Re: [R] [Tagged] Re: col.names in as.data.frame() ?
Jef, your terse reply was so constructive that you converted me! LOL!
That is an interesting point though that I remain a bit unclear on.
Both data.frame and as.data.frame can be used in some ways similarly as in:
> data.frame(matrix(1:12, nrow=3))
X1 X2 X3 X4
1 1 4 7 10
2 2 5 8 11
3 3 6 9 12
> as.data.frame(matrix(1:12, nrow=3))
V1 V2 V3 V4
1 1 4 7 10
2 2 5 8 11
3 3 6 9 12
But yes, the constructor accepts many arguments while the converter really
normally handles a single object.
Where do some other things like cbind() fit, not to mention a dplyr function
like tribble()?
I do wonder though, why asking a converter to convert a matrix to a data.frame
and perhaps adding column names, is considered changing the object contents.
The manual page for as.data.frame seems to include quite a few options to
specify the name of a single column, truncate column names, deal with row
names, and more as well as whether to convert strings to factors. Are those
things different enough than what we are discussing?
Of course, we may indeed be experiencing mission creep where something simple
keeps being improved with new features until the original simplicity and
clarity gets lost.
-Original Message-
From: R-help On Behalf Of Jeff Newmiller via
R-help
Sent: Saturday, October 28, 2023 2:54 PM
To: r-help@r-project.org; Boris Steipe ; R. Mailing
List
Subject: Re: [R] [Tagged] Re: col.names in as.data.frame() ?
as.data.frame is a _converter_, while data.frame is a _constructor_. Changing
the object contents is not what a conversion is for.
On October 28, 2023 11:39:22 AM PDT, Boris Steipe
wrote:
>Thanks Duncan and Avi!
>
>That you could use NULL in a matrix() dimnames = list(...) argument wasn't
>clear to me. I thought that would be equivalent to a one-element list - and
>thereby define rownames. So that's good to know.
>
>The documentation could be more explicit - but it is probably more work to do
>that than just patch the code to honour a col.names argument. (At least I
>can't see a reason not to.)
>
>Thanks again!
>:-)
>
>
>
>
>> On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
>>
>> Борис,
>>
>> Try this where you tell matrix the column names you want:
>>
>> nouns <- as.data.frame(
>> matrix(c(
>>"gaggle",
>>"geese",
>>
>>"dule",
>>"doves",
>>
>>"wake",
>>"vultures"
>> ),
>> ncol = 2,
>> byrow = TRUE,
>> dimnames=list(NULL, c("collective", "category"
>>
>> Result:
>>
>>> nouns
>> collective category
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>
>>
>> The above simply names the columns earlier when creating the matrix.
>>
>> There are other ways and the way you tried LOOKS like it should work but
>> fails for me with a message about it weirdly expecting three rows versus two
>> which seems to confuse rows and columns. My version of R is recent and I
>> wonder if there is a bug here.
>>
>> Consider whether you really need the data.frame created in a single
>> statement or can you change the column names next as in:
>>
>>
>>> nouns
>> V1 V2
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>> colnames(nouns)
>> [1] "V1" "V2"
>>> colnames(nouns) <- c("collective", "category")
>>> nouns
>> collective category
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>
>> Is there a known bug here or is the documentation wrong?
>>
>> -Original Message-
>> From: R-help On Behalf Of Boris Steipe
>> Sent: Saturday, October 28, 2023 1:54 PM
>> To: R. Mailing List
>> Subject: [R] col.names in as.data.frame() ?
>>
>> I have been trying to create a data frame from some structured text in a
>> single expression. Reprex:
>>
>> nouns <- as.data.frame(
>> matrix(c(
>>"gaggle",
>>"geese",
>>
>>"dule",
>>"doves",
>>
>>"wake",
>>"vultures"
>> ), ncol = 2, byrow = TRUE),
>> col.names = c("collective", "category")
>> )
>>
>> But ... :
>>
>>> str(nouns)
>> 'data.frame': 3 obs. of 2 variables:
>> $ V1: chr "gaggle" "dule" "wake"
>> $ V2: chr "geese" "doves" "vultures"
>>
>> i.e. the col.names argument does nothing. From my reading of ?as.data.frame,
>> my example should have worked.
>>
>> I know how to get the required result with colnames(), but I would like to
>> understand why the idiom as written didn't work, and how I could have known
>> that from the help file.
>>
>>
>> Thanks!
>> Boris
>>
>> __
>> 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-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting gui
Re: [R] [Tagged] Re: col.names in as.data.frame() ?
On 28/10/2023 4:45 p.m., Bert Gunter wrote:
Jeff, et. al. : but ...
Note that as.data.frame() *already* changes the matrix object by adding
column names of *its own choosing* when the matrix has none. So the issue
here is not *whether* col names should be added, but *what*/*how* they
should be. Unless you wish to extend your criticism to the current version
for its failure to adhere to your proscription.
Dataframes have to have column names. The function isn't modifying the
object any more than it has to.
Duncan Murdoch
Cheers,
Bert
On Sat, Oct 28, 2023 at 11:55 AM Jeff Newmiller via R-help <
r-help@r-project.org> wrote:
as.data.frame is a _converter_, while data.frame is a _constructor_.
