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POstrsplit into data.frame with incomplete input
 

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  1. POstrsplit into data.frame with incomplete input
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    <p>I try to split a vector of strings into a data.frame object and for a fixed order this isn't a problem (e.g. like written <a href="https://stackoverflow.com/questions/7069076/split-column-at-delimiter-in-data-frame">here</a>), but in my particular case the columns for the future data-frame are not complete in the string objects. This is how the output should look like for an toy input:</p> <pre><code>input &lt;- c("an=1;bn=3;cn=45", "bn=3.5;cn=76", "an=2;dn=5") res &lt;- do.something(input) &gt; res an bn cn dn [1,] 1 3 45 NA [2,] NA 3.5 76 NA [3,] 2 NA NA 5 </code></pre> <p>I am looking now for a function <code>do.something</code>that can do that in a efficient way. My naive solution at the moment would be to loop over the input objects, <code>strsplit</code> those for <code>;</code> then <code>strsplit</code> them again for <code>=</code> and then fill the <code>data.frame</code> result by result. Is there any way to do that more R-alike? I am afraid doing that element by element would take quite a long time for a long vector <code>input</code>. </p> <p>EDIT: Just for completeness, my naive solution looks like this:</p> <pre><code> do.something &lt;- function(x){ temp &lt;- strsplit(x,";") temp2 &lt;- sapply(temp,strsplit,"=") ul.temp2 &lt;- unlist(temp2) label &lt;- sort(unique(ul.temp2[seq(1,length(ul.temp2),2)])) res &lt;- data.frame(matrix(NA, nrow = length(x), ncol = length(label))) colnames(res) &lt;- label for(i in 1:length(temp)){ for(j in 1:length(label)){ curInfo &lt;- unlist(temp2[[i]]) if(sum(is.element(curInfo,label[j]))&gt;0){ res[i,j] &lt;- curInfo[which(curInfo==label[j])+1] } } } res } </code></pre> <p>EDIT2: Unfortunately my large input data looks like this (entries without '=' possible):</p> <pre><code>input &lt;- c("an=1;bn=3;cn=45", "an;bn=3.5;cn=76", "an=2;dn=5") </code></pre> <p>so I cannot compare the given answers to my problem at hand. My naive solution for that is</p> <pre><code>do.something &lt;- function(x){ temp &lt;- strsplit(x,";") tempNames &lt;- sort(unique(sapply(strsplit(unlist(temp),"="),"[",1))) res &lt;- data.frame(matrix(NA, nrow = length(x), ncol = length(tempNames))) colnames(res) &lt;- tempNames for(i in 1:length(temp)){ curSplit &lt;- strsplit(unlist(temp[[i]]),"=") curNames &lt;- sapply(curSplit,"[",1) curValues &lt;- sapply(curSplit,"[",2) for(j in 1:length(tempNames)){ if(is.element(colnames(res)[j],curNames)){ res[i,j] &lt;- curValues[curNames==colnames(res)[j]] } } } res } </code></pre>
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    1. USDaniel Fischer
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    1. COare your column names always two characters long?
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      1. USSimon O'Hanlon
    2. COOkay, sorry that was misleading. No, they aren't. They can be everything between 2 and 10 Characters.
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    3. COI edited my solution. It now uses just base package and should handle missing numbers efficiently.
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      1. USSimon O'Hanlon
 

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