Odd behavior of a function within apply - R-help

John Fox

Mon, Aug 8, 2022 10:37 AM #

Dear Erin,

The problem is that the data frame gets coerced to a character matrix, 
and the only column with "" entries is the 9th (the second one you 
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John

On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

______________________________________________
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.

John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://socialsciences.mcmaster.ca/jfox/

Erin Hodgess

Mon, Aug 8, 2022 10:41 AM #

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character matrix,
and the only column with "" entries is the 9th (the second one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L,
NA, 48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = c("December",
"June", "August", "", "", "August", "April", "February", "",
"December"), X1_1_HBM1_yr = c(2014L, 2018L, 2016L, NA, NA, 1985L,
2019L, 1993L, NA, 1990L)), row.names = c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it works fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
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.
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://socialsciences.mcmaster.ca/jfox/

Erin Hodgess

Mon, Aug 8, 2022 10:51 AM #

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

Error in FUN(X[[i]], ...) : object 'y' not found

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character matrix,
and the only column with "" entries is the 9th (the second one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L,
NA, 48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = c("December",
"June", "August", "", "", "August", "April", "February", "",
"December"), X1_1_HBM1_yr = c(2014L, 2018L, 2016L, NA, NA, 1985L,
2019L, 1993L, NA, 1990L)), row.names = c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
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.
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://socialsciences.mcmaster.ca/jfox/

Erin Hodgess

Mon, Aug 8, 2022 11:05 AM #

Nailed it!

There were a few "logical" columns in my data.frame.

Thanks for all of the help!
Sincerely,
Erin

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

On Mon, Aug 8, 2022 at 1:51 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

lapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

sapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com>
wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character matrix,
and the only column with "" entries is the 9th (the second one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L,
NA, 48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = c("December",
"June", "August", "", "", "August", "April", "February", "",
"December"), X1_1_HBM1_yr = c(2014L, 2018L, 2016L, NA, NA, 1985L,
2019L, 1993L, NA, 1990L)), row.names = c(NA, 10L), class =

"data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
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.
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://socialsciences.mcmaster.ca/jfox/

David L Carlson

Tue, Aug 9, 2022 8:33 AM #

Could you have columns that are not character or integer so that y is never
defined in the function?

count1a(1:5/3)
Error in count1a(1:5/3) : object 'y' not found

David Carlson

On Mon, Aug 8, 2022 at 1:35 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

OK.?? I'm back again.?? So my test1.??df is 236x390 If I put in the
following:?? lapply(test1.??df,count1a) Error in FUN(X[[i]], .??.??.??) :??
object 'y' not found > lapply(test1.??df,count1a) Error in FUN(X[[i]],
.??.??.??) :?? object 'y' not found > sapply(test1.??df,count1a)
ZjQcmQRYFpfptBannerStart
This Message Is From an External Sender
This message came from outside your organization.

ZjQcmQRYFpfptBannerEnd

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

lapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

sapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character matrix,
and the only column with "" entries is the 9th (the second one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L,
NA, 48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = c("December",
"June", "August", "", "", "August", "April", "February", "",
"December"), X1_1_HBM1_yr = c(2014L, 2018L, 2016L, NA, NA, 1985L,
2019L, 1993L, NA, 1990L)), row.names = c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide

and provide commented, minimal, self-contained, reproducible code.

______________________________________________R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, seehttps://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide https://urldefense.com/v3/__http://www.R-project.org/posting-guide.html__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MdYOqruE$
and provide commented, minimal, self-contained, reproducible code.

Erin Hodgess

Tue, Aug 9, 2022 8:35 AM #

My ?count1a? function was only looking for types of integers and
characters.  There were a few logical types, which were the source of the
error.

Thanks,
Erin

On Tue, Aug 9, 2022 at 11:33 AM David Carlson <dcarlson at tamu.edu> wrote:

Could you have columns that are not character or integer so that y is
never defined in the function?

count1a(1:5/3)
Error in count1a(1:5/3) : object 'y' not found

David Carlson


On Mon, Aug 8, 2022 at 1:35 PM Erin Hodgess <erinm.hodgess at gmail.com>
wrote:

OK.?? I'm back again.?? So my test1.??df is 236x390 If I put in the
following:?? lapply(test1.??df,count1a) Error in FUN(X[[i]], .??.??.??) :??
object 'y' not found > lapply(test1.??df,count1a) Error in FUN(X[[i]],
.??.??.??) :?? object 'y' not found > sapply(test1.??df,count1a)
ZjQcmQRYFpfptBannerStart
This Message Is From an External Sender
This message came from outside your organization.