Changing the object contents is not what a conversion is for.
On October 28, 2023 11:39:22 AM PDT, Boris Steipe <
boris.ste...@utoronto.ca> wrote:
Thanks Duncan and Avi!
That you could use NULL in a matrix() dimnames = list(...) argument
wasn't clear to me. I thought that would be equivalent to a one-element
list - and thereby define rownames. So that's good to know.
The documentation could be more explicit - but it is probably more work
to do that than just patch the code to honour a col.names argument. (At
least I can't see a reason not to.)
Thanks again!
:-)
On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
Борис,
Try this where you tell matrix the column names you want:
nouns <- as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
),
ncol = 2,
byrow = TRUE,
dimnames=list(NULL, c("collective", "category"
Result:
nouns
collective category
1 gagglegeese
2 duledoves
3 wake vultures
The above simply names the columns earlier when creating the matrix.
There are other ways and the way you tried LOOKS like it should work but
fails for me with a message about it weirdly expecting three rows
versus two
which seems to confuse rows and columns. My version of R is recent and I
wonder if there is a bug here.
Consider whether you really need the data.frame created in a single
statement or can you change the column names next as in:
nouns
V1 V2
1 gagglegeese
2 duledoves
3 wake vultures
colnames(nouns)
[1] "V1" "V2"
colnames(nouns) <- c("collective", "category")
nouns
collective category
1 gagglegeese
2 duledoves
3 wake vultures
Is there a known bug here or is the documentation wrong?
-Original Message-
From: R-help On Behalf Of Boris Steipe
Sent: Saturday, October 28, 2023 1:54 PM
To: R. Mailing List
Subject: [R] col.names in as.data.frame() ?
I have been trying to create a data frame from some structured text in a
single expression. Reprex:
nouns <- as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
), ncol = 2, byrow = TRUE),
col.names = c("collective", "category")
)
But ... :
str(nouns)
'data.frame': 3 obs. of 2 variables:
$ V1: chr "gaggle" "dule" "wake"
$ V2: chr "geese" "doves" "vultures"
i.e. the col.names argument does nothing. From my reading of
?as.data.frame,
my example should have worked.
I know how to get the required result with colnames(), but I would like
to
understand why the idiom as written didn't work, and how I could have
known
that from the help file.
Thanks!
Boris
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--
Sent from my phone. Please excuse my brevity.
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Re: [R] [Tagged] Re: col.names in as.data.frame() ?
Jeff, et. al. : but ...
Note that as.data.frame() *already* changes the matrix object by adding
column names of *its own choosing* when the matrix has none. So the issue
here is not *whether* col names should be added, but *what*/*how* they
should be. Unless you wish to extend your criticism to the current version
for its failure to adhere to your proscription.
Cheers,
Bert
On Sat, Oct 28, 2023 at 11:55 AM Jeff Newmiller via R-help <
r-help@r-project.org> wrote:
> as.data.frame is a _converter_, while data.frame is a _constructor_.
> Changing the object contents is not what a conversion is for.
>
> On October 28, 2023 11:39:22 AM PDT, Boris Steipe <
> boris.ste...@utoronto.ca> wrote:
> >Thanks Duncan and Avi!
> >
> >That you could use NULL in a matrix() dimnames = list(...) argument
> wasn't clear to me. I thought that would be equivalent to a one-element
> list - and thereby define rownames. So that's good to know.
> >
> >The documentation could be more explicit - but it is probably more work
> to do that than just patch the code to honour a col.names argument. (At
> least I can't see a reason not to.)
> >
> >Thanks again!
> >:-)
> >
> >
> >
> >
> >> On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
> >>
> >> Борис,
> >>
> >> Try this where you tell matrix the column names you want:
> >>
> >> nouns <- as.data.frame(
> >> matrix(c(
> >>"gaggle",
> >>"geese",
> >>
> >>"dule",
> >>"doves",
> >>
> >>"wake",
> >>"vultures"
> >> ),
> >> ncol = 2,
> >> byrow = TRUE,
> >> dimnames=list(NULL, c("collective", "category"
> >>
> >> Result:
> >>
> >>> nouns
> >> collective category
> >> 1 gagglegeese
> >> 2 duledoves
> >> 3 wake vultures
> >>
> >>
> >> The above simply names the columns earlier when creating the matrix.
> >>
> >> There are other ways and the way you tried LOOKS like it should work but
> >> fails for me with a message about it weirdly expecting three rows
> versus two
> >> which seems to confuse rows and columns. My version of R is recent and I
> >> wonder if there is a bug here.
> >>
> >> Consider whether you really need the data.frame created in a single
> >> statement or can you change the column names next as in:
> >>
> >>
> >>> nouns
> >> V1 V2
> >> 1 gagglegeese
> >> 2 duledoves
> >> 3 wake vultures
> >>> colnames(nouns)
> >> [1] "V1" "V2"
> >>> colnames(nouns) <- c("collective", "category")
> >>> nouns
> >> collective category
> >> 1 gagglegeese
> >> 2 duledoves
> >> 3 wake vultures
> >>
> >> Is there a known bug here or is the documentation wrong?
