Sorry,
still I don't get it:
```
# clean
df <- lapply(x, function(xx) {
+ xx[is.nan(xx)] <- NA
+ xx
+ })
NULL
```
On Thu, Sep 2, 2021 at 3:47 PM Andrew Simmons <akwsimmo at gmail.com> wrote:
You removed the second line 'xx' from the function, put it back and it
On Thu, Sep 2, 2021, 09:45 Luigi Marongiu <marongiu.luigi at gmail.com>
`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
})
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>
Hello,
I would use something like:
x <- c(1:5, NaN) |> sample(100, replace = TRUE) |> matrix(10, 10) |>
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
Is there an alternative for the global modification at once of all
instances?
I have seen from
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