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    <p>A solution with <code>combn</code> and <code>apply</code>:</p> <pre><code>cc &lt;- combn(colnames(ex)[1:4], 2) apply(cc, 2, function(x)ex[[x[1]]]-ex[[x[2]]]) </code></pre> <p>gives for the first 5 rows:</p> <pre><code> [,1] [,2] [,3] [,4] [,5] [,6] [1,] -0.43500930 -0.520148152 0.1602265 -0.08513885 0.59523580 0.680374655 [2,] -0.32964090 -0.153303302 -0.3807295 0.17633760 -0.05108855 -0.227426149 [3,] 0.25991705 -0.079679566 0.2040904 -0.33959662 -0.05582670 0.283769917 [4,] 0.35585252 0.153083047 0.2382553 -0.20276948 -0.11759719 0.085172292 [5,] -0.67081018 -0.116543468 -0.3413471 0.55426671 0.32946305 -0.224803663 </code></pre> <p><strong>Edit</strong> </p> <p>As Arun suggested, combn can take a function argument, so a better solution is </p> <pre><code>res &lt;- combn(colnames(ex)[1:4], 2, function(x) ex[[x[1]]] - ex[[x[2]]]) colnames(res) &lt;- combn(colnames(ex)[1:4], 2, paste, collapse="_") as.data.table(res) AAA_BBB AAA_CCC AAA_DDD BBB_CCC BBB_DDD CCC_DDD 1: -0.4350093 -0.52014815 0.16022650 -0.08513885 0.59523580 0.68037465 2: -0.3296409 -0.15330330 -0.38072945 0.17633760 -0.05108855 -0.22742615 3: 0.2599171 -0.07967957 0.20409035 -0.33959662 -0.05582670 0.28376992 4: 0.3558525 0.15308305 0.23825534 -0.20276948 -0.11759719 0.08517229 5: -0.6708102 -0.11654347 -0.34134713 0.55426671 0.32946305 -0.22480366 --- 999996: -0.8450458 -0.47951267 -0.30333929 0.36553310 0.54170648 0.17617338 999997: -0.5778393 -0.01784418 -0.24353237 0.55999516 0.33430697 -0.22568819 999998: 0.7127352 0.82554276 0.01258673 0.11280758 -0.70014846 -0.81295604 999999: -0.6693544 -0.42335069 -0.81080852 0.24600375 -0.14145408 -0.38745783 1000000: -0.8511655 -0.23341818 -0.15830584 0.61774732 0.69285966 0.07511234 </code></pre>
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    1. POSubtract every column from each other column in a R data.table
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    1. CO(+1) `combn` has a function argument as well: `combn(colnames(ex)[1:4], 2, function(x) ex[[x[1]]] - ex[[x[2]]])`
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