Re: [R] problem for strsplit function
This discussion has developed in such a way that it seems a better subject line would be "problem for the hairsplit function". :-) cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 __ R-help@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.
Re: [R] problem for strsplit function
A bit too fast there, Duncan... x[[c(1,2)]] is illegal.
On July 9, 2021 5:16:13 PM PDT, Duncan Murdoch wrote:
>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
> 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
> 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
> 1a
> 2b
> 3c
>> 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
Re: [R] problem for strsplit function
My mental model for the `[` vs `[[` behavior is that `[` indexes multiple
results while `[[` indexes only one item. If returning multiple items from a
list the result must be a list. For consistency, `[` always returns a list when
applied to a list. The double bracket drops the containing list.
The is.vector() behavior is not intuitive to me... I avoid that function, as I
think it is more useful to think of lists as vectors than as something "other".
On July 9, 2021 3:44:29 PM PDT, 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. 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
> 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
> 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
>> >> 1a
>> >> 2b
>> >> 3c
>> >>> 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
Re: [R] 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 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 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
1a
2b
3c
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,
Re: [R] problem for strsplit function
"But it takes me a while to get familiar R." Of course. That is true for all of us. Just keep on plugging away and you'll get it. Probably far better than I before too long. 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 3:45 PM Kai Yang wrote: > > 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 > 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 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 > >> 1a > >> 2b > >> 3c > >>> 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 > >> 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.
Re: [R] problem for strsplit function
Thanks Bert, I'm reading some books now. But it takes me a while to get familiar R. Best, KaiOn Friday, July 9, 2021, 03:06:11 PM PDT, Duncan Murdoch 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 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 >> 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
Re: [R] problem for strsplit function
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. 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 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 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
> >> 1a
> >> 2b
> >> 3c
> >>> 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
Re: [R] problem for strsplit function
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 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 1a 2b 3c 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 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")= a second level of indexing may or may not help, mostly
Re: [R] problem for strsplit function
"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. On July 9, 2021 2:36:19 PM PDT, Bert Gunter 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 >1a >2b >3c >> 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 >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")= >> >> >> 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")= >> > 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. >> >>
Re: [R] problem for strsplit function
"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 1a 2b 3c > 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 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")= > > > 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")= > > 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 > > __ > R-help@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,
Re: [R] How to create a matrix from a list without a for loop
On 09/07/2021 3:37 p.m., Laurent Rhelp wrote:
Very effective solution, I hope I remember that for the nex time.
The key things to remember are that in R a "list" object is a vector
whose elements are R objects, and that matrices are just vectors with
dimension attributes.
Those two ideas are each useful for other things, too!
Duncan Murdoch
Thank you
Le 09/07/2021 à 19:50, David Winsemius a écrit :
On 7/9/21 10:40 AM, Laurent Rhelp wrote:
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create
a matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to write nice R code to do this loop ?
Thank you very much
Laurent
##
## mock data, 16 dataframes in a list
##
init_l <- lapply( seq(1,16) , function(x) {
data.frame( V1 = rnorm(3),
V2 = rnorm(3),
V3 = rnorm(3)
)
})
##
## lists matrix creation with n = 4 columns and n = 4 rows
##
n <- 4
Just assign a dimension attribute and you will have your two
dimensional structure
dim(init_l) <- c(n,n)
init_l[ 2,2]
[[1]]
V1 V2 V3
1 -1.4103259 1.9214184 -0.1590919
2 0.1899490 0.3842191 2.4502078
3 -0.4282764 -0.9992190 1.5384344
is.matrix(init_l)
[1] TRUE
__
R-help@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.
Re: [R] How to create a matrix from a list without a for loop
Very effective solution, I hope I remember that for the nex time.
Thank you
Le 09/07/2021 à 19:50, David Winsemius a écrit :
On 7/9/21 10:40 AM, Laurent Rhelp wrote:
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create
a matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to write nice R code to do this loop ?
Thank you very much
Laurent
##
## mock data, 16 dataframes in a list
##
init_l <- lapply( seq(1,16) , function(x) {
data.frame( V1 = rnorm(3),
V2 = rnorm(3),
V3 = rnorm(3)
)
})
##
## lists matrix creation with n = 4 columns and n = 4 rows
##
n <- 4
Just assign a dimension attribute and you will have your two
dimensional structure
> dim(init_l) <- c(n,n)
> init_l[ 2,2]
[[1]]
V1 V2 V3
1 -1.4103259 1.9214184 -0.1590919
2 0.1899490 0.3842191 2.4502078
3 -0.4282764 -0.9992190 1.5384344
> is.matrix(init_l)
[1] TRUE
--
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antivirus Avast.
https://www.avast.com/antivirus
__
R-help@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.
Re: [R] How to create a matrix from a list without a for loop
You are right, this extra level disturbed me.
Very impressive this solution, thank you very much.
Le 09/07/2021 à 19:50, Bill Dunlap a écrit :
> Try
> matrix(init_l, nrow=4, ncol=4,
> dimnames=list(c("X1","X2","X3","X4"),c("X1","X2","X3","X4")))
> It doesn't give exactly what your code does, but your code introduces
> an extra level of "list", which you may not want.
>
> -Bill
>
> On Fri, Jul 9, 2021 at 10:40 AM Laurent Rhelp > wrote:
>
> Dear R-Help-list,
>
> I have a list init_l containing 16 dataframes and I want to
> create a
> matrix 4 x 4 from this list with a dataframe in every cell of the
> matrix. I succeeded to do that but my loop is very uggly (cf. below).
> Could somebody help me to write nice R code to do this loop ?
