Creating 4 back-to-back vectors of "A01 to A15" using paste0() in R?

Question

I'm wondering how to create 4 back-to-back, ordered vectors of characters each beginning from "A01" and ending in "A15" using paste0 or a better regular expression code (see the full vector below)?

I have tried the following (with no success):

rep(paste0("A", 0, 1:15), 4)

Here is the full vector I'm trying to achieve using paste0:

G = c("A01", "A02", "A03", "A04", "A05", "A06", "A07", "A08", "A09", "A10", "A11", "A12", 
      "A13", "A14", "A15", "A01", "A02", "A03", "A04", "A05", "A06", "A07", "A08", "A09", 
      "A11", "A12", "A13", "A14", "A15","A01", "A02", "A03", "A04", "A05", "A06", "A07", 
      "A08", "A09", "A12", "A13", "A14", "A15","A01", "A02", "A03", "A04", "A06", "A07", 
      "A08", "A09", "A12", "A13", "A14", "A15","A01", "A02", "A03", "A04", "A05", "A06", 
      "A07", "A08", "A09", "A10", "A11", "A12", "A13", "A14", "A15")

easier to use `sprintf` probably: `rep(sprintf("A02.0f", 1:15), 4)`. — lmo, Aug 29 '17 at 16:20
Oh no. I missed a % sign: `rep(sprintf("A%02.0f", 1:15), 4)` or `rep(sprintf("A%02d", 1:15), 4)`. — lmo, Aug 29 '17 at 16:37

score 0 · Accepted Answer · answered Aug 29 '17 at 16:25

0

sapply(1:15, (function(i) {paste0("A", rep(0, (2-nchar(as.character(i)))), i)} ))

answered Aug 29 '17 at 16:25

Damiano Fantini

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Creating 4 back-to-back vectors of "A01 to A15" using paste0() in R?

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