> >>
> >> -Original Message-
> >> From: R-help On Behalf Of Boris Steipe
> >> Sent: Saturday, October 28, 2023 1:54 PM
> >> To: R. Mailing List
> >> Subject: [R] col.names in as.data.frame() ?
> >>
> >> I have been trying to create a data frame from some structured text in a
> >> single expression. Reprex:
> >>
> >> nouns <- as.data.frame(
> >> matrix(c(
> >>"gaggle",
> >>"geese",
> >>
> >>"dule",
> >>"doves",
> >>
> >>"wake",
> >>"vultures"
> >> ), ncol = 2, byrow = TRUE),
> >> col.names = c("collective", "category")
> >> )
> >>
> >> But ... :
> >>
> >>> str(nouns)
> >> 'data.frame': 3 obs. of 2 variables:
> >> $ V1: chr "gaggle" "dule" "wake"
> >> $ V2: chr "geese" "doves" "vultures"
> >>
> >> i.e. the col.names argument does nothing. From my reading of
> ?as.data.frame,
> >> my example should have worked.
> >>
> >> I know how to get the required result with colnames(), but I would like
> to
> >> understand why the idiom as written didn't work, and how I could have
> known
> >> that from the help file.
> >>
> >>
> >> Thanks!
> >> Boris
> >>
> >> __
> >> 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.
>
> --
> Sent from my phone. Please excuse my brevity.
>
> __
> 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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Re: [R] Plot for 10 years extrapolation
Dear Rui,
I really thank you a lot for your precious R help. It is exactly what I was
trying to do! Once more, many thanks!
Best,
Sacha
Le vendredi 27 octobre 2023 à 09:36:18 UTC+2, Rui Barradas
a écrit :
Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
> Dear R-Experts,
>
> Here below my R code working but I don't know how to complete/finish my R
> code to get the final plot with the extrapolation for the10 more years.
>
> Indeed, I try to extrapolate my data with a linear fit over the next 10
> years. So I create a date sequence for the next 10 years and store as a
> dataframe to make the prediction possible.
> Now, I am trying to get the plot with the actual data (from year 2004 to
> 2018) and with the 10 more years extrapolation.
>
> Thanks for your help.
>
>
> date <-as.Date(c("2018-12-31", "2017-12-31", "2016-12-31", "2015-12-31",
> "2014-12-31", "2013-12-31", "2012-12-31", "2011-12-31", "2010-12-31",
> "2009-12-31", "2008-12-31", "2007-12-31", "2006-12-31", "2005-12-31",
> "2004-12-31"))
>
> value <-c(15348, 13136, 11733, 10737, 15674, 11098, 13721, 13209, 11099,
> 10087, 14987, 11098, 13421, 9023, 12098)
>
> model <- lm(value~date)
>
> plot(value~date ,col="grey",pch=20,cex=1.5,main="Plot")
> abline(model,col="darkorange",lwd=2)
>
> dfuture <- data.frame(date=seq(as.Date("2019-12-31"), by="1 year",
> length.out=10))
>
> predict(model,dfuture,interval="prediction")
>
>
> __
> 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.
Hello,
Here is a way with base R graphics. Explained in the code comments.
date <-as.Date(c("2018-12-31", "2017-12-31", "2016-12-31",
"2015-12-31", "2014-12-31", "2013-12-31",
"2012-12-31", "2011-12-31", "2010-12-31",
"2009-12-31", "2008-12-31", "2007-12-31",
"2006-12-31", "2005-12-31", "2004-12-31"))
value <-c(15348, 13136, 11733, 10737, 15674, 11098, 13721, 13209,
11099, 10087, 14987, 11098, 13421, 9023, 12098)
model <- lm(value ~ date)
dfuture <- data.frame(date = seq(as.Date("2019-12-31"), by="1 year",
length.out=10))
predfuture <- predict(model, dfuture, interval="prediction")
dfuture <- cbind(dfuture, predfuture)
# start the plot with the required x and y limits
xlim <- range(c(date, dfuture$date))
ylim <- range(c(value, dfuture$fit))
plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"
, xlim = xlim, ylim = ylim)
# abline extends the fitted line past the x value (date)
# limit making the next ten years line ugly and not even
# completely overplotting the abline drawn line
abline(model, col="darkorange", lwd=2)
lines(fit ~ date, dfuture
# , lty = "dashed"
, lwd=2
, col = "black")
# if lines() is used for both the interpolated and extrapolated
# values you will have a gap between both fitted and predicted lines
# but it is closer to what you want
# get the fitted values first (interpolated values)
ypred <- predict(model)
plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"
, xlim = xlim, ylim = ylim)
# plot the interpolated values
lines(ypred ~ date, col="darkorange", lwd = 2)
# and now the extrapolated values
# I use normal orange to make the difference more obvious
lines(fit ~ date, dfuture, lty = "dashed", lwd=2, col = "orange")
Hope this helps,
Rui Barradas
--
Este e-mail foi analisado pelo software antivírus AVG para verificar a presença
de vírus.
