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POusing R to select rows in data set with matching missing observations
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  1. POusing R to select rows in data set with matching missing observations
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    copied!<p>I have determined how to identify all unique patterns of missing observations in a data set. Now I would like to select all rows in that data set with a given pattern of missing observations. I would like to do this iteratively so that if there are n patterns of missing observations in the data set I end up with n data sets each containing only 1 pattern of missing observations. </p> <p>I know how to do this, but my method is not very efficient and is not general. I am hoping to learn a more efficient and more general approach because my real data sets are much larger and more variable than in the example below.</p> <p>Here is an example data set and the code I am using. I do not bother to include the code I used to create the matrix zzz from the matrix dd, but can add that code if it helps.</p> <pre><code>dd &lt;- matrix(c( 1, 0, 1, 1, NA, 1, 1, 0, NA, 0, 0, 0, NA, 1,NA, 1, NA, 1, 1, 1, 0, 0, 1, 0, NA, 0, 0, 0, 0,NA,NA,NA, 1,NA,NA,NA, 1, 1, 1, 1, NA, 1, 1, 0), nrow=11, byrow=T) zzz &lt;- matrix(c( 1, 1, 1, 1, NA, 1, 1, 1, NA, 1,NA, 1, 1,NA,NA,NA ), nrow=4, byrow=T) for(jj in 1:dim(zzz)[1]) { ddd &lt;- dd[ ((dd[, 1]%in%c(0,1) &amp; zzz[jj, 1]%in%c(0,1)) | (is.na(dd[, 1]) &amp; is.na(zzz[jj, 1]))) &amp; ((dd[, 2]%in%c(0,1) &amp; zzz[jj, 2]%in%c(0,1)) | (is.na(dd[, 2]) &amp; is.na(zzz[jj, 2]))) &amp; ((dd[, 3]%in%c(0,1) &amp; zzz[jj, 3]%in%c(0,1)) | (is.na(dd[, 3]) &amp; is.na(zzz[jj, 3]))) &amp; ((dd[, 4]%in%c(0,1) &amp; zzz[jj, 4]%in%c(0,1)) | (is.na(dd[, 4]) &amp; is.na(zzz[jj, 4]))),] print(ddd) } </code></pre> <p>The 4 resulting data sets in this example are: </p> <pre><code>a) 1 0 1 1 0 0 1 0 1 1 1 1 b) NA 1 1 0 NA 0 0 0 NA 1 1 1 NA 0 0 0 NA 1 1 0 c) NA 1 NA 1 d) 0 NA NA NA 1 NA NA NA </code></pre> <p>Is there a more general and more efficient method of doing the same thing? In the example above the 4 resulting data sets are not saved, but I do save them with my real data.</p> <p>Thank you for any advice. </p> <p>Mark Miller </p>
 

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