title: Seasonal decompose of monthly data including NA in r
body: <p>I need help to decompose my monthly data which have seasonality, but it does not work because NA values are not removed. There may be another problem. Please look at my data and errors as below.</p>
  <pre><code>    ts.monthly&lt;-ts(monthly$rBC.median, frequency=12, start=c(2006, 4))
      ts.monthly
          Jan        Feb        Mar        Apr        May        Jun
      2006                                   5.1656479  6.2847959 19.4833690
      2007  1.4252665  2.9127775  2.8912652  7.5326158  8.6182227 23.2129310
      2008         NA  1.8200842  1.3488755  2.0700927  5.3541366  8.6916708
      2009  1.2531161  1.5075780  2.4955524 10.6724704 10.1367162 16.0362127
      2010  0.8850190  2.4974866  1.8459976  9.2297697  3.8203789  7.1492986
      2011  2.6990434  0.4570701  1.3787403  5.8739804  4.1669501 13.2228535
      2012         NA  2.0670538  1.3758499 11.7306663  4.1248775 12.3604423
                  Jul        Aug        Sep        Oct        Nov        Dec
      2006  9.8028986  7.8167810  2.1333807  2.5777504  1.9022561  2.7254065
      2007  4.2121577  8.8604768 12.0017155  4.0978332  1.6053110         NA
      2008  5.7338211  9.7432563  4.6548508  1.3589789  0.9650082  1.2788504
      2009 11.7632775 11.2299683  1.6229679  1.0333217  1.0481580  1.0734208
      2010  3.5996501  4.3245873  4.4586863  1.6403104  2.8622518  1.2564256
      2011  3.0463918  7.1515472  6.5613683  1.3715623  1.9757217  5.4901524
      2012 11.1010563  3.6220968  2.2597341  
  
     ts.monthly=na.omit(ts.monthly)  
      Error in na.omit.ts(ts.monthly) : time series contains internal NAs
     ts.monthly.com&lt;-decompose(ts.monthly)
      Error in na.omit.ts(x) : time series contains internal NAs
     ts.monthly$seasonal
      Error in ts.monthly$seasonal : $ operator is invalid for atomic vectors
  </code></pre>
  <p>I do not understand why na.omit does not work. How can I treat this NA??</p>
  <p>Finally, after using a function &quot;decompose&quot;, I want to take only &quot;trend&quot; without seasonality, and then apply sen's slope estimator to get a slope for linear trend. Will it work?</p>
  <p>Thanks a lot for your help.</p>