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Re: [R] [Tagged] Re: col.names in as.data.frame() ?
Ah - that's an excellent point. Thanks.
> On Oct 28, 2023, at 14:54, Jeff Newmiller wrote:
>
> as.data.frame is a _converter_, while data.frame is a _constructor_.
> Changing the object contents is not what a conversion is for.
>
> On October 28, 2023 11:39:22 AM PDT, Boris Steipe
> wrote:
>> Thanks Duncan and Avi!
>>
>> That you could use NULL in a matrix() dimnames = list(...) argument wasn't
>> clear to me. I thought that would be equivalent to a one-element list - and
>> thereby define rownames. So that's good to know.
>>
>> The documentation could be more explicit - but it is probably more work to
>> do that than just patch the code to honour a col.names argument. (At least I
>> can't see a reason not to.)
>>
>> Thanks again!
>> :-)
>>
>>
>>
>>
>>> On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
>>>
>>> Борис,
>>>
>>> Try this where you tell matrix the column names you want:
>>>
>>> nouns <- as.data.frame(
>>> matrix(c(
>>> "gaggle",
>>> "geese",
>>>
>>> "dule",
>>> "doves",
>>>
>>> "wake",
>>> "vultures"
>>> ),
>>> ncol = 2,
>>> byrow = TRUE,
>>> dimnames=list(NULL, c("collective", "category"
>>>
>>> Result:
>>>
nouns
>>> collective category
>>> 1 gagglegeese
>>> 2 duledoves
>>> 3 wake vultures
>>>
>>>
>>> The above simply names the columns earlier when creating the matrix.
>>>
>>> There are other ways and the way you tried LOOKS like it should work but
>>> fails for me with a message about it weirdly expecting three rows versus two
>>> which seems to confuse rows and columns. My version of R is recent and I
>>> wonder if there is a bug here.
>>>
>>> Consider whether you really need the data.frame created in a single
>>> statement or can you change the column names next as in:
>>>
>>>
nouns
>>> V1 V2
>>> 1 gagglegeese
>>> 2 duledoves
>>> 3 wake vultures
colnames(nouns)
>>> [1] "V1" "V2"
colnames(nouns) <- c("collective", "category")
nouns
>>> collective category
>>> 1 gagglegeese
>>> 2 duledoves
>>> 3 wake vultures
>>>
>>> Is there a known bug here or is the documentation wrong?
>>>
>>> -Original Message-
>>> From: R-help On Behalf Of Boris Steipe
>>> Sent: Saturday, October 28, 2023 1:54 PM
>>> To: R. Mailing List
>>> Subject: [R] col.names in as.data.frame() ?
>>>
>>> I have been trying to create a data frame from some structured text in a
>>> single expression. Reprex:
>>>
>>> nouns <- as.data.frame(
>>> matrix(c(
>>> "gaggle",
>>> "geese",
>>>
>>> "dule",
>>> "doves",
>>>
>>> "wake",
>>> "vultures"
>>> ), ncol = 2, byrow = TRUE),
>>> col.names = c("collective", "category")
>>> )
>>>
>>> But ... :
>>>
str(nouns)
>>> 'data.frame': 3 obs. of 2 variables:
>>> $ V1: chr "gaggle" "dule" "wake"
>>> $ V2: chr "geese" "doves" "vultures"
>>>
>>> i.e. the col.names argument does nothing. From my reading of ?as.data.frame,
>>> my example should have worked.
>>>
>>> I know how to get the required result with colnames(), but I would like to
>>> understand why the idiom as written didn't work, and how I could have known
>>> that from the help file.
>>>
>>>
>>> Thanks!
>>> Boris
>>>
>>> __
>>> 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.
>
> --
> Sent from my phone. Please excuse my brevity.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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Re: [R] [Tagged] Re: col.names in as.data.frame() ?
as.data.frame is a _converter_, while data.frame is a _constructor_. Changing
the object contents is not what a conversion is for.
On October 28, 2023 11:39:22 AM PDT, Boris Steipe
wrote:
>Thanks Duncan and Avi!
>
>That you could use NULL in a matrix() dimnames = list(...) argument wasn't
>clear to me. I thought that would be equivalent to a one-element list - and
>thereby define rownames. So that's good to know.
>
>The documentation could be more explicit - but it is probably more work to do
>that than just patch the code to honour a col.names argument. (At least I
>can't see a reason not to.)
>
>Thanks again!
>:-)
>
>
>
>
>> On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
>>
>> Борис,
>>
>> Try this where you tell matrix the column names you want:
>>
>> nouns <- as.data.frame(
>> matrix(c(
>>"gaggle",
>>"geese",
>>
>>"dule",
>>"doves",
>>
>>"wake",
>>"vultures"
>> ),
>> ncol = 2,
>> byrow = TRUE,
>> dimnames=list(NULL, c("collective", "category"
>>
>> Result:
>>
>>> nouns
>> collective category
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>
>>
>> The above simply names the columns earlier when creating the matrix.
>>
>> There are other ways and the way you tried LOOKS like it should work but
>> fails for me with a message about it weirdly expecting three rows versus two
>> which seems to confuse rows and columns. My version of R is recent and I
>> wonder if there is a bug here.
