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    <p>R does have a function <code>addmargins</code> that lets you do something like this, but it expects a <code>table</code> or <code>matrix</code> as the input.</p> <pre><code>addmargins(as.matrix(mydf), 1, FUN = mean) # timetE4_1 timetE1_2 timetE2_2 timetE3_2 timetE4_2 eve_mean mor_mean tot_mean # 1 4048.605 59094.48 27675.59 26374.06 43310.01 7774.442 39113.53 23443.99 # 2 45729.986 139889.21 111309.64 129781.17 96924.62 43374.117 119476.16 81425.14 # 3 639686.154 1764684.16 1117027.29 1147967.45 1156442.48 585562.724 1296530.34 941046.53 # 4 4466.153 26250.32 20320.08 18413.54 29061.25 3866.547 23511.30 13688.92 # mean 173482.724 497479.54 319083.15 330634.05 331434.59 160144.458 369657.83 264901.15 </code></pre> <hr> <h3>Update</h3> <p>There is an <a href="https://stackoverflow.com/q/19792110/1270695">almost identical (conceptually) question here</a>, and I thought I would share my answer from there here too.</p> <p>Assume we're starting with:</p> <pre><code>mydf &lt;- structure(list(Description = c("A", "B", "C", "D"), timetE4_1 = c(4048.605, 45729.986, 639686.154, 4466.153), Boo = structure(1:4, .Label = c("a", "b", "c", "d"), class = "factor"), timetE1_2 = c(59094.48, 139889.21, 1764684.16, 26250.32), timetE2_2 = c(27675.59, 111309.64, 1117027.29, 20320.08), Baa = c(FALSE, FALSE, TRUE, NA)), .Names = c("Description", "timetE4_1", "Boo", "timetE1_2", "timetE2_2", "Baa"), row.names = c("1", "2", "3", "4"), class = "data.frame") mydf # Description timetE4_1 Boo timetE1_2 timetE2_2 Baa # 1 A 4048.605 a 59094.48 27675.59 FALSE # 2 B 45729.986 b 139889.21 111309.64 FALSE # 3 C 639686.154 c 1764684.16 1117027.29 TRUE # 4 D 4466.153 d 26250.32 20320.08 NA </code></pre> <p>@Jilber's solution won't work in that case and will lead to lots of misplaced columns. Instead, use <code>rbind.fill</code> from "plyr". I've used <code>sapply</code> to specify my function in this example to show that it is easy to use whatever function you want, not just the <code>col*</code> functions.</p> <pre><code>library(plyr) useme &lt;- sapply(mydf, is.numeric) rbind.fill(mydf, data.frame(t(sapply(mydf[useme], sum)))) # Description timetE4_1 Boo timetE1_2 timetE2_2 Baa # 1 A 4048.605 a 59094.48 27675.59 FALSE # 2 B 45729.986 b 139889.21 111309.64 FALSE # 3 C 639686.154 c 1764684.16 1117027.29 TRUE # 4 D 4466.153 d 26250.32 20320.08 NA # 5 &lt;NA&gt; 693930.898 &lt;NA&gt; 1989918.17 1276332.60 NA </code></pre>
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    2. CO@Rechlay, wow. You can post an error message! But seriously, have you played with the function at all to try different settings and see how it works or understand why you are getting an error message? Perhaps if you posted a reproducible example, you'll be able to get more meaningful help. See http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example for details.
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