Chris Evans
2016-Dec-06 21:26 UTC
[R] Odd behaviour of mean() with a numeric column in a tibble
I hope I am obeying the list rules here. I am using a raw R IDE for this and
running 3.3.2 (2016-10-31) on x86_64-w64-mingw32/x64 (64-bit)
Here is a reproducible example. Code only first
require(tibble)
tmpTibble <- tibble(ID=letters,num=1:26)
min(tmpTibble[,2]) # fine
max(tmpTibble[,2]) # fine
median(tmpTibble[,2]) # not fine
mean(tmpTibble[,2]) # not fine
newMeanFun <- function(x) {mean(as.numeric(unlist(x)))}
newMeanFun(tmpTibble[,2]) # solved problem but surely shouldn't be
necessary?!
newMedianFun <- function(x) {median(as.numeric(unlist(x)))}
newMedianFun(tmpTibble[,2]) # ditto
str(tmpTibble[,2])
### then I tried this to make sure it wasn't about having fed in integers
tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10)
tmpTibble2
mean(tmpTibble2[,3]) # not fine, not about integers!
### before I just created tmpTibble2 I found myself trying to add a column to
tmpTibble
tmpTibble$newNum <- tmpTibble[,2]/10 # NO!
tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO!
### and oddly enough ...
add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO!
Now here it is with the output:
> require(tibble)
Loading required package: tibble> tmpTibble <- tibble(ID=letters,num=1:26)
> min(tmpTibble[,2]) # fine
[1] 1> max(tmpTibble[,2]) # fine
[1] 26> median(tmpTibble[,2]) # not fine
Error in median.default(tmpTibble[, 2]) : need numeric
data> mean(tmpTibble[,2]) # not fine
[1] NA
Warning message:
In mean.default(tmpTibble[, 2]) :
argument is not numeric or logical: returning NA> newMeanFun <- function(x) {mean(as.numeric(unlist(x)))}
> newMeanFun(tmpTibble[,2]) # solved problem but surely shouldn't be
necessary?!
[1] 13.5> newMedianFun <- function(x) {median(as.numeric(unlist(x)))}
> newMedianFun(tmpTibble[,2]) # ditto
[1] 13.5> str(tmpTibble[,2])
Classes ?tbl_df?, ?tbl? and 'data.frame': 26 obs. of 1 variable:
$ num: int 1 2 3 4 5 6 7 8 9 10 ...>
> ### then I tried this to make sure it wasn't about having fed in
integers
>
> tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10)
> tmpTibble2
# A tibble: 26 ? 3
ID num num2
<chr> <int> <dbl>
1 a 1 0.1
2 b 2 0.2
3 c 3 0.3
4 d 4 0.4
5 e 5 0.5
6 f 6 0.6
7 g 7 0.7
8 h 8 0.8
9 i 9 0.9
10 j 10 1.0
# ... with 16 more rows> mean(tmpTibble2[,3]) # not fine, not about integers!
[1] NA
Warning message:
In mean.default(tmpTibble2[, 3]) :
argument is not numeric or logical: returning NA>
>
> ### before I just created tmpTibble2 I found myself trying to add a column
to tmpTibble
> tmpTibble$newNum <- tmpTibble[,2]/10 # NO!
> tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO!
> ### and oddly enough ...
> add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO!
Error: Each variable must be a 1d atomic vector or list.
Problem variables: 'newNum'>
>
I discovered this when I hit odd behaviour after using read_spss() from the
haven package for the first time as it seemed to be offering a step forward over
good old read.spss() from the excellent foreign package. I am reporting it here
not directly to Prof. Wickham as the issues seem rather general though I'm
guessing that it needs to be fixed with a fix to tibble. Or perhaps I've
completely missed something.