>>
>> Consider whether you really need the data.frame created in a single
>> statement or can you change the column names next as in:
>>
>>
>>> nouns
>> V1 V2
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>> colnames(nouns)
>> [1] "V1" "V2"
>>> colnames(nouns) <- c("collective", "category")
>>> nouns
>> collective category
>> 1 gagglegeese
>> 2 duledoves
>> 3 wake vultures
>>
>> Is there a known bug here or is the documentation wrong?
>>
>> -Original Message-
>> From: R-help On Behalf Of Boris Steipe
>> Sent: Saturday, October 28, 2023 1:54 PM
>> To: R. Mailing List
>> Subject: [R] col.names in as.data.frame() ?
>>
>> I have been trying to create a data frame from some structured text in a
>> single expression. Reprex:
>>
>> nouns <- as.data.frame(
>> matrix(c(
>>"gaggle",
>>"geese",
>>
>>"dule",
>>"doves",
>>
>>"wake",
>>"vultures"
>> ), ncol = 2, byrow = TRUE),
>> col.names = c("collective", "category")
>> )
>>
>> But ... :
>>
>>> str(nouns)
>> 'data.frame': 3 obs. of 2 variables:
>> $ V1: chr "gaggle" "dule" "wake"
>> $ V2: chr "geese" "doves" "vultures"
>>
>> i.e. the col.names argument does nothing. From my reading of ?as.data.frame,
>> my example should have worked.
>>
>> I know how to get the required result with colnames(), but I would like to
>> understand why the idiom as written didn't work, and how I could have known
>> that from the help file.
>>
>>
>> Thanks!
>> Boris
>>
>> __
>> 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.
--
Sent from my phone. Please excuse my brevity.
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Re: [R] col.names in as.data.frame() ?
Thanks Duncan and Avi!
That you could use NULL in a matrix() dimnames = list(...) argument wasn't
clear to me. I thought that would be equivalent to a one-element list - and
thereby define rownames. So that's good to know.
The documentation could be more explicit - but it is probably more work to do
that than just patch the code to honour a col.names argument. (At least I can't
see a reason not to.)
Thanks again!
:-)
> On Oct 28, 2023, at 14:24, avi.e.gr...@gmail.com wrote:
>
> Борис,
>
> Try this where you tell matrix the column names you want:
>
> nouns <- as.data.frame(
> matrix(c(
>"gaggle",
>"geese",
>
>"dule",
>"doves",
>
>"wake",
>"vultures"
> ),
> ncol = 2,
> byrow = TRUE,
> dimnames=list(NULL, c("collective", "category"
>
> Result:
>
>> nouns
> collective category
> 1 gagglegeese
> 2 duledoves
> 3 wake vultures
>
>
> The above simply names the columns earlier when creating the matrix.
>
> There are other ways and the way you tried LOOKS like it should work but
> fails for me with a message about it weirdly expecting three rows versus two
> which seems to confuse rows and columns. My version of R is recent and I
> wonder if there is a bug here.
>
> Consider whether you really need the data.frame created in a single
> statement or can you change the column names next as in:
>
>
>> nouns
> V1 V2
> 1 gagglegeese
> 2 duledoves
> 3 wake vultures
>> colnames(nouns)
> [1] "V1" "V2"
>> colnames(nouns) <- c("collective", "category")
>> nouns
> collective category
> 1 gagglegeese
> 2 duledoves
> 3 wake vultures
>
> Is there a known bug here or is the documentation wrong?
>
> -Original Message-
> From: R-help On Behalf Of Boris Steipe
> Sent: Saturday, October 28, 2023 1:54 PM
> To: R. Mailing List
> Subject: [R] col.names in as.data.frame() ?
>
> I have been trying to create a data frame from some structured text in a
> single expression. Reprex:
>
> nouns <- as.data.frame(
> matrix(c(
>"gaggle",
>"geese",
>
>"dule",
>"doves",
>
>"wake",
>"vultures"
> ), ncol = 2, byrow = TRUE),
> col.names = c("collective", "category")
> )
>
> But ... :
>
>> str(nouns)
> 'data.frame': 3 obs. of 2 variables:
> $ V1: chr "gaggle" "dule" "wake"
> $ V2: chr "geese" "doves" "vultures"
>
> i.e. the col.names argument does nothing. From my reading of ?as.data.frame,
> my example should have worked.
>
> I know how to get the required result with colnames(), but I would like to
> understand why the idiom as written didn't work, and how I could have known
> that from the help file.
>
>
> Thanks!
> Boris
>
> __
> 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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and provide commented, minimal, self-contained, reproducible code.
Re: [R] col.names in as.data.frame() ?
Борис,
Try this where you tell matrix the column names you want:
nouns <- as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
),
ncol = 2,
byrow = TRUE,
dimnames=list(NULL, c("collective", "category"
Result:
> nouns
collective category
1 gagglegeese
2 duledoves
3 wake vultures
The above simply names the columns earlier when creating the matrix.
