problem for strsplit function

On 09/07/2021 6:44 p.m., Bert Gunter wrote:

OK, I stand somewhat chastised.

But my point still is that what you get when you "extract" depends on
how you define "extract." Do note that ?"[" yields a help file titled
"Extract or Replace Parts of an object"; and afaics, the term

"subset"

is not explicitly used as Duncan prefers.

?"[[" gives you the same page, but I agree:  this part of the 
documentation isn't written very clearly. The "Introduction to R"
manual 
uses the terms I used (see section 2.7, "Index vectors; selecting and 
modifying subsets of a data set"), as does the source code (and the R 
Language Definition manual, though it's not as clear as the Intro).

But the point isn't to chastise you, it's to educate you (and the OP). 
Thinking of [] as subsetting is more helpful than thinking of it as 
extraction.  That way the result of x[c(1,2)] makes sense.  It's a 
little bit more of a stretch, but the result of x[[c(1,2)]] also makes 
sense when you think of it as extraction.

Duncan Murdoch

 The relevant part of the

Help file says for "[" for recursive objects says: "Indexing by [ is
similar to atomic vectors and selects a list of the specified
element(s)."  That a data.frame is a list is explicitly stated, as I
noted; that lists are in fact vectors is also explicitly stated

(?list

says: "Almost all lists in R internally are Generic Vectors") but

then

one is stuck with: a data.frame is a list and therefore a vector, but
is.vector(d3) is FALSE. The explanation is explicit again in
?is.vector ("is.vector returns TRUE if x is a vector of the specified
mode having no attributes other than names. It returns FALSE
otherwise."). But I would say these issues are sufficiently murky

that

my warning to be precise is not entirely inappropriate;

unfortunately,

I may have made them more so. Sigh....

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, Jul 9, 2021 at 3:05 PM Duncan Murdoch

<murdoch.duncan at gmail.com> wrote:

On 09/07/2021 5:51 p.m., Jeff Newmiller wrote:

"Strictly speaking", Greg is correct, Bert.

https://cran.r-project.org/doc/manuals/r-release/R-lang.html#List-objects

Lists in R are vectors. What we colloquially refer to as "vectors"

are more precisely referred to as "atomic vectors". And without a
doubt, this "vector" nature of lists is a key underlying concept that
explains why adding a dim attribute creates a matrix that can hold data
frames. It is also a stumbling block for programmers from other
languages that have things like linked lists.

I would also object to v3 (below) as "extracting" a column from d.
"d[2]" doesn't extract anything, it "subsets" the data frame, so the
result is a data frame, not what you get when you extract something

from

a data frame.

People don't realize that "x <- 1:10; y <- x[[3]]" is perfectly

legal.

That extracts the 3rd element (the number 3).  The problem is that R

has

no way to represent a scalar number, only a vector of numbers, so

x[[3]]

gets promoted to a vector containing that number when it is returned

and

assigned to y.

Lists are vectors of R objects, so if x is a list, x[[3]] is

something

that can be returned, and it is different from x[3].

Duncan Murdoch

On July 9, 2021 2:36:19 PM PDT, Bert Gunter

<bgunter.4567 at gmail.com> wrote:

"1.  a column, when extracted from a data frame, *is* a vector."
Strictly speaking, this is false; it depends on exactly what is

meant

by "extracted." e.g.:

d <- data.frame(col1 = 1:3, col2 = letters[1:3])
v1 <- d[,2] ## a vector
v2 <- d[[2]] ## the same, i.e
identical(v1,v2)

[1] TRUE

v3 <- d[2] ## a data.frame
v1

[1] "a" "b" "c"  ## a character vector

v3

   col2
1    a
2    b
3    c

is.vector(v1)

[1] TRUE

is.vector(v3)

[1] FALSE

class(v3)  ## data.frame

[1] "data.frame"
## but

is.list(v3)

[1] TRUE

which is simply explained in ?data.frame (where else?!) by:
"A data frame is a **list** [emphasis added] of variables of the

same

number of rows with unique row names, given class "data.frame". If

no

variables are included, the row names determine the number of

rows."