>
> Thank you very much
>
> Laurent
>
>
> ##
> ## mock data, 16 dataframes in a list
> ##
> init_l <- lapply( seq(1,16) , function(x) {
> data.frame( V1 = rnorm(3),
> V2 = rnorm(3),
> V3 = rnorm(3)
> )
> })
>
> ##
> ## lists matrix creation with n = 4 columns and n = 4 rows
> ##
> n <- 4
> ## an example of row to create the matrix with lists in the cells
> one_row <- rbind( rep( list(rep(list(1),3)) , n) )
> mymat <- do.call( "rbind" , rep( list(one_row) , n) )
>
> ##
> ## The UGGLY loop I would like to improve:
> ##
>
> ## populate the matrix
> k <- 1
> for( i in 1:n){
> for( j in 1:n){
> mymat[i,j][[1]] <- list( init_l[[ k ]] )
> k <- k+1
> }
> }
>
> colnames(mymat) <- c("X1", "X2", "X3", "X3")
> rownames(mymat) <- c("X1", "X2", "X3", "X4")
>
>
> mymat
>
> # X1 X2 X3 X3
> # X1 List,1 List,1 List,1 List,1
> # X2 List,1 List,1 List,1 List,1
> # X3 List,1 List,1 List,1 List,1
> # X4 List,1 List,1 List,1 List,1
>
>
> #
> # verification, it works
> #
> mymat[2,2]
> init_l[[6]]
>
> ##
> init_l[[6]]
>
> library(tidyverse)
> mymat.t <- as.tibble(mymat)
> mymat.t
> unnest(mymat.t[2,2],cols="X2")[[1]][[1]]
>
> mymat.df <- as.data.frame(mymat)
> mymat.df[2,2][[1]][[1]]
>
>
> thx
>
>
>
> --
> L'absence de virus dans ce courrier électronique a été vérifiée
> par le logiciel antivirus Avast.
> https://www.avast.com/antivirus
>
> __
> R-help@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.
>
--
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[[alternative HTML version deleted]]
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Re: [R] error message from read.csv in loop
Hi Migdonio,
I did try your code:
# Initialize the rr variable as a list.
rr <- as.list(rep(NA, nrow(ora)))
# Run the for-loop to store all the CSVs in rr.
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv")
rdcsv <- noquote(paste0("w:/project/_Joe.B/Oracle/data/", mycsv))
rr[[j]] <- read.csv(rdcsv)
}
this code is working, but rr is not a data frame, R said: Large list ( 20
elements .). how can I use it as a data frame one by one?
Thank you for your help
Kai
On Friday, July 9, 2021, 11:39:59 AM PDT, Migdonio González
wrote:
It seems that your problem is that you are using single quotes inside of the
double quotes. This is not necessary. Here is the corrected for-loop:
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv")
rdcsv <- noquote(paste0("w:/project/_Joe.B/Oracle/data/", mycsv))
rr <- read.csv(rdcsv)
}
Also note that the rr variable will only store the last CSV, not all CSV. You
will need to initialize the rr variable as a list to store all CSVs if that is
what you require. Something like this:
# Initialize the rr variable as a list.
rr <- as.list(rep(NA, nrow(ora)))
# Run the for-loop to store all the CSVs in rr.
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv")
rdcsv <- noquote(paste0("w:/project/_Joe.B/Oracle/data/", mycsv))
rr[[j]] <- read.csv(rdcsv)
}
RegardsMigdonio G.
On Fri, Jul 9, 2021 at 1:10 PM Kai Yang via R-help wrote:
Hello List,
I use for loop to read csv difference file into data frame rr. The data frame
rr will be deleted after a comparison and go to the next csv file. Below is my
code:
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv'")
rdcsv <- noquote(paste0("'w:/project/_Joe.B/Oracle/data/", mycsv))
rr <- read.csv(rdcsv)
}
but when I run this code, I got error message below:
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
cannot open file ''w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'': No
such file or directory
so, I checked the rdcsv and print it out, see below:
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DISCRETE_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_QUESTIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_RUNS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/DATA_ENTRY_PAGES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/DISCRETE_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ENTRY_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_LOT_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_SAMPLES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/MOLECULAR_WAREHOUSE.csv'
[1] 'w:/project/_Joe.B/Oracle/data/QUESTION_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/QUESTION_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/RESPONSES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST_NAMES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_PLATE_ADDRESSES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/STORAGE_UNITS.csv'
it seems correct. I copy and paste it into a code :
rr <- read.csv( 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv')
and it works fine.
Can someone help me debug where is the problem in my for loop code?
Thanks,
Kai
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Re: [R] problem for strsplit function
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")= 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")= > 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 __ R-help@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.
Re: [R] error message from read.csv in loop
it complained about ASSAY_DEFINITIONS not about RESPONDENTS.
Can you try with the ASSAY_DEFINITIONS file?
On Fri, Jul 9, 2021 at 9:10 PM Kai Yang via R-help
wrote:
> Hello List,
> I use for loop to read csv difference file into data frame rr. The data
> frame rr will be deleted after a comparison and go to the next csv file.
> Below is my code:
> for (j in 1:nrow(ora))
> {
> mycol <- ora[j,"fname"]
> mycsv <- paste0(mycol,".csv'")
> rdcsv <- noquote(paste0("'w:/project/_Joe.B/Oracle/data/", mycsv))
> rr <- read.csv(rdcsv)
> }
> but when I run this code, I got error message below:
> Error in file(file, "rt") : cannot open the connection
> In addition: Warning message:
> In file(file, "rt") :
> cannot open file
> ''w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'': No such file or
> directory
>
> so, I checked the rdcsv and print it out, see below:
> [1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DISCRETE_VALUES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/ASSAY_QUESTIONS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/ASSAY_RUNS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/DATA_ENTRY_PAGES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/DISCRETE_VALUES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/ENTRY_GROUPS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_GROUPS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_VALUES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/GEMD_LOT_DEFINITIONS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/GEMD_SAMPLES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/MOLECULAR_WAREHOUSE.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/QUESTION_DEFINITIONS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/QUESTION_GROUPS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/RESPONSES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST_NAMES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_PLATE_ADDRESSES.csv'
> [1] 'w:/project/_Joe.B/Oracle/data/STORAGE_UNITS.csv'
> it seems correct. I copy and paste it into a code :
> rr <- read.csv( 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv')
> and it works fine.
