Subscripting problem with is.na()

Good point. I did not think about factors. Also your example raises another issue since column c is logical, but gets silently converted to numeric. This would seem to get the job done assuming the conversion is intended for numeric columns only:

test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))
sapply(test, class)

        a         b         c
"numeric"  "factor" "logical"

num <- sapply(test, is.numeric)
test[, num][is.na(test[, num])] <- 0
test

  a    b  c
1 1    A NA
2 0    b NA
3 2 <NA> NA

David C

-----Original Message-----
From: Bert Gunter [mailto:bgunter.4567 at gmail.com]
Sent: Thursday, June 23, 2016 1:48 PM
To: David L Carlson
Cc: Ivan Calandra; R Help
Subject: Re: [R] Subscripting problem with is.na()

Not in general, David:

e.g.

test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))

is.na(test)

         a     b    c
[1,] FALSE FALSE TRUE
[2,]  TRUE FALSE TRUE
[3,] FALSE  TRUE TRUE

test[is.na(test)]

[1] NA NA NA NA NA

test[is.na(test)] <- 0

Warning message:
In `[<-.factor`(`*tmp*`, thisvar, value = 0) :
  invalid factor level, NA generated

test

  a    b c
1 1    A 0
2 0    b 0
3 2 <NA> 0


The problem is the default conversion to factors and the replacement
operation for factors. So:

test <- data.frame(a=c(1,NA,2), b = I(c("A","b",NA_character_)), c= rep(NA,3))
class(test$b)

[1] "AsIs"  ## so NOT a factor

test[is.na(test)] <- 0 # now works as you describe
test

  a b c
1 1 A 0
2 0 b 0
3 2 0 0

Of course the OP (and you) probably had a data frame of all numerics
in mind, so the problem doesn't arise. But I think one needs to make
the distinction and issue clear.

Cheers,
Bert





Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Jun 23, 2016 at 8:46 AM, David L Carlson <dcarlson at tamu.edu> wrote:

The function is.na() returns a matrix when applied to a data.frame so you can easily convert all the NAs to 0's:

ds_test

   var1 var2
1     1    1
2     2    2
3     3    3
4    NA   NA
5     5    5
6     6    6
7     7    7
8    NA   NA
9     9    9
10   10   10

is.na(ds_test)

       var1  var2
 [1,] FALSE FALSE
 [2,] FALSE FALSE
 [3,] FALSE FALSE
 [4,]  TRUE  TRUE
 [5,] FALSE FALSE
 [6,] FALSE FALSE
 [7,] FALSE FALSE
 [8,]  TRUE  TRUE
 [9,] FALSE FALSE
[10,] FALSE FALSE

ds_test[is.na(ds_test)] <- 0
ds_test

   var1 var2
1     1    1
2     2    2
3     3    3
4     0    0
5     5    5
6     6    6
7     7    7
8     0    0
9     9    9
10   10   10

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ivan Calandra
Sent: Thursday, June 23, 2016 10:14 AM
To: R Help
Subject: Re: [R] Subscripting problem with is.na()

Thank you Bert for this clarification. It is indeed an important point.

Ivan

--
Ivan Calandra, PhD
Scientific Mediator
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calandra at univ-reims.fr
--
https://www.researchgate.net/profile/Ivan_Calandra
https://publons.com/author/705639/

Le 23/06/2016 ? 17:06, Bert Gunter a ?crit :

Sorry, Ivan, your statement is incorrect:

"When you use a single bracket on a list with only one argument in
between, then R extracts "elements", i.e. columns in the case of a
data.frame. This explains your errors. "

e.g.

ex <- data.frame(a = 1:3, b = letters[1:3])
a <- 1:3
identical(ex[1], a)

[1] FALSE

class(ex[1])

[1] "data.frame"

class(a)

[1] "integer"

Compare:

identical(ex[[1]], a)

[1] TRUE

Why? Single bracket extraction on a list results in a list; double
bracket extraction results in the element of the list ( a "column" in
the case of a data frame, which is a specific kind of list). The
relevant sections of ?Extract are:

"Indexing by [ is similar to atomic vectors and selects a **list** of
the specified element(s).

Both [[ and $ select a **single element of the list**. "


Hope this clarifies this often-confused issue.


Cheers,
Bert
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Jun 23, 2016 at 7:34 AM, Ivan Calandra
<ivan.calandra at univ-reims.fr> wrote:

My statement "Using a single bracket '[' on a data.frame does the same as
for matrices: you need to specify rows and columns" was not correct.


When you use a single bracket on a list with only one argument in between,
then R extracts "elements", i.e. columns in the case of a data.frame. This
explains your errors.

But it is possible to use a single bracket on a data.frame with 2 arguments
(rows, columns) separated by a comma, as with matrices. This is the solution
you received.

Ivan


--
Ivan Calandra, PhD
Scientific Mediator
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calandra at univ-reims.fr
--
https://www.researchgate.net/profile/Ivan_Calandra
https://publons.com/author/705639/

Le 23/06/2016 ? 16:27, Ivan Calandra a ?crit :

Dear Georg,

You need to learn a bit more about the subsetting methods, depending on
the object structure you're trying to subset.

More specifically, when you run this: ds_test[is.na(ds_test$var1)]
you get this error: "Error in `[.data.frame`(ds_test, is.na(ds_test$var1))
: undefined columns selected"

This means that R does not understand which column you're trying to
select. But you're actually trying to select rows.

Using a single bracket '[' on a data.frame does the same as for matrices:
you need to specify rows and columns, like this:
ds_test[is.na(ds_test$var1), ] ## notice the last comma
ds_test[is.na(ds_test$var1), ] <- 0 ## works on all columns because you
didn't specify any after the comma

If you want it only for "var1", then you need to specify the column:
ds_test[is.na(ds_test$var1), "var1"] <- 0

It's the same problem with your 2nd and 4th tries (4th one has other
problems). Your 3rd try does not change ds_test at all.

HTH,
Ivan

--
Ivan Calandra, PhD
Scientific Mediator
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calandra at univ-reims.fr
--
https://www.researchgate.net/profile/Ivan_Calandra
https://publons.com/author/705639/

Le 23/06/2016 ? 15:57, G.Maubach at weinwolf.de a ?crit :

Hi All,

I would like to recode my NAs to 0. Using a single vector everything is
fine.

But if I use a data.frame things go wrong:

-- cut --

var1 <- c(1:3, NA, 5:7, NA, 9:10)
var2 <- c(1:3, NA, 5:7, NA, 9:10)
ds_test <-
    data.frame(var1, var2)

test <- var1
test[is.na(test)] <- 0
test  # NA recoded OK

# First try
ds_test[is.na(ds_test$var1)] <- 0  # duplicate subscripts WRONG

# Second try
ds_test[is.na("var1")] <- 0
ds_test$var1  # not recoded WRONG

# Third try: to me the most intuitive approach
is.na(ds_test["var1"]) <- 0  # attempt to select less than one element in
integerOneIndex WRONG

# Fourth try
ds_test[is.na(var1)] <- 0  # duplicate subscripts for columns WRONG

-- cut --
   How can I do it correctly?

Where could I have found something about it?

Kind regards

Georg