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    <p>Sorry for digging up an old question. I couldn't look up the function to do this job on the train, so I wrote one myself.</p> <p>I was proud to find out that it's a tiny bit faster.<br> It's less flexible though.</p> <p>But it plays nice with <code>ave</code>, which is what I needed.</p> <pre><code>repeat.before = function(x) { # repeats the last non NA value. Keeps leading NA ind = which(!is.na(x)) # get positions of nonmissing values if(is.na(x[1])) # if it begins with a missing, add the ind = c(1,ind) # first position to the indices rep(x[ind], times = diff( # repeat the values at these indices c(ind, length(x) + 1) )) # diffing the indices + length yields how often } # they need to be repeated x = c(NA,NA,'a',NA,NA,NA,NA,NA,NA,NA,NA,'b','c','d',NA,NA,NA,NA,NA,'e') xx = rep(x, 1000000) system.time({ yzoo = na.locf(xx,na.rm=F)}) ## user system elapsed ## 2.754 0.667 3.406 system.time({ yrep = repeat.before(xx)}) ## user system elapsed ## 0.597 0.199 0.793 </code></pre> <h2>Edit</h2> <p>As this became my most upvoted answer, I was reminded often that I don't use my own function, because I often need zoo's <code>maxgap</code> argument. Because zoo has some weird problems in edge cases when I use dplyr + dates that I couldn't debug, I came back to this today to improve my old function.</p> <p>I benchmarked my improved function and all the other entries here. For the basic set of features, <code>tidyr::fill</code> is fastest while also not failing the edge cases. The Rcpp entry by @BrandonBertelsen is faster still, but it's inflexible regarding the input's type (he tested edge cases incorrectly due to a misunderstanding of <code>all.equal</code>).</p> <p>If you need <code>maxgap</code>, my function below is faster than zoo (and doesn't have the weird problems with dates). </p> <p>I put up the <a href="http://rpubs.com/rubenarslan/repeat_last_na_locf" rel="noreferrer">documentation of my tests</a>.</p> <h3>new function</h3> <pre><code>repeat_last = function(x, forward = TRUE, maxgap = Inf, na.rm = FALSE) { if (!forward) x = rev(x) # reverse x twice if carrying backward ind = which(!is.na(x)) # get positions of nonmissing values if (is.na(x[1]) &amp;&amp; !na.rm) # if it begins with NA ind = c(1,ind) # add first pos rep_times = diff( # diffing the indices + length yields how often c(ind, length(x) + 1) ) # they need to be repeated if (maxgap &lt; Inf) { exceed = rep_times - 1 &gt; maxgap # exceeding maxgap if (any(exceed)) { # any exceed? ind = sort(c(ind[exceed] + 1, ind)) # add NA in gaps rep_times = diff(c(ind, length(x) + 1) ) # diff again } } x = rep(x[ind], times = rep_times) # repeat the values at these indices if (!forward) x = rev(x) # second reversion x } </code></pre> <p>I've also put the function in my <a href="https://github.com/rubenarslan/formr" rel="noreferrer">formr package</a> (Github only).</p>
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