There are other ways and the way you tried LOOKS like it should work but
fails for me with a message about it weirdly expecting three rows versus two
which seems to confuse rows and columns. My version of R is recent and I
wonder if there is a bug here.
Consider whether you really need the data.frame created in a single
statement or can you change the column names next as in:
> nouns
V1 V2
1 gagglegeese
2 duledoves
3 wake vultures
> colnames(nouns)
[1] "V1" "V2"
> colnames(nouns) <- c("collective", "category")
> nouns
collective category
1 gagglegeese
2 duledoves
3 wake vultures
Is there a known bug here or is the documentation wrong?
-Original Message-
From: R-help On Behalf Of Boris Steipe
Sent: Saturday, October 28, 2023 1:54 PM
To: R. Mailing List
Subject: [R] col.names in as.data.frame() ?
I have been trying to create a data frame from some structured text in a
single expression. Reprex:
nouns <- as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
), ncol = 2, byrow = TRUE),
col.names = c("collective", "category")
)
But ... :
> str(nouns)
'data.frame': 3 obs. of 2 variables:
$ V1: chr "gaggle" "dule" "wake"
$ V2: chr "geese" "doves" "vultures"
i.e. the col.names argument does nothing. From my reading of ?as.data.frame,
my example should have worked.
I know how to get the required result with colnames(), but I would like to
understand why the idiom as written didn't work, and how I could have known
that from the help file.
Thanks!
Boris
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Re: [R] weights vs. offset (negative binomial regression)
Using an offset of log(Effort) as in your second model is the more
standard way to approach this problem; it corresponds to assuming that
catch is strictly proportional to effort. Adding log(Effort) as a
covariate (as illustrated below) tests whether a power-law model (catch
propto (Effort)^(b+1), b!=0) is a better description of the data. (In
this case it is not, although the confidence intervals on b are very
wide, indicating that we have very little information -- this is not
surprising since the proportional range of effort is very small
(246-258) in this data set.
In general you should *not* check overdispersion of the raw data
(i.e., the *marginal distribution* of the data, you should check
overdispersion of a fitted (e.g. Poisson) model, as below.
cheers
Ben Bolker
edata <- data.frame(Catch, Effort, xx1, xx2, xx3)
## graphical exploration
library(ggplot2); theme_set(theme_bw())
library(tidyr)
edata_long <- edata |> pivot_longer(names_to="var", cols =-c("Catch",
"Effort"))
ggplot(edata_long, aes(value, Catch)) +
geom_point(alpha = 0.2, aes(size = Effort)) +
facet_wrap(~var, scale="free_x") +
geom_smooth(method = "glm", method.args = list(family =
"quasipoisson"))
#
library(MASS)
g1 <- glm.nb(Catch~xx1+xx2+xx3+offset(log(Effort)), data=edata)
g2 <- update(g1, . ~ . + log(Effort))
g0 <- glm(Catch~xx1+xx2+xx3+offset(log(Effort)), data=edata,
family = poisson)
performance::check_overdispersion(g0)
summary(g1)
summary(g2)
options(digits = 3)
confint(g2)
summary(g1)
On 2023-10-28 3:30 a.m., 유준택 wrote:
Colleagues,
I have a dataset that includes five variables.
- Catch: the catch number counted in some species (ind.)
- Effort: fishing effort (the number of fishing vessels)
- xx1, xx2, xx3: some environmental factors
As an overdispersion test on the “Catch” variable, I modeled with negative
binomial distribution using a GLM. The “Effort” variable showed a gradually
decreasing trend during the study period. I was able to get the results I
wanted when considered “Effort” function as a weights function in the
negative binomial regression as follows:
library(qcc)
Catch=c(25,2,7,6,75,5,1,4,66,15,9,25,40,8,7,4,36,11,1,14,141,9,74,38,126,3)
Effort=c(258,258,258,258,258,258,258,254,252,252,252,252,252,252,252,252,252,252,252,248,246,246,246,246,246,246)
xx1=c(0.8,0.5,1.2,0.5,1.1,1.1,1.0,0.6,0.9,0.5,1.2,0.6,1.2,0.7,1.0,0.6,1.6,0.7,0.8,0.6,1.7,0.9,1.1,0.5,1.4,0.5)
xx2=c(1.7,1.6,2.7,2.6,1.5,1.5,2.8,2.5,1.7,1.9,2.2,2.4,1.6,1.4,3.0,2.4,1.4,1.5,2.2,2.3,1.7,1.7,1.9,1.9,1.4,1.4)
xx3=c(188,40,2,10,210,102,117,14,141,28,48,15,220,115,10,14,320,20,3,10,400,150,145,160,460,66)
#
edata <- data.frame(Catch, Effort, xx1, xx2, xx3)
#
qcc.overdispersion.test(edata$Catch, type="poisson")
#
summary(glm.nb(Catch~xx1+xx2+xx3, weights=Effort, data=edata))
summary(glm.nb(Catch~xx1+xx2+xx3+offset(log(Effort)), data=edata))
I am not sure the application of the weights function to the negative
binomial regression is correct. Also I wonder if there is a better way
doing this. Can anyone help?