"2.  maybe your question is "is a given function for a vector, or

for a

     data frame/matrix/array?".  if so, i think the only way is

reading

     the help information (?foo)."

Indeed! Is this not what the Help system is for?! But note also

that

the S3 class system may somewhat blur the issue: foo() may work
appropriately and differently for different (S3) classes of

objects. A

detailed explanation of this behavior can be found in appropriate
resources or (more tersely) via ?UseMethod .

"you might find reading ?"[" and  ?"[.data.frame" useful"

Not just 'useful" -- **essential** if you want to work in R,

unless

one gets this information via any of the numerous online

tutorials,

courses, or books that are available. The Help system is accurate

and

authoritative, but terse. I happen to like this mode of

documentation,

but others may prefer more extended expositions. I stand by this

claim

even if one chooses to use the "Tidyverse", data.table package, or
other alternative frameworks for handling data. Again, others may
disagree, but R is structured around these basics, and imo one

remains

ignorant of them at their peril.

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, Jul 9, 2021 at 11:57 AM Greg Minshall <minshall at umich.edu>
wrote:

Kai,

one more question, how can I know if the function is for column
manipulations or for vector?

i still stumble around R code.  but, i'd say the following (and

look

forward to being corrected! :):

1.  a column, when extracted from a data frame, *is* a vector.

2.  maybe your question is "is a given function for a vector, or

for

a

      data frame/matrix/array?".  if so, i think the only way is

reading

      the help information (?foo).

3.  sometimes, extracting the column as a vector from a data

frame-like

      object might be non-intuitive.  you might find reading ?"["

and

      ?"[.data.frame" useful (as well as ?"[.data.table" if you

use

that

      package).  also, the str() command can be helpful in

understanding

      what is happening.  (the lobstr:: package's sxp() function,

as

well

      as more verbose .Internal(inspect()) can also give you

insight.)

      with the data.table:: package, for example, if "DT" is a

data.table

      object, with "x2" as a column, adding or leaving off

quotation

marks

      for the column name can make all the difference between

ending up

      with a vector, or with a (much reduced) data table:
----

is.vector(DT[, x2])

[1] TRUE

str(DT[, x2])

   num [1:9] 32 32 32 32 32 32 32 32 32

is.vector(DT[, "x2"])

[1] FALSE

str(DT[, "x2"])

Classes ?data.table? and 'data.frame':  9 obs. of  1 variable:
   $ x2: num  32 32 32 32 32 32 32 32 32
   - attr(*, ".internal.selfref")=<externalptr>
----

      a second level of indexing may or may not help, mostly

depending

on

      the use of '[' versus of '[['.  this can sometimes cause

confusion

      when you are learning the language.
----

str(DT[, "x2"][1])

Classes ?data.table? and 'data.frame':  1 obs. of  1 variable:
   $ x2: num 32
   - attr(*, ".internal.selfref")=<externalptr>

str(DT[, "x2"][[1]])

   num [1:9] 32 32 32 32 32 32 32 32 32
----

      the tibble:: package (used in, e.g., the dplyr:: package)

also

      (always?) returns a single column as a non-vector.  again,

a

      second indexing with double '[[]]' can produce a vector.
----

DP <- tibble(DT)
is.vector(DP[, "x2"])

[1] FALSE

is.vector(DP[, "x2"][[1]])

[1] TRUE
----

      but, note that a list of lists is also a vector:

is.vector(list(list(1), list(1,2,3)))

[1] TRUE

str(list(list(1), list(1,2,3)))

List of 2
   $ :List of 1
    ..$ : num 1
   $ :List of 3
    ..$ : num 1
    ..$ : num 2
    ..$ : num 3

      etc.

hth.  good luck learning!

cheers, Greg

code.