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PO
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20413599
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2013-12-06T00:22:10.937
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You have not provided a reproducible example, so i'm guessing a little: but it seems that the problem is that <code>example1</code> is made up of factors. Here's a basic guess at <code>example1</code> <pre><code>example1 <- as.factor(LETTERS[1:9]) </code></pre> when you print your <code>factor1</code> you probably see something like the following: <pre><code>R> example1 [1] A B C D E F G H I Levels: A B C D E F G H I </code></pre> Now if we try and replace any item with a non-factor (something not listed in <code>levels</code> above), we will get the following error (which is similar to yours): <pre><code>R> example1[8] <- "KK" Warning message: In `[<-.factor`(`*tmp*`, 8, value = "KK") : invalid factor level, NA generated </code></pre> but note that you could make a substitution of one listed factor for another -- meaning that <code>example1[8] <- "A"</code> is valid. My guess is that you don't want factors -- you want characters. So you need to coerce <code>example1</code> to character. Do this as follows <pre><code>R> example1.ch <- as.character(example1) </code></pre> No your substitution will work: <pre><code>R> example1.ch[8] <- 'kk' R> example1.ch [1] "A" "B" "C" "D" "E" "F" "G" "kk" "I" </code></pre> In general, you can use the command <code>str()</code> to learn about what your data object is comprised of -- which will help when you get odd errors like this one. <pre><code>R> str(example1) Factor w/ 9 levels "A","B","C","D",..: 1 2 3 4 5 6 7 1 9 </code></pre>
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1. POreplace a character element of a data frame
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1. USricardo
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1. POreplace a character element of a data frame
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