[[alternative HTML version deleted]]
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Re: [R] col.names in as.data.frame() ?
Sent a slightly shorter version of this to your email, this one is to
the list:
On 28/10/2023 1:54 p.m., Boris Steipe wrote:
> > I have been trying to create a data frame from some structured text
in a single expression. Reprex:
> >
> > nouns <- as.data.frame(
> >matrix(c(
> > "gaggle",
> > "geese",
> >
> > "dule",
> > "doves",
> >
> > "wake",
> > "vultures"
> >), ncol = 2, byrow = TRUE),
> >col.names = c("collective", "category")
> > )
> >
You are calling it on a matrix, so the as.data.frame.matrix method is
what matters. It doesn't have a col.names argument, only row.names.
The docs are vague about what ... does, but if you look at the method,
you can see any unnamed arguments are ignored completely.
If you want to specify the column names in a single call, you'll need to
put them in the matrix, e.g.
as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
), ncol = 2, byrow = TRUE,
dimnames = list(NULL, c("collective", "category"))
)
)
Duncan Murdoch
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[R] col.names in as.data.frame() ?
I have been trying to create a data frame from some structured text in a single
expression. Reprex:
nouns <- as.data.frame(
matrix(c(
"gaggle",
"geese",
"dule",
"doves",
"wake",
"vultures"
), ncol = 2, byrow = TRUE),
col.names = c("collective", "category")
)
But ... :
> str(nouns)
'data.frame': 3 obs. of 2 variables:
$ V1: chr "gaggle" "dule" "wake"
$ V2: chr "geese" "doves" "vultures"
i.e. the col.names argument does nothing. From my reading of ?as.data.frame, my
example should have worked.
I know how to get the required result with colnames(), but I would like to
understand why the idiom as written didn't work, and how I could have known
that from the help file.
Thanks!
Boris
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Re: [R] How to Reformat a dataframe
You can also use the pivot_longer to do it:
library(tidyverse)
input <- structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465,
28.00536465, 120.0068103, 31.9980405, 85.0071837, 40.1532933,
19.3120917, 113.12581575, 28.45843425, 114.400074, 143.925,
46.439634, 20.7845679, 50.82874575, 36.9818061, 44.6273556, 40.57804605,
:
:
:
row.names = c(NA, -126L), class = "data.frame")
> input$row <- seq(nrow(input)) # add row number for reference> head(input)
>...1 ...2 ...3 ...4 ...5 ...6 ...7 ...8
1 92.99254 34.99963 24.04101 43.01330 53.00914 62.01390 91.01036 88.99986
2 76.00243 22.00219 22.00219 25.00378 44.99547 60.99449 63.00499 92.99254
3 44.99547 22.99328 15.00793 15.99902 38.00121 44.00438 62.01390 79.99510
4 28.00536 19.00061 12.99743 12.99743 44.00438 49.01647 55.95410 69.00816
5 120.00681 35.99072 22.99328 47.99706 60.00341 62.01390 60.00341 66.00658
6 31.99804 23.95606 13.98852 15.99902 38.99230 54.99132 88.00877 89.99095
...9...10 ...11 ...12 row
1 54.00023 75.01134 111.99314 49.01647 1
2 68.01707 75.01134 82.99669 63.99608 2
3 60.99449 91.01036 84.01609 65.01549 3
4 82.99669 78.01292 135.01474 85.99827 4
5 66.99767 91.01036 88.99986 51.98974 5
6 79.00401 78.01292 113.52225 155.00644 6> > x <-
pivot_longer(input,names_to = "col",cols=1:12)> head(x, 20)# A tibble:
20 × 3
row col value
1 1 ...1 93.0 2 1 ...2 35.0 3 1
...3 24.0 4 1 ...4 43.0 5 1 ...5 53.0 6 1 ...6
62.0 7 1 ...7 91.0 8 1 ...8 89.0 9 1 ...9 54.010
1 ...10 75.011 1 ...11 112. 12 1 ...12 49.013 2 ...1
76.014 2 ...2 22.015 2 ...3 22.016 2 ...4 25.017
2 ...5 45.018 2 ...6 61.019 2 ...7 63.020 2 ...8
93.0
>
Thanks
Jim Holtman
*Data Munger Guru*
*What is the problem that you are trying to solve?Tell me what you want to
do, not how you want to do it.*
On Fri, Oct 27, 2023 at 10:41 PM Paul Bernal wrote:
> Hi Iris,
>
> Thank you so much for your valuable feedback. I wonder why your code gives
> you 1512 rows, given that the original structure has 12 columns and 126
> rows, so I would expect (125*12)+ 9=1,509 total rows.
