Clash Royale CLAN TAG#URR8PPP

All strings in an alpha-numeric range in R

In R, how can I get a vector of strings given a "range" of strings? (equivalent of x:y, but allowing x and y to be strings of the same length) A couple examples...

"A001":"A003" == c("A001","A002","A003")

"A99":"B02" == c("A99","B00","B01","B02")

update:
I can get "A01":"A10" using

paste0('A',sprintf("%02d",1:10))
[1] "A01" "A02" "A03" "A04" "A05" "A06" "A07" "A08" "A09" "A10"


But I'm not sure how to get from A to B (ie "A99:B02") in a seamless way.

3 Answers
3

Yep! Here's a working example for ya. The outer function that generates a matrix, but you can always fix with nested list and unlist wrappers

outer

list

unlist

You could do this with a single nums assignment, except you've specified you want two-digit codes, hence the extra paste0 for 1:10

nums

paste0

Note that this won't work for your "wraparound" case where you go from "A99" to "B02", but it'd probably be easier to generate a larger list and then subset it

letts <- c("a", "b", "c")
nums <- c(paste0("0", c(0:9)), 10:99)
sort(unlist(list(outer(letts, nums, paste0))))


Punintended's answer inspired me to figure out a solution that works (including the "wraparound" case where you go from "A99" to "B02")

stringrange = function(x,y)
full =unlist(lapply(LETTERS[which(LETTERS==substr(x,1,1)):which(LETTERS==substr(y,1,1))],
function(x)paste0(x,gettextf(paste0("%02d"),0:99))))
full[which(full==x):which(full==y)]


>stringrange("A98","C03")
[1] "A98" "A99" "B00" "B01" "B02" "B03" "B04" "B05" "B06" "B07" "B08" "B09" "B10" "B11" "B12" "B13" "B14" "B15" "B16" "B17" "B18"
[22] "B19" "B20" "B21" "B22" "B23" "B24" "B25" "B26" "B27" "B28" "B29" "B30" "B31" "B32" "B33" "B34" "B35" "B36" "B37" "B38" "B39"
[43] "B40" "B41" "B42" "B43" "B44" "B45" "B46" "B47" "B48" "B49" "B50" "B51" "B52" "B53" "B54" "B55" "B56" "B57" "B58" "B59" "B60"
[64] "B61" "B62" "B63" "B64" "B65" "B66" "B67" "B68" "B69" "B70" "B71" "B72" "B73" "B74" "B75" "B76" "B77" "B78" "B79" "B80" "B81"
[85] "B82" "B83" "B84" "B85" "B86" "B87" "B88" "B89" "B90" "B91" "B92" "B93" "B94" "B95" "B96" "B97" "B98" "B99" "C00" "C01" "C02"
[106] "C03"


You can do:

str_seq=function(X)
a = toupper(strsplit(X, ':')[[1]])# split while ensuring the letters are uppercase
nums = as.numeric(sub('[A-Z]', '', a))# Obtain the numbers
stopifnot(nums < 100) # If the number for a range is greater than 100 produce an error
a = as.numeric(paste0(setNames(1:26, LETTERS)[sub('\d+', '', a)],sprintf("%02d", nums)))
b = do.call(seq, as.list(c(a, by = if (diff(a) > 0) 1 else -1))) # Ensure you can go forward or backward
b = b[!!b%%100] # Remove A00,B00, etc
paste0(LETTERS[b %/% 100], sprintf("%02d", b %% 100))


str_seq('i89:j23')
[1] "I89" "I90" "I91" "I92" "I93" "I94" "I95" "I96" "I97" "I98" "I99" "J01" "J02" "J03" "J04"
[16] "J05" "J06" "J07" "J08" "J09" "J10" "J11" "J12" "J13" "J14" "J15" "J16" "J17" "J18" "J19"
[31] "J20" "J21" "J22" "J23"
> str_seq('j23:i89')
[1] "J23" "J22" "J21" "J20" "J19" "J18" "J17" "J16" "J15" "J14" "J13" "J12" "J11" "J10" "J09"
[16] "J08" "J07" "J06" "J05" "J04" "J03" "J02" "J01" "I99" "I98" "I97" "I96" "I95" "I94" "I93"
[31] "I92" "I91" "I90" "I89"


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