ZjQcmQRYFpfptBannerEnd

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

lapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

sapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character matrix,
and the only column with "" entries is the 9th (the second one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L,
NA, 48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = c("December",
"June", "August", "", "", "August", "April", "February", "",
"December"), X1_1_HBM1_yr = c(2014L, 2018L, 2016L, NA, NA, 1985L,
2019L, 1993L, NA, 1990L)), row.names = c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org

 mailing list -- To UNSUBSCRIBE and more, see
https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide
https://urldefense.com/v3/__http://www.R-project.org/posting-guide.html__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MdYOqruE$
and provide commented, minimal, self-contained, reproducible code.
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://urldefense.com/v3/__https://socialsciences.mcmaster.ca/jfox/__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MRU4wu3o$



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PLEASE do read the posting guide https://urldefense.com/v3/__http://www.R-project.org/posting-guide.html__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MdYOqruE$
and provide commented, minimal, self-contained, reproducible code.

--

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

@vi@e@gross m@iii@g oii gm@ii@com

Tue, Aug 9, 2022 9:55 AM #

Yes, David, the function described seems to insist it be of type integer or type character and if the type was double or others might well fail as y would never be initialized.

The goal seems to be to count how many "missing" values are found as in NA if a numeric type or an empty string if character.

But you can have some form of NA in all kinds of object types including character as in this construct:

[1] "a"   NA    ""    "b"   "NA)"

The above has three useless elements if both NA and "" are considered empty. So logically the condition could be to count NA and IF it is of type character, also count "". 

So rather than play games testing not just is.integer, is.double (or just is.numeric) as well as is.logical and is.raw, all the above can be tested with is.na() first to add up how many Na they contain. If then it is of type character, you can add any blank strings. 

So the algorithm would initialize y to sum(is.na(vec)) and then if the vec is character, add the sum of how many empty strings.

Alternately, the function should deal with what it wants to do if any other type is encountered. You can internally converts many things to integer or character and then operate on them. Or you can return a zero or raise an alarm when given something else.

In this case, simply setting y to zero before using it would make it defined and avoid the error, albeit report nothing found if it was a double or Boolean vector even if it did contain NA.


-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of David Carlson via R-help
Sent: Tuesday, August 9, 2022 11:33 AM
To: Erin Hodgess <erinm.hodgess at gmail.com>
Cc: r-help at r-project.org
Subject: Re: [R] Odd behavior of a function within apply

Could you have columns that are not character or integer so that y is never defined in the function?

count1a(1:5/3)
Error in count1a(1:5/3) : object 'y' not found

David Carlson

On Mon, Aug 8, 2022 at 1:35 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

OK.?? I'm back again.?? So my test1.??df is 236x390 If I put in the 
following:?? lapply(test1.??df,count1a) Error in FUN(X[[i]], 
.??.??.??) :?? object 'y' not found > lapply(test1.??df,count1a) Error 
in FUN(X[[i]],
.??.??.??) :?? object 'y' not found > sapply(test1.??df,count1a) 
ZjQcmQRYFpfptBannerStart This Message Is From an External Sender This 
message came from outside your organization.

ZjQcmQRYFpfptBannerEnd

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

lapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

sapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com> wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character 
matrix, and the only column with "" entries is the 9th (the second 
one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L, NA, 
48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon = 
c("December", "June", "August", "", "", "August", "April", 
"February", "", "December"), X1_1_HBM1_yr = c(2014L, 2018L, 
2016L, NA, NA, 1985L, 2019L, 1993L, NA, 1990L)), row.names = 
c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it 
works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

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______________________________________________
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RHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide

and provide commented, minimal, self-contained, reproducible code.

______________________________________________R-help at r-project.org 
mailing list -- To UNSUBSCRIBE and more, 
seehttps://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r
-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJ
Jdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide 
https://urldefense.com/v3/__http://www.R-project.org/posting-guide.htm
l__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6
Uq0p4rpBa4E3DkmQ65UImH48MdYOqruE$ and provide commented, minimal, 
self-contained, reproducible code.

______________________________________________
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.

Erin Hodgess

Tue, Aug 9, 2022 10:04 AM #

Avi, that?s great!

Thanks

On Tue, Aug 9, 2022 at 12:56 PM <avi.e.gross at gmail.com> wrote:

Yes, David, the function described seems to insist it be of type integer
or type character and if the type was double or others might well fail as y
would never be initialized.

The goal seems to be to count how many "missing" values are found as in NA
if a numeric type or an empty string if character.

But you can have some form of NA in all kinds of object types including
character as in this construct:

x <- c("a", NA, "", "b", "NA)")
x

[1] "a"   NA    ""    "b"   "NA)"

The above has three useless elements if both NA and "" are considered
empty. So logically the condition could be to count NA and IF it is of type
character, also count "".

So rather than play games testing not just is.integer, is.double (or just
is.numeric) as well as is.logical and is.raw, all the above can be tested
with is.na() first to add up how many Na they contain. If then it is of
type character, you can add any blank strings.

So the algorithm would initialize y to sum(is.na(vec)) and then if the
vec is character, add the sum of how many empty strings.

Alternately, the function should deal with what it wants to do if any
other type is encountered. You can internally converts many things to
integer or character and then operate on them. Or you can return a zero or
raise an alarm when given something else.

