`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 <akwsimmo at 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.luigi at
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 at 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
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.luigi at 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 <akwsimmo at 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.luigi at 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 at 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 >[[alternative HTML version deleted]]
Hi Luigi.
Weird. But maybe it is the desired behaviour of summary when calculating
mean of numeric column full of NAs.
See example
dat <- data.frame(x=rep(NA, 110), y=rep(1, 110), z= rnorm(110))
# change all values in second column to NA
dat[,2] <- NA
# change some of them to NAN
dat[5:6, 2:3] <- 0/0
# see summary
summary(dat)
x y z
Mode:logical Min. : NA Min. :-1.9798
NA's:110 1st Qu.: NA 1st Qu.:-0.4729
Median : NA Median : 0.1745
Mean :NaN Mean : 0.1856
3rd Qu.: NA 3rd Qu.: 0.8017
Max. : NA Max. : 2.5075
NA's :110 NA's :2
# change NAN values to NA
dat[sapply(dat, is.nan)] <- NA
*************************
#summary is same
summary(dat)
x y z
Mode:logical Min. : NA Min. :-1.9798
NA's:110 1st Qu.: NA 1st Qu.:-0.4729
Median : NA Median : 0.1745
Mean :NaN Mean : 0.1856
3rd Qu.: NA 3rd Qu.: 0.8017
Max. : NA Max. : 2.5075
NA's :110 NA's :2
# but no NAN value in data
dat[1:10,]
x y z
1 NA NA -0.9148696
2 NA NA 0.7110570
3 NA NA -0.1901676
4 NA NA 0.5900650
5 NA NA NA
6 NA NA NA
7 NA NA 0.7987658
8 NA NA -0.5225229
9 NA NA 0.7673103
10 NA NA -0.5263897
So my "nice compact command"
dat[sapply(dat, is.nan)] <- NA
works as expected, but summary gives as mean NAN.
Cheers
Petr
> -----Original Message-----
> From: R-help <r-help-bounces at r-project.org> On Behalf Of Luigi
Marongiu
> Sent: Thursday, September 2, 2021 3:46 PM
> To: Andrew Simmons <akwsimmo at gmail.com>
> Cc: r-help <r-help at r-project.org>
> Subject: Re: [R] How to globally convert NaN to NA in dataframe?
>
> `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 <akwsimmo at 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.luigi at
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 at 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
>
> ______________________________________________
> R-help at 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.