> Can someone help me debug where is the problem in my for loop code?
> Thanks,
> Kai
>
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.
>
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[R] error message from read.csv in loop
Hello List,
I use for loop to read csv difference file into data frame rr. The data frame
rr will be deleted after a comparison and go to the next csv file. Below is my
code:
for (j in 1:nrow(ora))
{
mycol <- ora[j,"fname"]
mycsv <- paste0(mycol,".csv'")
rdcsv <- noquote(paste0("'w:/project/_Joe.B/Oracle/data/", mycsv))
rr <- read.csv(rdcsv)
}
but when I run this code, I got error message below:
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
cannot open file ''w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'': No
such file or directory
so, I checked the rdcsv and print it out, see below:
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_DISCRETE_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_QUESTIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ASSAY_RUNS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/DATA_ENTRY_PAGES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/DISCRETE_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/ENTRY_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_CODELIST_VALUES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_LOT_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/GEMD_SAMPLES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/MOLECULAR_WAREHOUSE.csv'
[1] 'w:/project/_Joe.B/Oracle/data/QUESTION_DEFINITIONS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/QUESTION_GROUPS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv'
[1] 'w:/project/_Joe.B/Oracle/data/RESPONSES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_LIST_NAMES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/SAMPLE_PLATE_ADDRESSES.csv'
[1] 'w:/project/_Joe.B/Oracle/data/STORAGE_UNITS.csv'
it seems correct. I copy and paste it into a code :
rr <- read.csv( 'w:/project/_Joe.B/Oracle/data/RESPONDENTS.csv')
and it works fine.
Can someone help me debug where is the problem in my for loop code?
Thanks,
Kai
[[alternative HTML version deleted]]
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Re: [R] How to create a matrix from a list without a for loop
On 7/9/21 10:40 AM, Laurent Rhelp wrote:
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create a
matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to write nice R code to do this loop ?
Thank you very much
Laurent
##
## mock data, 16 dataframes in a list
##
init_l <- lapply( seq(1,16) , function(x) {
data.frame( V1 = rnorm(3),
V2 = rnorm(3),
V3 = rnorm(3)
)
})
##
## lists matrix creation with n = 4 columns and n = 4 rows
##
n <- 4
Just assign a dimension attribute and you will have your two dimensional
structure
> dim(init_l) <- c(n,n)
> init_l[ 2,2]
[[1]]
V1 V2 V3
1 -1.4103259 1.9214184 -0.1590919
2 0.1899490 0.3842191 2.4502078
3 -0.4282764 -0.9992190 1.5384344
> is.matrix(init_l)
[1] TRUE
--
David
## an example of row to create the matrix with lists in the cells
one_row <- rbind( rep( list(rep(list(1),3)) , n) )
mymat <- do.call( "rbind" , rep( list(one_row) , n) )
##
## The UGGLY loop I would like to improve:
##
## populate the matrix
k <- 1
for( i in 1:n){
for( j in 1:n){
mymat[i,j][[1]] <- list( init_l[[ k ]] )
k <- k+1
}
}
colnames(mymat) <- c("X1", "X2", "X3", "X3")
rownames(mymat) <- c("X1", "X2", "X3", "X4")
mymat
# X1 X2 X3 X3
# X1 List,1 List,1 List,1 List,1
# X2 List,1 List,1 List,1 List,1
# X3 List,1 List,1 List,1 List,1
# X4 List,1 List,1 List,1 List,1
#
# verification, it works
#
mymat[2,2]
init_l[[6]]
##
init_l[[6]]
library(tidyverse)
mymat.t <- as.tibble(mymat)
mymat.t
unnest(mymat.t[2,2],cols="X2")[[1]][[1]]
mymat.df <- as.data.frame(mymat)
mymat.df[2,2][[1]][[1]]
thx
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Re: [R] How to create a matrix from a list without a for loop
Try
matrix(init_l, nrow=4, ncol=4,
dimnames=list(c("X1","X2","X3","X4"),c("X1","X2","X3","X4")))
It doesn't give exactly what your code does, but your code introduces an
extra level of "list", which you may not want.
-Bill
On Fri, Jul 9, 2021 at 10:40 AM Laurent Rhelp wrote:
> Dear R-Help-list,
>
>I have a list init_l containing 16 dataframes and I want to create a
> matrix 4 x 4 from this list with a dataframe in every cell of the
> matrix. I succeeded to do that but my loop is very uggly (cf. below).
> Could somebody help me to write nice R code to do this loop ?
