Next time send your email using plain text format rather than HTML so we see
what you saw.
Try
idx <- is.na( df$mean )
df[ idx, "mean" ] <- df[ idx, "offset" ]
BTW there is a commonly-used function called df, so you might improve clarity
by using DF for your temporary data frame name.
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On October 18, 2015 10:48:14 AM PDT, Jonathan Reardon
<jonathanrear...@outlook.com> wrote:
>Hi everyone,
>A simple question, but i cannot figure this out.
>
>I have a data-frame with 4 columns (onset, offset, outcome, mean):
>onset offset outcome mean8 72071 72503 1 7244615 142598
>143030 1 NaN30 293729 294161 1 294080
>For each 'NaN' in the mean column, i want to replace that NaN with the
>'offset' value in the same row.
>Intended outcome:
>onset offset outcome mean8 72071 72503 1 7244615 142598
>143030 1 14303030 293729 294161 1 294080
>I have tried:
> df$mean <- replace(df$mean, is.na(df$mean), df$offset)
>but i get the error message: 'number of items to replace is not a
>multiple of replacement length'. I'm assuming because this is trying to
>insert the whole 'offset' column into my one NaN cell. Is this a
>correct interpretation of the error message?
>Can anyone tell me how to replace any mean row NaN's with the offset
>value from that very same row?
>I don't want to use any pasting etc as this needs to be used as part of
>a function working over a large dataset than the one shown here.
>Cheers
>Jonathan
> [[alternative HTML version deleted]]
>
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>and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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and provide commented, minimal, self-contained, reproducible code.
