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    <p>I don't have a full answer to why the times for lapply are not linear, but notice that you are using lapply as a glorified loop by applying over a list of variable names (and then using these names to index the actual values) rather than applying over a list of values. I'm not sure how to solve this using lapply, but I can offer an alternative that is (roughly) a liner function of the number of variables. </p> <p>Start by creating the example data:</p> <pre><code>Genedata &lt;- structure(list(time = c(120, 120, 28.96, 119.21, 59.53, 68.81, 82.29, 110.82, 65.88, 84.13, 16.47, 89.75, 76.07, 67.82, 52.24, 64.1, 55.13, 57.79, 50.1, 43.79, 30.27, 3.64, 36.59, 20.02, 118.58, 55.33, 120, 120, 120, 106.94, 11.7, 28.79, 26.82, 52.5, 24.03, 88.99, 102.44, 33.73, 85.28, 26.53, 37.31, 86.43, 55.92, 70.65, 76.04, 79.13, 83.99, 80.25, 40.6, 95.37, 31.06, 37.31, 22.02, 63.25, 34.09, 52.14, 66.04, 59.96, 47.86, 58.45, 45.99, 60.42, 37.67, 15.71, 39.25, 49.87, 25.24, 44.97, 35.9, 26.66, 36.78, 41.52, 22.22, 33.2, 39.68, 25.61, 83.99, 15.05, 8.38, 18.18, 27.15, 24.26, 105.17, 11.76, 70.45, 95.07, 112.33, 27.51, 45.92, 54.04, 103.98, 6.11, 99.51, 20.44, 76.6, 10.02, 41.45, 96.26, 85.61, 78.87, 22.25, 74.53, 59.07, 47.8, 5.68, 79.39, 74.07, 50.76, 67.82, 70.84, 50.59, 34.58, 38.72, 54.9, 67.92, 58.45, 59.34, 54.54, 19.03, 26.26, 52.86, 32.05, 55.95, 56.67, 50.43, 4.24, 49.11, 49.57, 50.49, 63.05, 49.24, 0.92, 31.36, 40.3, 116.64, 31.92, 19.98, 24.62, 18.54, 120, 17.62, 5.32, 2.36, 5.72, 15.28, 95.07, 4.96, 28.89, 3.25, 0.92, 18.77, 57.73, 14.39, 5.12, 4.99, 33.17, 50.53, 5.72, 8.02, 34.02, 1.44, 36.95, 60.75, 37.44, 23.07, 19.85, 31.85, 8.61, 42.27, 15.25, 14.56, 9.96, 8.94, 32.67, 2.07, 22.78, 22.35), cens = c(0L, 0L, NA, 0L, 0L, 1L, 0L, 0L, NA, 0L, NA, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, NA, 1L, 1L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, NA, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, NA, 1L, 0L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 1L, 0L, NA, 1L, 1L, 1L, 0L, 0L, 0L, NA, 0L, NA, NA, 1L, 0L, 0L, 1L, 1L, 0L, 0L, 1L, NA, NA, 0L, 0L, 0L, 0L, 1L, NA, 0L, 1L, 0L, 0L, 0L, NA, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, NA, 1L, 1L, 1L, 1L, NA, 0L, NA, 1L, 1L, 0L, 1L, 0L, 1L, 1L, NA, 1L, 0L, 1L, 1L, 0L)), .Names = c("time", "cens"), class = "data.frame", row.names = c(NA, -177L)) data &lt;- replicate(10000, abs(rnorm(177))) Genedata &lt;- data.frame(Genedata, data) </code></pre> <p>My approach is the reshape the data using the <code>melt</code> function in the reshape2 package, then use <code>dlply</code> from the plyr package to run the model on each subset.</p> <pre><code>ibrary(maxstat) library(plyr) library(reshape2) system.time({ dat.m &lt;- melt(Genedata, id.vars=1:3) maxstat.estimate &lt;- dlply(dat.m, .(variable), function(DF) { maxstat.test(Surv(time, cens) ~ value, smethod="LogRank", pmethod="Lau94", data=DF)})}) </code></pre> <p>Some timings from this approach are as follows:</p> <pre><code>## 1000 variables ## user system elapsed ## 11.613 0.090 11.780 ## 2000 variables ## user system elapsed ## 23.847 0.060 24.043 ## 3000 variables ## user system elapsed ## 35.316 0.093 35.905 ## 10000 variables ## user system elapsed ## 120.654 0.594 122.140 </code></pre> <p>showing that system time is roughly a linear function of the number of variables.</p> <p>The full 43000 variables takes just over 10 minutes, just enough time for a cup of coffee:</p> <pre><code>## 43000 ## user system elapsed ## 612.587 3.134 630.464 </code></pre>
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    1. COOk problem solved :) cheers. It works well, surprising considering lapply is usually suggested for large datasets.
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