>
> Thank you very much
>
> Laurent
>
>
> ##
> ## mock data, 16 dataframes in a list
> ##
> init_l <- lapply( seq(1,16) , function(x) {
>data.frame( V1 = rnorm(3),
>V2 = rnorm(3),
>V3 = rnorm(3)
> )
> })
>
> ##
> ## lists matrix creation with n = 4 columns and n = 4 rows
> ##
> n <- 4
> ## an example of row to create the matrix with lists in the cells
> one_row <- rbind( rep( list(rep(list(1),3)) , n) )
> mymat <- do.call( "rbind" , rep( list(one_row) , n) )
>
> ##
> ## The UGGLY loop I would like to improve:
> ##
>
> ## populate the matrix
> k <- 1
> for( i in 1:n){
>for( j in 1:n){
> mymat[i,j][[1]] <- list( init_l[[ k ]] )
> k <- k+1
>}
> }
>
> colnames(mymat) <- c("X1", "X2", "X3", "X3")
> rownames(mymat) <- c("X1", "X2", "X3", "X4")
>
>
> mymat
>
> # X1 X2 X3 X3
> # X1 List,1 List,1 List,1 List,1
> # X2 List,1 List,1 List,1 List,1
> # X3 List,1 List,1 List,1 List,1
> # X4 List,1 List,1 List,1 List,1
>
>
> #
> # verification, it works
> #
> mymat[2,2]
> init_l[[6]]
>
> ##
> init_l[[6]]
>
> library(tidyverse)
> mymat.t <- as.tibble(mymat)
> mymat.t
> unnest(mymat.t[2,2],cols="X2")[[1]][[1]]
>
> mymat.df <- as.data.frame(mymat)
> mymat.df[2,2][[1]][[1]]
>
>
> thx
>
>
>
> --
> L'absence de virus dans ce courrier électronique a été vérifiée par le
> logiciel antivirus Avast.
> https://www.avast.com/antivirus
>
> __
> R-help@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]]
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and provide commented, minimal, self-contained, reproducible code.
[R] How to create a matrix from a list without a for loop
Dear R-Help-list,
I have a list init_l containing 16 dataframes and I want to create a
matrix 4 x 4 from this list with a dataframe in every cell of the
matrix. I succeeded to do that but my loop is very uggly (cf. below).
Could somebody help me to write nice R code to do this loop ?
Thank you very much
Laurent
##
## mock data, 16 dataframes in a list
##
init_l <- lapply( seq(1,16) , function(x) {
data.frame( V1 = rnorm(3),
V2 = rnorm(3),
V3 = rnorm(3)
)
})
##
## lists matrix creation with n = 4 columns and n = 4 rows
##
n <- 4
## an example of row to create the matrix with lists in the cells
one_row <- rbind( rep( list(rep(list(1),3)) , n) )
mymat <- do.call( "rbind" , rep( list(one_row) , n) )
##
## The UGGLY loop I would like to improve:
##
## populate the matrix
k <- 1
for( i in 1:n){
for( j in 1:n){
mymat[i,j][[1]] <- list( init_l[[ k ]] )
k <- k+1
}
}
colnames(mymat) <- c("X1", "X2", "X3", "X3")
rownames(mymat) <- c("X1", "X2", "X3", "X4")
mymat
# X1 X2 X3 X3
# X1 List,1 List,1 List,1 List,1
# X2 List,1 List,1 List,1 List,1
# X3 List,1 List,1 List,1 List,1
# X4 List,1 List,1 List,1 List,1
#
# verification, it works
#
mymat[2,2]
init_l[[6]]
##
init_l[[6]]
library(tidyverse)
mymat.t <- as.tibble(mymat)
mymat.t
unnest(mymat.t[2,2],cols="X2")[[1]][[1]]
mymat.df <- as.data.frame(mymat)
mymat.df[2,2][[1]][[1]]
thx
--
L'absence de virus dans ce courrier électronique a été vérifiée par le logiciel
antivirus Avast.
https://www.avast.com/antivirus
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] How to estimate the parameter for many variable?
Dear Rui ang Jim,
Thank you very much.
Thank you Rui Barradas, I already tried using your coding and I'm grateful
I got the answer.
ok now, I have some condition on the location parameter which is cyclic
condition.
So, I will add another 2 variables for the column.
and the condition for location is this one.
ti[,1] = seq(1, 888, 1)
ti[,2]=sin(2*pi*(ti[,1])/52)
ti[,3]=cos(2*pi*(ti[,1])/52)
fit0<-gev.fit(x[,i], ydat = ti, mul=c(2, 3))
Thank you again.
On Fri, 9 Jul 2021 at 14:38, Rui Barradas wrote:
> Hello,
>
> The following lapply one-liner fits a GEV to each column vector, there
> is no need for the double for loop. There's also no need to create a
> data set x.
>
>
> library(ismev)
> library(mgcv)
> library(EnvStats)
>
> Ozone_weekly2 <- read.table("~/tmp/Ozone_weekly2.txt", header = TRUE)
>
> # fit a GEV to each column
> gev_fit_list <- lapply(Ozone_weekly2, gev.fit, show = FALSE)
>
> # extract the parameters MLE estimates
> mle_params <- t(sapply(gev_fit_list, '[[', 'mle'))
>
> # assign column names
> colnames(mle_params) <- c("location", "scale", "shape")
>
> # see first few rows
> head(mle_params)
>
>
>
> The OP doesn't ask for plots but, here they go.
>
>
> y_vals <- function(x, params){
>loc <- params[1]
>scale <- params[2]
>shape <- params[3]
>EnvStats::dgevd(x, loc, scale, shape)
> }
> plot_fit <- function(data, vec, verbose = FALSE){
>fit <- gev.fit(data[[vec]], show = verbose)
>x <- sort(data[[vec]])
>hist(x, freq = FALSE)
>lines(x, y_vals(x, params = fit$mle))
> }
>
> # seems a good fit
> plot_fit(Ozone_weekly2, 1) # column number
> plot_fit(Ozone_weekly2, "CA01") # col name, equivalent
>
> # the data seems gaussian, not a good fit
> plot_fit(Ozone_weekly2, 4) # column number
> plot_fit(Ozone_weekly2, "CA08") # col name, equivalent
>
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 00:59 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
> > Dear all,
> >
> > Thank you very much for the feedback.
