- c()
n <- 0
for(i in 1:length(V)){
for (j in 1:length(V)){
for (k in 1:length(V)){
C <- c(paste0(V[i], V[j], V[k]))
n <- n + 1
print(c(paste0("Number of combination: ", n), paste0("Combination: ", C)))
}
}
}
}
FE2(V)
The function does work. The problem is that for each element of the original vector, I have to add another for(). That is, if I wanted to compute all possible combinations for a vector of 10 elements, say [1, 2, ..., 10] I would have to create 10 for()-loops in my function. I wonder if there's another, more efficient way to do it, as long as it does not significantly differs from my actual solution.
With expand.grid you then need to "collapse" the rows:
apply( expand.grid( 1:2, 1:2), 1, paste0, collapse="")
[1] "11" "21" "12" "22"
I'm not sure if I understood the goals but here's that method applied to 1:10 with 2 and then 3 instances of the vector.
> str(apply( expand.grid( 1:10, 1:10), 1, paste0, collapse="") )
chr [1:100] "11" "21" "31" "41" "51" "61" "71" "81" "91" "101" "12" "22" "32" "42" "52" "62" "72" "82" "92" ...
> str(apply( expand.grid( 1:10, 1:10, 1:10), 1, paste0, collapse="") )
chr [1:1000] "111" "211" "311" "411" "511" "611" "711" "811" "911" "1011" "121" "221" "321" "421" "521" "621" ...
x <- 1:3
Use Reduce to iteratively apply the same operation
op <- function (a, b) {
paste(rep(a, times = length(b)), rep(b, each = length(a)), sep = "-")
}
Reduce(op, rep(list(x), length(x)))
# [1] "1-1-1" "2-1-1" "3-1-1" "1-2-1" "2-2-1" "3-2-1" "1-3-1" "2-3-1" "3-3-1"
#[10] "1-1-2" "2-1-2" "3-1-2" "1-2-2" "2-2-2" "3-2-2" "1-3-2" "2-3-2" "3-3-2"
#[19] "1-1-3" "2-1-3" "3-1-3" "1-2-3" "2-2-3" "3-2-3" "1-3-3" "2-3-3" "3-3-3"
Generate permutations
library(RcppAlgos)
permuteGeneral(x, length(x), repetition = TRUE,
FUN = function (x) paste0(x, collapse = "-"))
data <- c("Demand = 001 979", "Demand = -08 976 (154)", "Demand = -01 975 (359)")
data <- str_match(data, pattern = ("Demand = (.*) (.*)"))
I need to extract the first 2 sets of numbers (including the - sign) into columns using str_match.
Exclude 3rd set of numbers in bracket ().
Any help is welcomed.
Output:
## [1] "001" "-08" "-01"
## [2] "979" "976" "975"
How about removing everything else?
data <- c("Demand = 001 979", "Demand = -08 976 (154)", "Demand = -01 975 (359)")
data <- gsub("Demand = ", "", x = data)
data <- trimws(gsub("\\(.*\\)", "", x = data))
out <- list()
out[[1]] <- sapply(data, "[", 1)
out[[2]] <- sapply(data, "[", 2)
out
[[1]]
[1] "001" "-08" "-01"
[[2]]
[1] "979" "976" "975"
A possibility with str_extract_all() from stringr:
sapply(str_extract_all(x, "-?[0-9]+?[0-9]*"), function(x) x[1])
[1] "001" "-08" "-01"
sapply(str_extract_all(x, "-?[0-9]+?[0-9]*"), function(x) x[2])
[1] "979" "976" "975"
Or using the idea of #Roman Luštrik with strsplit():
sapply(strsplit(gsub("Demand = ", "", x), " "), function(x) x[1])
[1] "001" "-08" "-01"
I have a vector of strings, similar to this one, but with many more elements:
s <- c("CGA-DV-558_T_90.67.0_DV_1541_07", "TC-V-576_T_90.0_DV_151_0", "TCA-DV-X_T_6.0_D_A2_07", "T-V-Z_T_2_D_A_0", "CGA-DV-AW0_T.1_24.4.0_V_A6_7", "ACGA-DV-A4W0_T_274.46.0_DV_A266_07")
And I would like to use a function that extracts the string between the nth and ith instances of the delimiter "_". For example, the string between the 2nd (n = 2) and 3rd (i = 3) instances, to get this:
[1] "90.67.0" "90.0" "6.0" "2" "24.4.0" "274.46.0"
Or if n = 4 and i = 5"
[1] "1541" "151" "A2" "A" "A" "A266"
Any suggestions? Thank you for your help!
