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    <p>If you really want the levels of the factor to be used, you're either doing something very wrong or too clever for its own good.</p> <p>If what you have is a factor containing numbers stored in the levels of the factor, then you want to coerce it to numeric first using <code>as.numeric(as.character(...))</code>:</p> <pre><code>dat &lt;- data.frame(f=as.character(runif(10))) </code></pre> <p>You can see the difference between accessing the factor indices and assigning the factor contents here:</p> <pre><code>&gt; as.numeric(dat$f) [1] 9 7 2 1 4 6 5 3 10 8 &gt; as.numeric(as.character(dat$f)) [1] 0.6369432 0.4455214 0.1204000 0.0336245 0.2731787 0.4219241 0.2910194 [8] 0.1868443 0.9443593 0.5784658 </code></pre> <p>Timings vs. an alternative approach which only does the conversion on the levels shows it's faster if levels are not unique to each element:</p> <pre><code>dat &lt;- data.frame( f = sample(as.character(runif(10)),10^4,replace=TRUE) ) library(microbenchmark) microbenchmark( as.numeric(as.character(dat$f)), as.numeric( levels(dat$f) )[dat$f] , as.numeric( levels(dat$f)[dat$f] ), times=50 ) expr min lq median uq max 1 as.numeric(as.character(dat$f)) 7835865 7869228 7919699 7998399 9576694 2 as.numeric(levels(dat$f))[dat$f] 237814 242947 255778 270321 371263 3 as.numeric(levels(dat$f)[dat$f]) 7817045 7905156 7964610 8121583 9297819 </code></pre> <p>Therefore, if <code>length(levels(dat$f)) &lt; length(dat$f)</code>, use <code>as.numeric(levels(dat$f))[dat$f]</code> for a substantial speed gain.</p> <p>If <code>length(levels(dat$f))</code> is approximately equal to <code>length(dat$f)</code>, there is no speed gain:</p> <pre><code>dat &lt;- data.frame( f = as.character(runif(10^4) ) ) library(microbenchmark) microbenchmark( as.numeric(as.character(dat$f)), as.numeric( levels(dat$f) )[dat$f] , as.numeric( levels(dat$f)[dat$f] ), times=50 ) expr min lq median uq max 1 as.numeric(as.character(dat$f)) 7986423 8036895 8101480 8202850 12522842 2 as.numeric(levels(dat$f))[dat$f] 7815335 7866661 7949640 8102764 15809456 3 as.numeric(levels(dat$f)[dat$f]) 7989845 8040316 8122012 8330312 10420161 </code></pre>
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    1. POArithmetic operations on R factors
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    1. USAri B. Friedman
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    1. POArithmetic operations on R factors
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    1. CO+1 illustrates the point I was trying to make.
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      1. USBrandon Bertelsen
    2. COAlthough, R is smart about sorting before factoring, so if they are wholenumbers this problem is irrelevant.
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    3. CO@Brandon: Unless someone has used `relevel` or the integer sequence is not continuous. Assuming the level indices are the same as the level contents seems like a dangerous assumption to make.
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