CODEWITHSUNDEEP
r
S. Jeon
S. Jeon

Reputation: 5

How can I replace %s to vector using sprintf()?

I want to replace %s as a vector.

So, I coded as below:

items <- c("a", "b", "c", "d")
items.txt <- sprintf("y <- c(%s)", items)

My expected result is:

"y <- c("a", "b", "c", "d")"

But real result is:

"y <- c(a)"   "y <- c(b)"   "y <- c(c)"   "y <- c(d)"

Thus I tried as follows:

  1. items.txt <- sprintf("y <- c(%s)", paste(items, collapse - ","))
  2. items.txt <- sprintf("y <- c(%s)", paste(items, collapse - '","'))

But these are not working.

Are there are any ideas for solving this problem?

Upvotes: 0

Views: 352

Answers (2)

Ma&#235;l
Ma&#235;l

Reputation: 51974

What about this?

items <- 'c("a", "b", "c", "d")'
items.txt <- sprintf('"y <- %s"', items)
cat(items.txt)
# "y <- c("a", "b", "c", "d")"

Upvotes: 1

Ben Bolker
Ben Bolker

Reputation: 226172

This seems easiest to me:

s <- sprintf("y <- c(%s)", paste(sprintf('"%s"',items), collapse=","))

Note that print(s) will look weird because of the backslashes protecting the quotation marks. cat(s,"\n") looks more normal:

y <- c("a","b","c","d")

dput(items, textConnection("s", "w")) might also be useful.

Upvotes: 1

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