It seems like you might've missed one more thing, you need the brackets
next to 'x' to get it to work.
x[] <- lapply(x, function(xx) {
xx[is.nan(xx)] <- NA_real_
xx
})
is different from
x <- lapply(x, function(xx) {
xx[is.nan(xx)] <- NA_real_
xx
})
Also, if all of your data is numeric, it might be better to convert to a
matrix before doing your calculations. For example:
x <- as.matrix(x)
x[is.nan(x)] <- NA_real_
I'd also suggest this same solution for the other question you posted,
x[x == 0] <- NA
On Thu, Sep 2, 2021 at 10:01 AM Luigi Marongiu <marongiu.lu...@gmail.com>
wrote:
> Sorry,
> still I don't get it:
> ```
> > dim(df)
> [1] 302 626
> > # clean
> > df <- lapply(x, function(xx) {
> + xx[is.nan(xx)] <- NA
> + xx
> + })
> > dim(df)
> NULL
> ```
>
> On Thu, Sep 2, 2021 at 3:47 PM Andrew Simmons <akwsi...@gmail.com> wrote:
> >
> > You removed the second line 'xx' from the function, put it back and it
> should work
> >
> > On Thu, Sep 2, 2021, 09:45 Luigi Marongiu <marongiu.lu...@gmail.com>
> wrote:
> >>
> >> `data[sapply(data, is.nan)] <- NA` is a nice compact command, but I
> >> still get NaN when using the summary function, for instance one of the
> >> columns give:
> >> ```
> >> Min. : NA
> >> 1st Qu.: NA
> >> Median : NA
> >> Mean :NaN
> >> 3rd Qu.: NA
> >> Max. : NA
> >> NA's :110
> >> ```
> >> I tried to implement the second solution but:
> >> ```
> >> df <- lapply(x, function(xx) {
> >> xx[is.nan(xx)] <- NA
> >> })
> >> > str(df)
> >> List of 1
> >> $ sd_ef_rash_loc___palm: logi NA
> >> ```
> >> What am I getting wrong?
> >> Thanks
> >>
> >> On Thu, Sep 2, 2021 at 3:30 PM Andrew Simmons <akwsi...@gmail.com>
> wrote:
> >> >
> >> > Hello,
> >> >
> >> >
> >> > I would use something like:
> >> >
> >> >
> >> > x <- c(1:5, NaN) |> sample(100, replace = TRUE) |> matrix(10, 10) |>
> as.data.frame()
> >> > x[] <- lapply(x, function(xx) {
> >> > xx[is.nan(xx)] <- NA_real_
> >> > xx
> >> > })
> >> >
> >> >
> >> > This prevents attributes from being changed in 'x', but accomplishes
> the same thing as you have above, I hope this helps!
> >> >
> >> > On Thu, Sep 2, 2021 at 9:19 AM Luigi Marongiu <
> marongiu.lu...@gmail.com> wrote:
> >> >>
> >> >> Hello,
> >> >> I have some NaN values in some elements of a dataframe that I would
> >> >> like to convert to NA.
> >> >> The command `df1$col[is.nan(df1$col)]<-NA` allows to work
> column-wise.
> >> >> Is there an alternative for the global modification at once of all
> >> >> instances?
> >> >> I have seen from
> >> >>
> https://stackoverflow.com/questions/18142117/how-to-replace-nan-value-with-zero-in-a-huge-data-frame/18143097#18143097
> >> >> that once could use:
> >> >> ```
> >> >>
> >> >> is.nan.data.frame <- function(x)
> >> >> do.call(cbind, lapply(x, is.nan))
> >> >>
> >> >> data123[is.nan(data123)] <- 0
> >> >> ```
> >> >> replacing o with NA, but I got
> >> >> ```
> >> >> str(df)
> >> >> > logi NA
> >> >> ```
> >> >> when modifying my dataframe df.
> >> >> What would be the correct syntax?
> >> >> Thank you
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >> Best regards,
> >> >> Luigi
> >> >>
> >> >> ______________________________________________
> >> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> >> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >>
> >>
> >> --
> >> Best regards,
> >> Luigi
>
>
>
> --
> Best regards,
> Luigi
>
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