> >
> > Sorry for the lack of information about this problem.
> >
> > Here, I explain again.
> >
> > I use this package to run my coding.
> >
> > library(ismev)
> > library(mgcv)
> > library(nlme)
> >
> > The purpose of this is I want to get the value of parameter estimation
> > using MLE by applying the GEV distribution.
> >
> > x <- data.matrix(Ozone_weekly2) x refers to my data
> > that consists of 19 variables. I will attach the data together.
> > x
> > head(gev.fit)[1:4]
> > ti = matrix(ncol = 3, nrow = 888)
> > ti[,1] = seq(1, 888, 1)
> > ti[,2]=sin(2*pi*(ti[,1])/52)
> > ti[,3]=cos(2*pi*(ti[,1])/52)
> >
> > /for(i in 1:nrow(x))
> >+ { for(j in 1:ncol(x))the problem in
> > here, i don't no to create the coding. i target my output will come out
> > in matrix that
> > + {x[i,j] = 1}} show the
> > parameter estimation for 19 variable which have 19 row and 3 column/
> > / row --
> > refer to variable (station) ; column -- refer to parameter estimation
> > for GEV distribution
> >
> > /thank you.
> >
> > On Thu, 8 Jul 2021 at 18:40, Rui Barradas > > wrote:
> >
> > Hello,
> >
> > Also, in the code
> >
> > x <- data.matrix(Ozone_weekly)
> >
> > [...omited...]
> >
> > for(i in 1:nrow(x))
> > + { for(j in 1:ncol(x))
> > + {x[i,j] = 1}}
> >
> > not only you rewrite x but the double for loop is equivalent to
> >
> >
> > x[] <- 1
> >
> >
> > courtesy R's vectorised behavior. (The square parenthesis are needed
> to
> > keep the dimensions, the matrix form.)
> > And, I'm not sure but isn't
> >
> > head(gev.fit)[1:4]
> >
> > equivalent to
> >
> > head(gev.fit, n = 4)
> >
> > ?
> >
> > Like Jim says, we need more information, can you post Ozone_weekly2
> and
> > the code that produced gev.fit? But in the mean time you can revise
> > your
> > code.
> >
> > Hope this helps,
> >
> > Rui Barradas
> >
> >
> > Às 11:08 de 08/07/21, Jim Lemon escreveu:
> > > Hi Siti,
> > > I think we need a bit more information to respond helpfully. I
> > have no
> > > idea what "Ozone_weekly2" is and Google is also ignorant.
> > "gev.fit" is
> > > also unknown. The name suggests that it is the output of some
> > > regression or similar. What function produced it, and from what
> > > library? "ti" is known as you have defined it. However, I don't
> know
> > > what you want to do with it. Finally, as this is a text mailing
> list,
> > > we don't get any highlighting, so the text to which you refer
> cannot
> > > be identified. I can see you have a problem, but cannot offer any
> > help
> > > right now.
> > >
> > > Jim
> > >
> > > On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
> > > >
Re: [R] How to estimate the parameter for many variable?
Hello,
With the condition for the location it can be estimated like the following.
fit_list2 <- gev_fit_list <- lapply(Ozone_weekly2, gev.fit, ydat = ti,
mul = c(2, 3), show = FALSE)
mle_params2 <- t(sapply(fit_list2, '[[', 'mle'))
# assign column names
colnames(mle_params2) <- c("location", "scale", "shape", "mul2", "mul3")
head(mle_params2)
Hope this helps,
Rui Barradas
Às 09:53 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
Dear Rui ang Jim,
Thank you very much.
Thank you Rui Barradas, I already tried using your coding and I'm
grateful I got the answer.
ok now, I have some condition on the location parameter which is cyclic
condition.
So, I will add another 2 variables for the column.
and the condition for location is this one.
ti[,1] = seq(1, 888, 1)
ti[,2]=sin(2*pi*(ti[,1])/52)
ti[,3]=cos(2*pi*(ti[,1])/52)
fit0<-gev.fit(x[,i], ydat = ti, mul=c(2, 3))
Thank you again.
On Fri, 9 Jul 2021 at 14:38, Rui Barradas > wrote:
Hello,
The following lapply one-liner fits a GEV to each column vector, there
is no need for the double for loop. There's also no need to create a
data set x.
library(ismev)
library(mgcv)
library(EnvStats)
Ozone_weekly2 <- read.table("~/tmp/Ozone_weekly2.txt", header = TRUE)
# fit a GEV to each column
gev_fit_list <- lapply(Ozone_weekly2, gev.fit, show = FALSE)
# extract the parameters MLE estimates
mle_params <- t(sapply(gev_fit_list, '[[', 'mle'))
# assign column names
colnames(mle_params) <- c("location", "scale", "shape")
# see first few rows
head(mle_params)
The OP doesn't ask for plots but, here they go.
y_vals <- function(x, params){
loc <- params[1]
scale <- params[2]
shape <- params[3]
EnvStats::dgevd(x, loc, scale, shape)
}
plot_fit <- function(data, vec, verbose = FALSE){
fit <- gev.fit(data[[vec]], show = verbose)
x <- sort(data[[vec]])
hist(x, freq = FALSE)
lines(x, y_vals(x, params = fit$mle))
}
# seems a good fit
plot_fit(Ozone_weekly2, 1) # column number
plot_fit(Ozone_weekly2, "CA01") # col name, equivalent
# the data seems gaussian, not a good fit
plot_fit(Ozone_weekly2, 4) # column number
plot_fit(Ozone_weekly2, "CA08") # col name, equivalent
Hope this helps,
Rui Barradas
Às 00:59 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
> Dear all,
>
> Thank you very much for the feedback.