You can do this with gsub
n = 2
i = 3
pattern1 = paste0("(.*?_){", n, "}")
temp = gsub(pattern1, "", s)
pattern2 = paste0("((.*?_){", i-n, "}).*")
temp = gsub(pattern2, "\\1", temp)
temp = gsub("_$", "", temp)
[1] "1541" "151" "A2" "A" "A6" "A266"
#FUNCTION
foo = function(x, n, i){
do.call(c, lapply(x, function(X)
paste(unlist(strsplit(X, "_"))[(n+1):(i)], collapse = "_")))
}
#USAGE
foo(x = s, n = 3, i = 5)
#[1] "DV_1541" "DV_151" "D_A2" "D_A" "V_A6" "DV_A266"
A third method, that uses substring for the extraction and gregexpr to find the positions is
# extract postions of "_" from each vector element, returns a list
spots <- gregexpr("_", s, fixed=TRUE)
# extract text in between third and fifth underscores
substring(s, sapply(spots, "[", 3) + 1, sapply(spots, "[", 5) - 1)
"DV_1541" "DV_151" "D_A2" "D_A" "V_A6" "DV_A266"
Given is vector:
vec <- c(LETTERS[1:10])
I would like to be able to combine it in a following manner:
resA <- c("AB", "CD", "EF", "GH", "IJ")
resB <- c("ABCDEF","GHIJ")
where elements of the vector vec are merged together according to the desired size of a new element constituting the resulting vector. This is 2 in case of resA and 5 in case of resB.
Desired solution characteristics
The solution should allow for flexibility with respect to the element sizes, i.e. I may want to have vectors with elements of size 2 or 20
There may be not enough elements in the vector to match the desired chunk size, in that case last element should be shortened accordingly (as shown)
This is shouldn't make a difference but the solution should work on words as well
Attempts
Initially, I was thinking of using something on the lines:
c(
paste0(vec[1:2], collapse = ""),
paste0(vec[3:4], collapse = ""),
paste0(vec[5:6], collapse = "")
# ...
)
but this would have to be adapted to jump through the remaining pairs/bigger groups of the vec and handle last group which often would be of a smaller size.
Here is what I came up with. Using Harlan's idea in this question, you can split the vector in different number of chunks. You also want to use your paste0() idea in lapply() here. Finally, you unlist a list.
unlist(lapply(split(vec, ceiling(seq_along(vec)/2)), function(x){paste0(x, collapse = "")}))
# 1 2 3 4 5
#"AB" "CD" "EF" "GH" "IJ"
unlist(lapply(split(vec, ceiling(seq_along(vec)/5)), function(x){paste0(x, collapse = "")}))
# 1 2
#"ABCDE" "FGHIJ"
unlist(lapply(split(vec, ceiling(seq_along(vec)/3)), function(x){paste0(x, collapse = "")}))
# 1 2 3 4
#"ABC" "DEF" "GHI" "J"
vec <- c(LETTERS[1:10])
f1 <- function(x, n){
f <- function(x) paste0(x, collapse = '')
regmatches(f(x), gregexpr(f(rep('.', n)), f(x)))[[1]]
}
f1(vec, 2)
# [1] "AB" "CD" "EF" "GH" "IJ"
or
f2 <- function(x, n)
apply(matrix(x, nrow = n), 2, paste0, collapse = '')
f2(vec, 5)
# [1] "ABCDE" "FGHIJ"
or
f3 <- function(x, n) {
f <- function(x) paste0(x, collapse = '')
strsplit(gsub(sprintf('(%s)', f(rep('.', n))), '\\1 ', f(x)), '\\s+')[[1]]
}
f3(vec, 4)
# [1] "ABCD" "EFGH" "IJ"
I would say the last is best of these since n for the others must be a factor or you will get warnings or recycling
edit - more
f4 <- function(x, n) {
f <- function(x) paste0(x, collapse = '')
Vectorize(substring, USE.NAMES = FALSE)(f(x), which((seq_along(x) %% n) == 1),
which((seq_along(x) %% n) == 0))
}
f4(vec, 2)
# [1] "AB" "CD" "EF" "GH" "IJ"
or
f5 <- function(x, n)
mapply(function(x) paste0(x, collapse = ''),
split(x, c(0, head(cumsum(rep_len(sequence(n), length(x)) %in% n), -1))),
USE.NAMES = FALSE)
f5(vec, 4)
# [1] "ABCD" "EFGH" "IJ"
Here is another way, working with the original array.
A side note, working with words is not straightforward, since there is at least two ways to understand it: you can either keep each word separately or collapse them first an get individual characters. The next function can deal with both options.
vec <- c(LETTERS[1:10])
vec2 <- c("AB","CDE","F","GHIJ")
cuts <- function(x, n, bychar=F) {
if (bychar) x <- unlist(strsplit(paste0(x, collapse=""), ""))
ii <- seq_along(x)
li <- split(ii, ceiling(ii/n))
return(sapply(li, function(y) paste0(x[y], collapse="")))
}
cuts(vec2,2,F)
# 1 2
# "ABCDE" "FGHIJ"
cuts(vec2,2,T)
# 1 2 3 4 5
# "AB" "CD" "EF" "GH" "IJ"
I use LETTERS most of the time for my factors but today I tried to go beyond 26 characters:
LETTERS[1:32]
Expecting there to be an automatic recursive factorization AA, AB, AC... But was disappointed. Is this simply a limitation of LETTERS or is there a way to get what I'm looking for using another function?
Would 702 be enough?
LETTERS702 <- c(LETTERS, sapply(LETTERS, function(x) paste0(x, LETTERS)))
If not, how about 18,278?
