Subscripting problem with is.na()

As Petr and Don have shown you, changing NA to 0 is unnecessary to get
what you want. However, recoding to 0 may be OK, as NA has a specific
meaning in this context, and you are just adding an extra code to a
factor for a different level.

But it still might cause you trouble later. One of R's strengths is
it's ability to simply deal with NA's -- most of the time anyway .For
example note that you would have to make sure these columns are
factors (*not numerics*), if you wanted to, say, investigate how
category of closing related to other covariates via e.g. multinomial
logistic regression or even just to tabulate the "closed" categories.
Keeping NA as NA allows R's built-in facilities to simply handle (e.g.
omit) the data for the "still open" cases, but you will have to do it
explicitly yourself if you code to 0. That seems to be asking for
trouble to me.

As always, contrary views welcome. This discussion still seems on
(r-help) topic to me, but if not, please say so.

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 Fri, Jun 24, 2016 at 12:14 AM,  <G.Maubach at gmx.de> wrote:

Hi Bert,

many thanks for all your help and your comments. I learn at lot this way.

My question was about is.na() at the first sight but the actual task

looks like this:

I have two variables in my customer data that signal if the customer

accout was closed by master data management or by sales. Say these
variables are closed_mdm and closed_sls. They contain NA if the customer
account is still open or a closing code from "01" to "08" if the customer
account was closed and why.

For my analysis I need a variable that combines the two variables

closed_mdm and closed_sls to set a filter easily on those who are closed
not matter what the reason was nor who closed the account.

As I always encounter problems when dealing with ifelse statements and

NA I decided to merge these two variables to one variable containing 0 =
not closed and 1 = closed. In my context this seems to be - at least to me
- a reasonable approach.

Replacement of missing values and merging the variables is the easiest

way for me.

-- cut --

cust_id <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,

18, 19, 20)

closed_mdm <- c("01", NA, NA, NA, "08", "07", NA, NA, "05", NA, NA, NA,

"04", NA, NA, NA, NA, NA, NA, NA)

closed_sls <- c(NA, "08", NA, NA, "08", "07", NA, NA, NA, NA, "03", NA,

NA, NA, "05", NA, NA, NA, NA, NA)

# 1st try
ds_temp1 <- data.frame(cust_id, closed_mdm, closed_sls)
ds_temp1

ds_temp1$closed <- closed_mdm | closed_sls  # WRONG

# 2nd try
closed_mdm_fac1 <- as.factor(closed_mdm)
closed_sls_fac1 <- as.factor(closed_sls)

ds_temp2 <- data.frame(cust_id, closed_mdm_fac1, closed_sls_fac1)
ds_temp2

ds_temp2$closed <- ds_temp$closed_mdm_fac1 | ds_temp$closed_sls_fac1  #

WRONG

# 3rd try
closed_mdm_num1 <- as.numeric(closed_mdm)  # OK
closed_sls_num1 <- as.numeric(closed_sls)  # OK

ds_temp3 <- data.frame(cust_id, closed_mdm_num1, closed_sls_num1)
ds_temp3

ds_temp3$closed <- ds_temp$closed_mdm_num1 | ds_temp$closed_sls_num1  #

WRONG

# 4th try
ds_temp4 <- ds_temp3
ds_temp4

# Does not run due to not allowed NA in subscripts
ds_temp4[is.na(ds_temp4$closed_mdm_num1), ds_temp4$closed_mdm_num1] <- 0
ds_temp4[is.na(ds_temp4$closed_sls_num1), ds_temp4$closed_sls_num1] <- 0

# 5th try
ds_temp4$closed_mdm_num1 <- ifelse(is.na(ds_temp4$closed_mdm_num1), 1,

0)

ds_temp4$closed_sls_num1 <- ifelse(is.na(ds_temp4$closed_sls_num1), 1,

0)

ds_temp4

ds_temp4$closed <- ifelse(ds_temp4$closed_mdm_num1 == 1 |

ds_temp4$closed_sls_num1 == 1, 1, 0)

ds_temp4

-- cut --

Is there a better way to do it?

Kind regards

Georg

Gesendet: Donnerstag, 23. Juni 2016 um 23:55 Uhr
Von: "Bert Gunter" <bgunter.4567 at gmail.com>
An: "David L Carlson" <dcarlson at tamu.edu>
Cc: "R Help" <r-help at r-project.org>
Betreff: Re: [R] Subscripting problem with is.na()

... actually, FWIW, I would say that this little discussion mostly
demonstrates why the OP's request is probably not a good idea in the
first place. Usually, NA's should be left as NA's to be dealt with
properly by R and packages. In biological measurements, for example,
NA's often mean "below the ability to reliably measure." Biologists
with whom I've worked over many years often want to convert these to 0
or omit the cases, both of which lead to biased estimates and/or
underestimates of variability and excess claims of "statistical
significance" (for those who belong to this religious persuasion). One
should never say never, but I suspect that there are relatively few
circumstances where the conversion the OP requested is actually wise.

Feel free to ignore/reject such extraneous comments of course.

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 12:14 PM, David L Carlson <dcarlson at tamu.edu>

wrote:

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

see

code.