>
> Sorry for the lack of information about this problem.
>
> Here, I explain again.
>
> I use this package to run my coding.
>
> library(ismev)
> library(mgcv)
> library(nlme)
>
> The purpose of this is I want to get the value of parameter
estimation
> using MLE by applying the GEV distribution.
>
> x <- data.matrix(Ozone_weekly2) x refers to
my data
> that consists of 19 variables. I will attach the data together.
> x
> head(gev.fit)[1:4]
> ti = matrix(ncol = 3, nrow = 888)
> ti[,1] = seq(1, 888, 1)
> ti[,2]=sin(2*pi*(ti[,1])/52)
> ti[,3]=cos(2*pi*(ti[,1])/52)
>
> /for(i in 1:nrow(x))
> + { for(j in 1:ncol(x)) the problem in
> here, i don't no to create the coding. i target my output will
come out
> in matrix that
> + {x[i,j] = 1}} show the
> parameter estimation for 19 variable which have 19 row and 3 column/
> /
row --
> refer to variable (station) ; column -- refer to parameter
estimation
> for GEV distribution
>
> /thank you.
>
> On Thu, 8 Jul 2021 at 18:40, Rui Barradas mailto:ruipbarra...@sapo.pt>
> >> wrote:
>
> Hello,
>
> Also, in the code
>
> x <- data.matrix(Ozone_weekly)
>
> [...omited...]
>
> for(i in 1:nrow(x))
> + { for(j in 1:ncol(x))
> + {x[i,j] = 1}}
>
> not only you rewrite x but the double for loop is equivalent to
>
>
> x[] <- 1
>
>
> courtesy R's vectorised behavior. (The square parenthesis are
needed to
> keep the dimensions, the matrix form.)
> And, I'm not sure but isn't
>
> head(gev.fit)[1:4]
>
> equivalent to
>
> head(gev.fit, n = 4)
>
> ?
>
> Like Jim says, we need more information, can you post
Ozone_weekly2 and
> the code that produced gev.fit? But in the mean time you can
revise
> your
> code.
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 11:08
Re: [R] How to estimate the parameter for many variable?
Hello,
The following lapply one-liner fits a GEV to each column vector, there
is no need for the double for loop. There's also no need to create a
data set x.
library(ismev)
library(mgcv)
library(EnvStats)
Ozone_weekly2 <- read.table("~/tmp/Ozone_weekly2.txt", header = TRUE)
# fit a GEV to each column
gev_fit_list <- lapply(Ozone_weekly2, gev.fit, show = FALSE)
# extract the parameters MLE estimates
mle_params <- t(sapply(gev_fit_list, '[[', 'mle'))
# assign column names
colnames(mle_params) <- c("location", "scale", "shape")
# see first few rows
head(mle_params)
The OP doesn't ask for plots but, here they go.
y_vals <- function(x, params){
loc <- params[1]
scale <- params[2]
shape <- params[3]
EnvStats::dgevd(x, loc, scale, shape)
}
plot_fit <- function(data, vec, verbose = FALSE){
fit <- gev.fit(data[[vec]], show = verbose)
x <- sort(data[[vec]])
hist(x, freq = FALSE)
lines(x, y_vals(x, params = fit$mle))
}
# seems a good fit
plot_fit(Ozone_weekly2, 1) # column number
plot_fit(Ozone_weekly2, "CA01") # col name, equivalent
# the data seems gaussian, not a good fit
plot_fit(Ozone_weekly2, 4) # column number
plot_fit(Ozone_weekly2, "CA08") # col name, equivalent
Hope this helps,
Rui Barradas
Às 00:59 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
Dear all,
Thank you very much for the feedback.
Sorry for the lack of information about this problem.
Here, I explain again.
I use this package to run my coding.
library(ismev)
library(mgcv)
library(nlme)
The purpose of this is I want to get the value of parameter estimation
using MLE by applying the GEV distribution.
x <- data.matrix(Ozone_weekly2) x refers to my data
that consists of 19 variables. I will attach the data together.
x
head(gev.fit)[1:4]
ti = matrix(ncol = 3, nrow = 888)
ti[,1] = seq(1, 888, 1)
ti[,2]=sin(2*pi*(ti[,1])/52)
ti[,3]=cos(2*pi*(ti[,1])/52)
/for(i in 1:nrow(x))
+ { for(j in 1:ncol(x)) the problem in
here, i don't no to create the coding. i target my output will come out
in matrix that
+ {x[i,j] = 1}} show the
parameter estimation for 19 variable which have 19 row and 3 column/
/ row --
refer to variable (station) ; column -- refer to parameter estimation
for GEV distribution
/thank you.
On Thu, 8 Jul 2021 at 18:40, Rui Barradas > wrote:
Hello,
Also, in the code
x <- data.matrix(Ozone_weekly)
[...omited...]
for(i in 1:nrow(x))
+ { for(j in 1:ncol(x))
+ {x[i,j] = 1}}
not only you rewrite x but the double for loop is equivalent to
x[] <- 1
courtesy R's vectorised behavior. (The square parenthesis are needed to
keep the dimensions, the matrix form.)
And, I'm not sure but isn't
head(gev.fit)[1:4]
equivalent to
head(gev.fit, n = 4)
?
Like Jim says, we need more information, can you post Ozone_weekly2 and
the code that produced gev.fit? But in the mean time you can revise
your
code.
Hope this helps,
Rui Barradas
Às 11:08 de 08/07/21, Jim Lemon escreveu:
> Hi Siti,
> I think we need a bit more information to respond helpfully. I
have no
> idea what "Ozone_weekly2" is and Google is also ignorant.
