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)] <- 0Warning message: In `[<-.factor`(`*tmp*`, thisvar, value = 0) : invalid factor level, NA generated> testa 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 > testa 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 >>>>> >>>>> ______________________________________________ >>>>> 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. >>>> >>> ______________________________________________ >>> 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. > ______________________________________________ > 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.
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 > testa 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)] <- 0Warning message: In `[<-.factor`(`*tmp*`, thisvar, value = 0) : invalid factor level, NA generated> testa 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 > testa 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 >>>>> >>>>> ______________________________________________ >>>>> 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. >>>> >>> ______________________________________________ >>> 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. > ______________________________________________ > 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.
... 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 >>>>>> >>>>>> ______________________________________________ >>>>>> 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. >>>>> >>>> ______________________________________________ >>>> 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. >> ______________________________________________ >> 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.
G.Maubach at weinwolf.de
2016-Jun-27 08:45 UTC
[R] Antwort: Fw: Re: Subscripting problem with is.na()
Hi David,
Hi Bert,
many thanks for the valuable discussion on NA in R (please see extract
below). I follow your arguments leaving NA as they are for most of the
time. In special occasions however I want to replace the NA with another
value. To preserve the newly acquired knowledge for me I wrote this
function:
-- cut --
t_replace_na <- function(dataset, variable, value) {
if(inherits(dataset[[variable]], "factor") == TRUE) {
dataset[variable] <- as.character(dataset[variable])
print(class(dataset[variable]))
dataset[, variable][is.na(dataset[, variable])] <- value
dataset[variable] <- as.factor(dataset[variable])
print(class(dataset[variable]))
} else {
dataset[, variable][is.na(dataset[, variable])] <- value
}
return(dataset)
}
ds_test <- data.frame(a=c(1,NA,2), b = rep(NA,3), c =
c("A","b",NA))
print(sapply(ds_test, class))
t_replace_na(ds_test, "a", value = -1)
t_replace_na(ds_test, "b", value = -2)
t_replace_na(ds_test, "c", value = -3)
-- cut --
Unfortunately the if-statement does not work due to a wrong class
definition within the function. When finding out what is going on I did
this:
-- cut --
test_class <- function(dataset, variable) {
if(inherits(dataset[, variable], "factor") == TRUE) {
return(c(class(dataset[variable]), TRUE))
} else {
return(c(class(dataset[variable]), FALSE))
}
}
ds_test <- data.frame(a=c(1,NA,2), b = rep(NA,3), c =
c("A","b",NA))
print(sapply(ds_test, class))
# -- Test a --
class(ds_test[, "a"])
if(inherits(ds_test[, "a"], "factor")) {
print(c(class(ds_test[, "a"]), "TRUE"))
} else {
print(c(class(ds_test[, "a"]), "FALSE"))
}
test_class(ds_test, "a")
warning("'a' should be numeric NOT data.frame!")
# -- Test b --
if(inherits(ds_test[, "b"], "factor")) {
print(c(class(ds_test[, "b"]), "TRUE"))
} else {
print(c(class(ds_test[, "b"]), "FALSE"))
}
class(ds_test[, "b"])
test_class(ds_test, "b")
warning("'b' should be logical NOT data.frame!")
# -- Test c --
if(inherits(ds_test[, "c"], "factor")) {
print(c(class(ds_test[, "c"]), "TRUE"))
} else {
print(c(class(ds_test[, "c"]), "FALSE"))
}
class(ds_test[, "c"])
test_class(ds_test, "c")
warning("'c' should be factor NOT data.frame.
In addition data.frame != factor")
-- cut --
Why do I get different results for the same function if it is inside or
outside my own function definition?
Kind regards
Georg
--------------------------------
> Gesendet: Donnerstag, 23. Juni 2016 um 21:14 Uhr
> Von: "David L Carlson" <dcarlson at tamu.edu>
> An: "Bert Gunter" <bgunter.4567 at gmail.com>
> Cc: "R Help" <r-help at r-project.org>
> Betreff: Re: [R] 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
Hi see in line> -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of > G.Maubach at weinwolf.de > Sent: Monday, June 27, 2016 10:45 AM > To: David L Carlson <dcarlson at tamu.edu>; Bert Gunter > <bgunter.4567 at gmail.com> > Cc: r-help at r-project.org > Subject: [R] Antwort: Fw: Re: Subscripting problem with is.na() > > Hi David, > Hi Bert, > > many thanks for the valuable discussion on NA in R (please see extract > below). I follow your arguments leaving NA as they are for most of the > time. In special occasions however I want to replace the NA with another > value. To preserve the newly acquired knowledge for me I wrote this > function: > > -- cut -- > t_replace_na <- function(dataset, variable, value) { > if(inherits(dataset[[variable]], "factor") == TRUE) { > dataset[variable] <- as.character(dataset[variable]) > print(class(dataset[variable])) > dataset[, variable][is.na(dataset[, variable])] <- value > dataset[variable] <- as.factor(dataset[variable]) > print(class(dataset[variable])) > } else { > dataset[, variable][is.na(dataset[, variable])] <- value > } > return(dataset) > } ><snip>> class(ds_test[, "c"]) > test_class(ds_test, "c") > warning("'c' should be factor NOT data.frame. > In addition data.frame != factor") > -- cut -- > > Why do I get different results for the same function if it is inside or > outside my own function definition?Because you still are missing the way how to subscript data frames. test_class <- function(dataset, variable) { if(inherits(dataset[, variable], "factor") == TRUE) { return(c(class(dataset[,variable]), TRUE)) #### ^^^^ } else { return(c(class(dataset[,variable]), FALSE)) ###### ^^^^ } }> test_class(ds_test, "a")[1] "numeric" "FALSE"> test_class(ds_test, "c")[1] "factor" "TRUE">If you properly arrange commas in your function you get desired result p_replace_na <- function(dataset, variable, value) { if(inherits(dataset[,variable], "factor") == TRUE) { dataset[,variable] <- as.character(dataset[,variable]) print(class(dataset[,variable])) dataset[, variable][is.na(dataset[, variable])] <- value dataset[, variable] <- as.factor(dataset[, variable]) print(class(dataset[, variable])) } else { dataset[, variable][is.na(dataset[, variable])] <- value } return(dataset) }> p_replace_na(ds_test, "c", value = -3)[1] "character" [1] "factor" a b c 1 1 NA A 2 NA NA b 3 2 NA -3> t_replace_na(ds_test, "c", value = -3)[1] "data.frame" Error in sort.list(y) : 'x' must be atomic for 'sort.list' Have you called 'sort' on a list?>Cheers Petr> > Kind regards > > Georg > > -------------------------------- > > > Gesendet: Donnerstag, 23. Juni 2016 um 21:14 Uhr > > Von: "David L Carlson" <dcarlson at tamu.edu> > > An: "Bert Gunter" <bgunter.4567 at gmail.com> > > Cc: "R Help" <r-help at r-project.org> > > Betreff: Re: [R] 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 > > ______________________________________________ > 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.________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. 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