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POAvoid for loop in string replacement?
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copied!<p>I've got data, a character vector (eventually I'll collapse it, so I don't care if it stays a vector or if it's treated as a single string), a vector of patterns, and a vector of replacements. I want each pattern in the data to be replaced by its respective replacement. I got it done with a <code>stringr</code> and a for loop, but is there a more R-like way to do it?</p> <pre><code>require(stringr) start_string <- sample(letters[1:10], 10) my_pattern <- c("a", "b", "c", "z") my_replacement <- c("[this was an a]", "[this was a b]", "[this was a c]", "[no z!]") str_replace(start_string, pattern = my_pattern, replacement = my_replacement) # bad lengths, doesn't work str_replace(paste0(start_string, collapse = ""), pattern = my_pattern, replacement = my_replacement) # vector output, not what I want in this case my_result <- start_string for (i in 1:length(my_pattern)) { my_result <- str_replace(my_result, pattern = my_pattern[i], replacement = my_replacement[i]) } > my_result [1] "[this was a c]" "[this was an a]" "e" "g" "h" "[this was a b]" [7] "d" "j" "f" "i" # This is what I want, but is there a better way? </code></pre> <p>In my case, I know each pattern will occur at most once, but not every pattern will occur. I know I could use <code>str_replace_all</code> if patterns might occur more than once; I hope a solution would also provide that option. I'd also like a solution that uses <code>my_pattern</code> and <code>my_replacement</code> so that it could be part of a function with those vectors as arguments.</p>

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