"gev.fit" is
> also unknown. The name suggests that it is the output of some
> regression or similar. What function produced it, and from what
> library? "ti" is known as you have defined it. However, I don't know
> what you want to do with it. Finally, as this is a text mailing list,
> we don't get any highlighting, so the text to which you refer cannot
> be identified. I can see you have a problem, but cannot offer any
help
> right now.
>
> Jim
>
> On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
> mailto:aisyahzaka...@unimap.edu.my>> wrote:
>>
>> Dear all,
>>
>> Can I ask something about programming in marginal distribution
for spatial
>> extreme?
>> I really stuck on my coding to obtain the parameter estimation for
>> univariate or marginal distribution for new model in spatial
extreme.
>>
>> I want to run my data in order to get the parameter estimation
value for 25
>> stations in one table. But I really didn't get the idea of the
correct
>> coding. Here I attached my coding
>>
>> x <- data.matrix(Ozone_weekly2)
>> x
>> head(gev.fit)[1:4]
>> ti = matrix(ncol = 3, nrow = 888)
>> ti[,1] = seq(1, 888, 1)
>> ti[,2]=sin(2*pi*(ti[,1])/52)
>> ti[,3]=cos(2*pi*(ti[,1])/52)
>> for(i in 1:nrow(x))
>> + { for(j in 1:ncol(x))
>> + {x[i,j] = 1}}
>>
>> My problem is highlighted in red color.
>> And if are not hesitate to all. Can someone share with me the
Re: [R] How to estimate the parameter for many variable?
Dear all,
Thank you very much for the feedback.
Sorry for the lack of information about this problem.
Here, I explain again.
I use this package to run my coding.
library(ismev)
library(mgcv)
library(nlme)
The purpose of this is I want to get the value of parameter estimation
using MLE by applying the GEV distribution.
x <- data.matrix(Ozone_weekly2) x refers to my data
that consists of 19 variables. I will attach the data together.
x
head(gev.fit)[1:4]
ti = matrix(ncol = 3, nrow = 888)
ti[,1] = seq(1, 888, 1)
ti[,2]=sin(2*pi*(ti[,1])/52)
ti[,3]=cos(2*pi*(ti[,1])/52)
*for(i in 1:nrow(x)) + { for(j in 1:ncol(x))
the problem in here, i don't no to create the coding. i target my
output will come out in matrix that + {x[i,j] = 1}}
show the parameter estimation for 19 variable which have
19 row and 3 column*
* row -- refer
to variable (station) ; column -- refer to parameter estimation for GEV
distribution*thank you.
On Thu, 8 Jul 2021 at 18:40, Rui Barradas wrote:
> Hello,
>
> Also, in the code
>
> x <- data.matrix(Ozone_weekly)
>
> [...omited...]
>
> for(i in 1:nrow(x))
>+ { for(j in 1:ncol(x))
> + {x[i,j] = 1}}
>
> not only you rewrite x but the double for loop is equivalent to
>
>
> x[] <- 1
>
>
> courtesy R's vectorised behavior. (The square parenthesis are needed to
> keep the dimensions, the matrix form.)
> And, I'm not sure but isn't
>
> head(gev.fit)[1:4]
>
> equivalent to
>
> head(gev.fit, n = 4)
>
> ?
>
> Like Jim says, we need more information, can you post Ozone_weekly2 and
> the code that produced gev.fit? But in the mean time you can revise your
> code.
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 11:08 de 08/07/21, Jim Lemon escreveu:
> > Hi Siti,
> > I think we need a bit more information to respond helpfully. I have no
> > idea what "Ozone_weekly2" is and Google is also ignorant. "gev.fit" is
> > also unknown. The name suggests that it is the output of some
> > regression or similar. What function produced it, and from what
> > library? "ti" is known as you have defined it. However, I don't know
> > what you want to do with it. Finally, as this is a text mailing list,
> > we don't get any highlighting, so the text to which you refer cannot
> > be identified. I can see you have a problem, but cannot offer any help
> > right now.
> >
> > Jim
> >
> > On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
> > wrote:
> >>
> >> Dear all,
> >>
> >> Can I ask something about programming in marginal distribution for
> spatial
> >> extreme?
> >> I really stuck on my coding to obtain the parameter estimation for
> >> univariate or marginal distribution for new model in spatial extreme.
> >>
> >> I want to run my data in order to get the parameter estimation value
> for 25
> >> stations in one table. But I really didn't get the idea of the correct
> >> coding. Here I attached my coding
> >>
> >> x <- data.matrix(Ozone_weekly2)
> >> x
> >> head(gev.fit)[1:4]
> >> ti = matrix(ncol = 3, nrow = 888)
> >> ti[,1] = seq(1, 888, 1)
> >> ti[,2]=sin(2*pi*(ti[,1])/52)
> >> ti[,3]=cos(2*pi*(ti[,1])/52)
> >> for(i in 1:nrow(x))
> >>+ { for(j in 1:ncol(x))
> >> + {x[i,j] = 1}}
> >>
> >> My problem is highlighted in red color.
> >> And if are not hesitate to all. Can someone share with me the procedure,
> >> how can I map my data using spatial extreme.
> >> For example:
> >> After I finish my marginal distribution, what the next procedure. It is
> I
> >> need to get the spatial independent value.
> >>
> >> That's all
> >> Thank you.
> >>
> >> --
> >>
> >>
> >>
> >>
> >>
> >> "..Millions of trees are used to make papers, only to be thrown away
> >> after a couple of minutes reading from them. Our planet is at stake.
> Please
> >> be considerate. THINK TWICE BEFORE PRINTING THIS.."
> >>
> >> DISCLAIMER: This email \ and any files transmitte...{{dropped:24}}
> >>
> >> __
> >> R-help@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.