TIA,
Chris
Ista Zahn
2016-Dec-06 21:40 UTC
[R] Odd behaviour of mean() with a numeric column in a tibble
Not at a computer to check right now, but I believe single bracket indexing a tibble always returns a tibble. To extract a vector use [[ On Dec 6, 2016 4:28 PM, "Chris Evans" <chrishold at psyctc.org> wrote:> > I hope I am obeying the list rules here. I am using a raw R IDE for thisand running 3.3.2 (2016-10-31) on x86_64-w64-mingw32/x64 (64-bit)> > Here is a reproducible example. Code only first > > require(tibble) > tmpTibble <- tibble(ID=letters,num=1:26) > min(tmpTibble[,2]) # fine > max(tmpTibble[,2]) # fine > median(tmpTibble[,2]) # not fine > mean(tmpTibble[,2]) # not fineI think you want mean(tmpTibble[[2]]> newMeanFun <- function(x) {mean(as.numeric(unlist(x)))} > newMeanFun(tmpTibble[,2]) # solved problem but surely shouldn't benecessary?!> newMedianFun <- function(x) {median(as.numeric(unlist(x)))} > newMedianFun(tmpTibble[,2]) # ditto > str(tmpTibble[,2]) > > ### then I tried this to make sure it wasn't about having fed in integers > > tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10) > tmpTibble2 > mean(tmpTibble2[,3]) # not fine, not about integers! > > > ### before I just created tmpTibble2 I found myself trying to add acolumn to tmpTibble> tmpTibble$newNum <- tmpTibble[,2]/10 # NO! > tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO! > ### and oddly enough ... > add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO! > > Now here it is with the output: > > > require(tibble) > Loading required package: tibble > > tmpTibble <- tibble(ID=letters,num=1:26) > > min(tmpTibble[,2]) # fine > [1] 1 > > max(tmpTibble[,2]) # fine > [1] 26 > > median(tmpTibble[,2]) # not fine > Error in median.default(tmpTibble[, 2]) : need numeric data > > mean(tmpTibble[,2]) # not fine > [1] NA > Warning message: > In mean.default(tmpTibble[, 2]) : > argument is not numeric or logical: returning NA > > newMeanFun <- function(x) {mean(as.numeric(unlist(x)))} > > newMeanFun(tmpTibble[,2]) # solved problem but surely shouldn't benecessary?!> [1] 13.5 > > newMedianFun <- function(x) {median(as.numeric(unlist(x)))} > > newMedianFun(tmpTibble[,2]) # ditto > [1] 13.5 > > str(tmpTibble[,2]) > Classes ?tbl_df?, ?tbl? and 'data.frame': 26 obs. of 1 variable: > $ num: int 1 2 3 4 5 6 7 8 9 10 ... > > > > ### then I tried this to make sure it wasn't about having fed inintegers> > > > tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10) > > tmpTibble2 > # A tibble: 26 ? 3 > ID num num2 > <chr> <int> <dbl> > 1 a 1 0.1 > 2 b 2 0.2 > 3 c 3 0.3 > 4 d 4 0.4 > 5 e 5 0.5 > 6 f 6 0.6 > 7 g 7 0.7 > 8 h 8 0.8 > 9 i 9 0.9 > 10 j 10 1.0 > # ... with 16 more rows > > mean(tmpTibble2[,3]) # not fine, not about integers! > [1] NA > Warning message: > In mean.default(tmpTibble2[, 3]) : > argument is not numeric or logical: returning NA > > > > > > ### before I just created tmpTibble2 I found myself trying to add acolumn to tmpTibble> > tmpTibble$newNum <- tmpTibble[,2]/10 # NO! > > tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO! > > ### and oddly enough ... > > add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO! > Error: Each variable must be a 1d atomic vector or list. > Problem variables: 'newNum' > > > > > > I discovered this when I hit odd behaviour after using read_spss() fromthe haven package for the first time as it seemed to be offering a step forward over good old read.spss() from the excellent foreign package. I am reporting it here not directly to Prof. Wickham as the issues seem rather general though I'm guessing that it needs to be fixed with a fix to tibble. Or perhaps I've completely missed something.> > TIA, > > Chris > > ______________________________________________ > 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 guidehttp://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]]
Chris Evans
2016-Dec-06 22:10 UTC
[R] Odd behaviour of mean() with a numeric column in a tibble
{{SIGH}}
You are absolutely right.