In this case, simply setting y to zero before using it would make it
defined and avoid the error, albeit report nothing found if it was a double
or Boolean vector even if it did contain NA.


-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of David Carlson
via R-help
Sent: Tuesday, August 9, 2022 11:33 AM
To: Erin Hodgess <erinm.hodgess at gmail.com>
Cc: r-help at r-project.org
Subject: Re: [R] Odd behavior of a function within apply

Could you have columns that are not character or integer so that y is
never defined in the function?

count1a(1:5/3)
Error in count1a(1:5/3) : object 'y' not found

David Carlson


On Mon, Aug 8, 2022 at 1:35 PM Erin Hodgess <erinm.hodgess at gmail.com>
wrote:

OK.?? I'm back again.?? So my test1.??df is 236x390 If I put in the
following:?? lapply(test1.??df,count1a) Error in FUN(X[[i]],
.??.??.??) :?? object 'y' not found > lapply(test1.??df,count1a) Error
in FUN(X[[i]],
.??.??.??) :?? object 'y' not found > sapply(test1.??df,count1a)
ZjQcmQRYFpfptBannerStart This Message Is From an External Sender This
message came from outside your organization.

ZjQcmQRYFpfptBannerEnd

OK.  I'm back again.

So my test1.df is 236x390

If I put in the following:
 lapply(test1.df,count1a)
Error in FUN(X[[i]], ...) : object 'y' not found

lapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

sapply(test1.df,count1a)

Error in FUN(X[[i]], ...) : object 'y' not found

What am I doing wrong, please?
Thanks,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:41 PM Erin Hodgess <erinm.hodgess at gmail.com>

wrote:

Awesome, thanks so much!!

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com


On Mon, Aug 8, 2022 at 1:38 PM John Fox <jfox at mcmaster.ca> wrote:

Dear Erin,

The problem is that the data frame gets coerced to a character
matrix, and the only column with "" entries is the 9th (the second
one you
supplied):

as.matrix(test1.df)
    X1_1_HZP1 X1_1_HBM1_mon X1_1_HBM1_yr
1  "48160"   "December"    "2014"
2  "48198"   "June"        "2018"
3  "80027"   "August"      "2016"
4  "48161"   ""            NA
5  NA        ""            NA
6  "48911"   "August"      "1985"
7  NA        "April"       "2019"
8  "48197"   "February"    "1993"
9  "48021"   ""            NA
10 "11355"   "December"    "1990"

(Here, test1.df only contains the three columns you provided.)

A solution is to use sapply:

 > sapply(test1.df, count1a)

     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
             2             3             3


I hope this helps,
  John


On 2022-08-08 1:22 p.m., Erin Hodgess wrote:

Hello!

I have the following data.frame
  dput(test1.df[1:10,8:10])
structure(list(X1_1_HZP1 = c(48160L, 48198L, 80027L, 48161L, NA,
48911L, NA, 48197L, 48021L, 11355L), X1_1_HBM1_mon =
c("December", "June", "August", "", "", "August", "April",
"February", "", "December"), X1_1_HBM1_yr = c(2014L, 2018L,
2016L, NA, NA, 1985L, 2019L, 1993L, NA, 1990L)), row.names =
c(NA, 10L), class = "data.frame")

And the following function:

dput(count1a)

function (x)
{
     if (typeof(x) == "integer")
         y <- sum(is.na(x))
     if (typeof(x) == "character")
         y <- sum(x == "")
     return(y)
}
When I use the apply function with count1a, I get the following:
  apply(test1.df[1:10,8:10],2,count1a)
     X1_1_HZP1 X1_1_HBM1_mon  X1_1_HBM1_yr
            NA             3            NA
However, when I do use columns 8 and 10, I get the correct response:
  apply(test1.df[1:10,c(8,10)],2,count1a)
    X1_1_HZP1 X1_1_HBM1_yr
            2            3

I am really baffled.  If I use count1a on a single column, it
works

fine.

Any suggestions much appreciated.
Thanks,
Sincerely,
Erin


Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

      [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo
/r-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kd
RHAy8SJJdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide

and provide commented, minimal, self-contained, reproducible code.

______________________________________________R-help at r-project.org
mailing list -- To UNSUBSCRIBE and more,
seehttps://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r
-help__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJ
Jdx6Uq0p4rpBa4E3DkmQ65UImH48MBvSbrfE$
PLEASE do read the posting guide
https://urldefense.com/v3/__http://www.R-project.org/posting-guide.htm
l__;!!KwNVnqRv!CHx9JKnbOObpAt0LltEogLSxDUEl9qJDI6FgqMJBG_kdRHAy8SJJdx6
Uq0p4rpBa4E3DkmQ65UImH48MdYOqruE$ and provide commented, minimal,
self-contained, reproducible code.

______________________________________________
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.

______________________________________________
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.

Erin Hodgess, PhD
mailto: erinm.hodgess at gmail.com

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