> >
> > __
> > R-help@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.
> >
>
--
"..Millions of trees are used to make papers, only to be thrown away
after a couple of minutes reading from them. Our planet is at stake. Please
be considerate. THINK TWICE BEFORE PRINTING THIS.."
DISCLAIMER: This email
and any files transmitted with it are confidential and intended solely for
the use of the individual orentity to whom they are addressed. If you
Re: [R] How to estimate the parameter for many variable?
Hi Siti,
I think some progress has been made. You have a data set with 888 rows
and 19 columns:
ow2<-read.table("Ozone_weekly2.txt",
header=TRUE)
dim(ow2)
[1] 888 19
The values may be parts per million ozone in the atmosphere. The
columns may represent different measuring locations and my guess is
that rows are times of measurement, perhaps in weeks given the name.
If so, you might have run ismev::gev.fit() on all of the columns,
looking for estimated maxima at each location. Is this at all close to
what you are doing?
Jim
On Fri, Jul 9, 2021 at 9:59 AM SITI AISYAH ZAKARIA
wrote:
>
> Dear all,
>
> Thank you very much for the feedback.
>
> Sorry for the lack of information about this problem.
>
> Here, I explain again.
>
> I use this package to run my coding.
>
> library(ismev)
> library(mgcv)
> library(nlme)
>
> The purpose of this is I want to get the value of parameter estimation using
> MLE by applying the GEV distribution.
>
> x <- data.matrix(Ozone_weekly2) x refers to my data that
> consists of 19 variables. I will attach the data together.
> x
> head(gev.fit)[1:4]
> ti = matrix(ncol = 3, nrow = 888)
> ti[,1] = seq(1, 888, 1)
> ti[,2]=sin(2*pi*(ti[,1])/52)
> ti[,3]=cos(2*pi*(ti[,1])/52)
>
> for(i in 1:nrow(x))
> + { for(j in 1:ncol(x))the problem in here, i
> don't no to create the coding. i target my output will come out in matrix that
> + {x[i,j] = 1}} show the parameter
> estimation for 19 variable which have 19 row and 3 column
> row -- refer to
> variable (station) ; column -- refer to parameter estimation for GEV
> distribution
>
> thank you.
>
> On Thu, 8 Jul 2021 at 18:40, Rui Barradas wrote:
>>
>> Hello,
>>
>> Also, in the code
>>
>> x <- data.matrix(Ozone_weekly)
>>
>> [...omited...]
>>
>> for(i in 1:nrow(x))
>>+ { for(j in 1:ncol(x))
>> + {x[i,j] = 1}}
>>
>> not only you rewrite x but the double for loop is equivalent to
>>
>>
>> x[] <- 1
>>
>>
>> courtesy R's vectorised behavior. (The square parenthesis are needed to
>> keep the dimensions, the matrix form.)
>> And, I'm not sure but isn't
>>
>> head(gev.fit)[1:4]
>>
>> equivalent to
>>
>> head(gev.fit, n = 4)
>>
>> ?
>>
>> Like Jim says, we need more information, can you post Ozone_weekly2 and
>> the code that produced gev.fit? But in the mean time you can revise your
>> code.
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>>
>> Às 11:08 de 08/07/21, Jim Lemon escreveu:
>> > Hi Siti,
>> > I think we need a bit more information to respond helpfully. I have no
>> > idea what "Ozone_weekly2" is and Google is also ignorant. "gev.fit" is
>> > also unknown. The name suggests that it is the output of some
>> > regression or similar. What function produced it, and from what
>> > library? "ti" is known as you have defined it. However, I don't know
>> > what you want to do with it. Finally, as this is a text mailing list,
>> > we don't get any highlighting, so the text to which you refer cannot
>> > be identified. I can see you have a problem, but cannot offer any help
>> > right now.
>> >
>> > Jim
>> >
>> > On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
>> > wrote:
>> >>
>> >> Dear all,
>> >>
>> >> Can I ask something about programming in marginal distribution for spatial
>> >> extreme?
>> >> I really stuck on my coding to obtain the parameter estimation for
>> >> univariate or marginal distribution for new model in spatial extreme.
>> >>
>> >> I want to run my data in order to get the parameter estimation value for
>> >> 25
>> >> stations in one table. But I really didn't get the idea of the correct
>> >> coding. Here I attached my coding
>> >>
>> >> x <- data.matrix(Ozone_weekly2)
>> >> x
>> >> head(gev.fit)[1:4]
>> >> ti = matrix(ncol = 3, nrow = 888)
>> >> ti[,1] = seq(1, 888, 1)
>> >> ti[,2]=sin(2*pi*(ti[,1])/52)
>> >> ti[,3]=cos(2*pi*(ti[,1])/52)
>> >> for(i in 1:nrow(x))
>> >>+ { for(j in 1:ncol(x))
>> >> + {x[i,j] = 1}}
>> >>
>> >> My problem is highlighted in red color.
>> >> And if are not hesitate to all. Can someone share with me the procedure,
>> >> how can I map my data using spatial extreme.
>> >> For example:
>> >> After I finish my marginal distribution, what the next procedure. It is I
>> >> need to get the spatial independent value.
>> >>
>> >> That's all
>> >> Thank you.
>> >>
>> >> --
>> >>
>> >>
>> >>
>> >>
>> >>
>> >> "..Millions of trees are used to make papers, only to be thrown away
>> >> after a couple of minutes reading from them. Our planet is at stake.
>> >> Please
>> >> be considerate. THINK TWICE BEFORE PRINTING THIS.."
>> >>
>> >> DISCLAIMER: This email \ and any files transmitte...{{dropped:24}}
>> >>
>> >> __
>> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >>
