Message-ID: <812547.1625856998@apollo2.minshall.org>
Date: 2021-07-09T18:56:38Z
From: Greg Minshall
Subject: problem for strsplit function
In-Reply-To: Your message of "Thu, 08 Jul 2021 15:24:25 +0000." <762548966.2994133.1625757865539@mail.yahoo.com>
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