On 3 Sep 2021, at 09:59 , Luigi Marongiu <marongiu.luigi at gmail.com> wrote:
Fair enough, I'll check the actual data to see if there are indeed any
NaN (which should not, since the data are categories, not generated by
math).
Thanks!
On Fri, Sep 3, 2021 at 8:26 AM PIKAL Petr <petr.pikal at precheza.cz> wrote:
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
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
})
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>
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-
-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)
```
when modifying my dataframe df.
What would be the correct syntax?
Thank you
--
Best regards,
Luigi