I wonder if I am losing some cognitive capacities that are needed to be part of
the evolving R community. It seems to me that if a tibble is designed to be an
enhanced replacement for a dataframe then it shouldn't quite so radically
change things.
I notice that the documentation on tibble says "[ Never simplifies (drops),
so always returns data.frame"
That is much less explicit than I would have liked and actually doesn't seem
to be true. In fact, as you rightly say, it generally, but not quite always,
returns a tibble. In fact it can be fooled into a vector of length 1.
> tmpTibble[[1,]]
Error in `[[.data.frame`(tmpTibble, 1, ) :
argument "..2" is missing, with no default
> tmpTibble[1]
# A tibble: 26 ? 1
ID
<chr>
1 a
2 b
3 c
4 d
5 e
6 f
7 g
8 h
9 i
10 j
# ... with 16 more rows > tmpTibble[,1]
# A tibble: 26 ? 1
ID
<chr>
1 a
2 b
3 c
4 d
5 e
6 f
7 g
8 h
9 i
10 j
# ... with 16 more rows > tmpTibble[1,]
Error in `[<-.data.frame`(`*tmp*`, , value = list(ID = c("a",
"a", "a", :
replacement element 3 is a matrix/data frame of 26 rows, need 1
In addition: Warning messages:
1: In `[<-.data.frame`(`*tmp*`, , value = list(ID = c("a",
"a", "a", :
replacement element 1 has 26 rows to replace 1 rows
2: In `[<-.data.frame`(`*tmp*`, , value = list(ID = c("a",
"a", "a", :
replacement element 2 has 26 rows to replace 1 rows > tmpTibble[1,1:26]
Error: Invalid column indexes: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26 > tmpTibble[[1,2]]
[1] 1 > str(tmpTibble[[1,2]])
int 1 > str(tmpTibble[[1:2,2]])
Error in col[[i, exact = exact]] :
attempt to select more than one element in vectorIndex >
> tmpTibble[[1,1:2]]
[1] "b" >
So [[a,b]] works if a and b are legal with the dimensions of the tibble and if a
is of length 1 but returns NOT a tibble but a vector of length 1 (I think), I
can see that's logical but not what it says in the documentation.
[[a]] and [[,a]] return the same result, that seems excessively tolerant to me.
[[a,b:c]] actually returns [[a,c]] and again as a single value, NOT a tibble.
And row subsetting/indexing has gone.
Why create replacement for a dataframe that has no row indexing and so radically
redefines column indexing, in fact redefines the whole of indexing and
subsetting?
OK. I will go to sleep now and hope to feel less dumb(ed) when I wake. Perhaps
Prof. Wickham or someone can spell out a bit less tersely, and I think
incompletely, than the tibble documentation does, why all this is good.
Thanks anyway Ista, you certainly hit the issue!
Very best all,
Chris
> From: "Ista Zahn" <istazahn at gmail.com>
> To: "Chris Evans" <chrishold at psyctc.org>
> Cc: "r-helpr-project.org" <r-help at r-project.org>
> Sent: Tuesday, 6 December, 2016 21:40:41
> Subject: Re: [R] Odd behaviour of mean() with a numeric column in a tibble
> Not at a computer to check right now, but I believe single bracket indexing
a
> tibble always returns a tibble. To extract a vector use [[
> On Dec 6, 2016 4:28 PM, "Chris Evans" < chrishold at
psyctc.org > wrote:
>> I hope I am obeying the list rules here. I am using a raw R IDE for
this and
> > running 3.3.2 (2016-10-31) on x86_64-w64-mingw32/x64 (64-bit)
> > Here is a reproducible example. Code only first
> > require(tibble)
> > tmpTibble <- tibble(ID=letters,num=1:26)
> > min(tmpTibble[,2]) # fine
> > max(tmpTibble[,2]) # fine
> > median(tmpTibble[,2]) # not fine
> > mean(tmpTibble[,2]) # not fine
> I think you want
> mean(tmpTibble[[2]]
> > newMeanFun <- function(x) {mean(as.numeric(unlist(x)))}
> > newMeanFun(tmpTibble[,2]) # solved problem but surely shouldn't be
necessary?!