MOAR_LETTERS <- function(n=2) {
n <- as.integer(n[1L])
if(!is.finite(n) || n < 2)
stop("'n' must be a length-1 integer >= 2")
res <- vector("list", n)
res[[1]] <- LETTERS
for(i in 2:n)
res[[i]] <- c(sapply(res[[i-1L]], function(y) paste0(y, LETTERS)))
unlist(res)
}
ml <- MOAR_LETTERS(3)
str(ml)
# chr [1:18278] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K" "L" "M" "N" "O" ...
This solution uses recursion. Usage is a bit different in the sense MORELETTERS is not a long vector you will have to store and possibly expand as your inputs get larger. Instead, it is a function that converts your numbers into the new base.
extend <- function(alphabet) function(i) {
base10toA <- function(n, A) {
stopifnot(n >= 0L)
N <- length(A)
j <- n %/% N
if (j == 0L) A[n + 1L] else paste0(Recall(j - 1L, A), A[n %% N + 1L])
}
vapply(i-1L, base10toA, character(1L), alphabet)
}
MORELETTERS <- extend(LETTERS)
MORELETTERS(1:1000)
# [1] "A" "B" ... "ALL"
MORELETTERS(c(1, 26, 27, 1000, 1e6, .Machine$integer.max))
# [1] "A" "Z" "AA" "ALL" "BDWGN" "FXSHRXW"
You can make what you want like this:
LETTERS2<-c(LETTERS[1:26], paste0("A",LETTERS[1:26]))
Another solution for excel style column names, generalized to any number of letters
#' Excel Style Column Names
#'
#' #param n maximum number of letters in column name
excel_style_colnames <- function(n){
unlist(Reduce(
function(x, y) as.vector(outer(x, y, 'paste0')),
lapply(1:n, function(x) LETTERS),
accumulate = TRUE
))
}
A variant on eipi10's method (ordered correctly) using data.table:
library(data.table)
BIG_LETTERS <- c(LETTERS,
do.call("paste0",CJ(LETTERS,LETTERS)),
do.call("paste0",CJ(LETTERS,LETTERS,LETTERS)))
Yet another option:
l2 = c(LETTERS, sort(do.call("paste0", expand.grid(LETTERS, LETTERS[1:3]))))
Adjust the two instances of LETTERS inside expand.grid to get the number of letter pairs you'd like.
A function to produce Excel-style column names, i.e.
# A, B, ..., Z, AA, AB, ..., AZ, BA, BB, ..., ..., ZZ, AAA, ...
letterwrap <- function(n, depth = 1) {
args <- lapply(1:depth, FUN = function(x) return(LETTERS))
x <- do.call(expand.grid, args = list(args, stringsAsFactors = F))
x <- x[, rev(names(x)), drop = F]
x <- do.call(paste0, x)
if (n <= length(x)) return(x[1:n])
return(c(x, letterwrap(n - length(x), depth = depth + 1)))
}
letterwrap(26^2 + 52) # through AAZ
## This will take a few seconds:
# x <- letterwrap(1e6)
It's probably not the fastest, but it extends indefinitely and is nicely predictable. Took about 20 seconds to produce through 1 million, BDWGN.
(For a few more details, see here: https://stackoverflow.com/a/21689613/903061)
A little late to the party, but I want to play too.
You can also use sub, and sprintf in place of paste0 and get a length 702 vector.
c(LETTERS, sapply(LETTERS, sub, pattern = " ", x = sprintf("%2s", LETTERS)))
Here's another addition to the list. This seems a bit faster than Gregor's (comparison done on my computer - using length.out = 1e6 his took 12.88 seconds, mine was 6.2), and can also be extended indefinitely. The flip side is that it's 2 functions, not just 1.
make.chars <- function(length.out, case, n.char = NULL) {
if(is.null(n.char))
n.char <- ceiling(log(length.out, 26))
m <- sapply(n.char:1, function(x) {
rep(rep(1:26, each = 26^(x-1)) , length.out = length.out)
})
m.char <- switch(case,
'lower' = letters[m],
'upper' = LETTERS[m]
)
m.char <- LETTERS[m]
dim(m.char) <- dim(m)
apply(m.char, 1, function(x) paste(x, collapse = ""))
}
get.letters <- function(length.out, case = 'upper'){
max.char <- ceiling(log(length.out, 26))
grp <- rep(1:max.char, 26^(1:max.char))[1:length.out]
unlist(lapply(unique(grp), function(n) make.chars(length(grp[grp == n]), case = case, n.char = n)))
}
##
make.chars(5, "lower", 2)
#> [1] "AA" "AB" "AC" "AD" "AE"
make.chars(5, "lower")
#> [1] "A" "B" "C" "D" "E"
make.chars(5, "upper", 4)
#> [1] "AAAA" "AAAB" "AAAC" "AAAD" "AAAE"
tmp <- get.letters(800)
head(tmp)
#> [1] "A" "B" "C" "D" "E" "F"
tail(tmp)
#> [1] "ADO" "ADP" "ADQ" "ADR" "ADS" "ADT"
Created on 2019-03-22 by the reprex package (v0.2.1)