problem for strsplit function

Thanks Bert,

I'm reading some books now. But it takes me a while to get familiar R.


Best,

Kai
On Friday, July 9, 2021, 03:06:11 PM PDT, 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