> > newMedianFun <- function(x) {median(as.numeric(unlist(x)))}
> > newMedianFun(tmpTibble[,2]) # ditto
> > str(tmpTibble[,2])
> > ### then I tried this to make sure it wasn't about having fed in
integers
> > tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10)
> > tmpTibble2
> > mean(tmpTibble2[,3]) # not fine, not about integers!
>> ### before I just created tmpTibble2 I found myself trying to add a
column to
> > tmpTibble
> > tmpTibble$newNum <- tmpTibble[,2]/10 # NO!
> > tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO!
> > ### and oddly enough ...
> > add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO!
> > Now here it is with the output:
> > > require(tibble)
> > Loading required package: tibble
> > > tmpTibble <- tibble(ID=letters,num=1:26)
> > > min(tmpTibble[,2]) # fine
> > [1] 1
> > > max(tmpTibble[,2]) # fine
> > [1] 26
> > > median(tmpTibble[,2]) # not fine
> > Error in median.default(tmpTibble[, 2]) : need numeric data
> > > mean(tmpTibble[,2]) # not fine
> > [1] NA
> > Warning message:
> > In mean.default(tmpTibble[, 2]) :
> > argument is not numeric or logical: returning NA
> > > newMeanFun <- function(x) {mean(as.numeric(unlist(x)))}
> > > newMeanFun(tmpTibble[,2]) # solved problem but surely
shouldn't be necessary?!
> > [1] 13.5
> > > newMedianFun <- function(x) {median(as.numeric(unlist(x)))}
> > > newMedianFun(tmpTibble[,2]) # ditto
> > [1] 13.5
> > > str(tmpTibble[,2])
> > Classes ?tbl_df?, ?tbl? and 'data.frame': 26 obs. of 1
variable:
> > $ num: int 1 2 3 4 5 6 7 8 9 10 ...
> > > ### then I tried this to make sure it wasn't about having fed
in integers
> > > tmpTibble2 <- tibble(ID=letters,num=1:26,num2=(1:26)/10)
> > > tmpTibble2
> > # A tibble: 26 ? 3
> > ID num num2
> > <chr> <int> <dbl>
> > 1 a 1 0.1
> > 2 b 2 0.2
> > 3 c 3 0.3
> > 4 d 4 0.4
> > 5 e 5 0.5
> > 6 f 6 0.6
> > 7 g 7 0.7
> > 8 h 8 0.8
> > 9 i 9 0.9
> > 10 j 10 1.0
> > # ... with 16 more rows
> > > mean(tmpTibble2[,3]) # not fine, not about integers!
> > [1] NA
> > Warning message:
> > In mean.default(tmpTibble2[, 3]) :
> > argument is not numeric or logical: returning NA
>> > ### before I just created tmpTibble2 I found myself trying to add
a column to
> > > tmpTibble
> > > tmpTibble$newNum <- tmpTibble[,2]/10 # NO!
> > > tmpTibble[["newNum"]] <- tmpTibble[,2]/10 # NO!
> > > ### and oddly enough ...
> > > add_column(tmpTibble,newNum = tmpTibble[,2]/10) # NO!
> > Error: Each variable must be a 1d atomic vector or list.
> > Problem variables: 'newNum'
>> I discovered this when I hit odd behaviour after using read_spss() from
the
>> haven package for the first time as it seemed to be offering a step
forward
>> over good old read.spss() from the excellent foreign package. I am
reporting it
>> here not directly to Prof. Wickham as the issues seem rather general
though I'm
>> guessing that it needs to be fixed with a fix to tibble. Or perhaps
I've
> > completely missed something.
> > TIA,
> > Chris
> > ______________________________________________
> > 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.
[[alternative HTML version deleted]]