"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 <minsh...@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
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