>
> Cheers,
> Paul
> El El vie, 27 de oct. de 2023 a la(s) 10:40 p. m., Iris Simmons <
> ikwsi...@gmail.com> escribió:
>
> > You are not getting the structure you want because the indexes are
> > wrong. They should be something more like this:
> >
> > i <- 0
> > for (row in 1:nrow(alajuela_df)){
> > for (col in 1:ncol(alajuela_df)){
> > i <- i + 1
> > df[i,1]=alajuela_df[row,col]
> > }
> > }
> >
> > but I think what you are doing can be written much shorter and will run
> > faster:
> >
> > ## transpose here matches your original code
> > df <- data.frame(aportes_alajuela = c(t(alajuela_df)))
> >
> > ## but if you do not want to transpose, then do this
> > df <- data.frame(aportes_alajuela = unlist(alajuela_df, use.names =
> FALSE))
> >
> > However, you said you expected 1509 observations, but this gives you
> > 1512 observations. If you want to exclude the 3 NA observations, do
> > something like:
> >
> > df <- df[!is.na(df$aportes_alajuela), , drop = FALSE]
> >
> > On Fri, Oct 27, 2023 at 11:14 PM Paul Bernal
> > wrote:
> > >
> > > Dear friends,
> > >
> > > I have the following dataframe:
> > > dim(alajuela_df)
> > > [1] 126 12
> > >
> > > dput(alajuela_df)
> > > structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465,
> > > 28.00536465, 120.0068103, 31.9980405, 85.0071837, 40.1532933,
> > > 19.3120917, 113.12581575, 28.45843425, 114.400074, 143.925,
> > > 46.439634, 20.7845679, 50.82874575, 36.9818061, 44.6273556,
> 40.57804605,
> > > 30.38398005, 47.94042705, 36.38715225, 28.06199835, 28.4867511,
> > > 122.86681215, 56.4071652, 35.9057658, 52.669341, 24.94714485,
> > > 54.4249857, 61.164396, 47.88379335, 30.582198, 26.051502, 43.041612,
> > > 64.59073485, 51.6499344, 78.8202902886201, 35.2390173175627,
> > > 82.2394568898745, 47.760850180466, 54.3654763342294, 49.4878058854839,
> > > 32.8813266149642, 38.9301880693548, 51.9506275455197, 55.4404001992832,
> > > 50.7979761262545, 37.1198211413082, 36.9144309425627, 33.7829493281362,
> > > 32.8647492475806, 42.892686344, 63.9814428257048, 39.219040238172,
> > > 88.7557324417563, 42.0964144925627, 129.15973304991, 117.872998635484,
> > > 35.4004098300179, 83.4102757505377, 38.6443638074373, 100.491764259319,
> > > 40.219162961828, 35.901029409319, 85.2814714674731, 26.5821299974014,
> > > 33.0141992892473, 52.4006692386201, 62.7450310643369, 33.3732853655914,
> > > 36.4616827405018, 146.501706130466, 59.9869292767025, 132.659282967294,
> > > 29.9692075517921, 45.8436438112007, 29.3028312143369, 49.9042699517921,
> > > 27.467868886, 29.8483957580645, 26.1192323867384, 62.1371491517921,
> > > 53.92489050681, 64.5840869873656, 28.4476420455197, 52.9893180218638,
> > > 36.4730500781362, 40.8595060662186, 33.4571194806452, 35.6201445262545,
> > >
Re: [R] How to Reformat a dataframe
Paul,
I have snipped away your long message and want to suggest another approach
or way of thinking to consider.
You have received other good suggestions and I likely would have used
something like that, probably within the dplyr/tidyverse but consider
something simpler.
You seem to be viewing a data.frame as similar to a matrix that you want to
reformat. There are similarities but a data.frame is also different. A
Matrix actually may be the right way for you to deal with your data. Can you
read it in as a matrix or must it be a data.frame?
The thing about a matrix is that underneath, it is just a linear vector
which you really seem to want. All your columns seem to be the same kind of
numeric and perhaps the order does not matter whether it is row major or
column major. So consider my smaller example. I am making a data.frame that
is smaller for illustration:
> small <- data.frame(A=1:4, B=5:8, C=9:12)
> small
A B C
1 1 5 9
2 2 6 10
3 3 7 11
4 4 8 12
Now I am making it a matrix and keeping the columns the same:
> small.mat <- as.matrix(small)
> small.mat
A B C
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
This can be linearized into a vector in many ways such as this:
> small.vec <- as.vector(small.mat)
> small.vec
[1] 1 2 3 4 5 6 7 8 9 10 11 12
You can make that into a data.frame if you like:
> revised <- data.frame(colname=small.vec)
> revised
colname
11
22
33
44
55
66
77
88
99
10 10
11 11
12 12
Of course, the above can be combined into more of a one-liner or made more
efficient. But in some cases, if you know the exact details of your
data.frame, you can spell out a way to combine the columns trivially. In my
example, I have three columns that can simply be concatenated into a vector
like so:
> small.onecol <- data.frame(onecol=c(small$A, small$B, small$C))
> small.onecol
onecol
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
This is not a generalized solution but is simple enough even with the number
of columns you have. You are simply consolidating the vectors into one
bigger one. If you want to connect many, there are shorter loops that can do
it as in:
> cols <- colnames(small)
> cols
[1] "A" "B" "C"
>
> new <- vector(mode="numeric", length=0)
> for (col in cols) {
+ new <- append(new, small[[col]])
+ }
> new
[1] 1 2 3 4 5 6 7 8 9 10 11 12
>
> new.df <- data.frame(newname=new)
> new.df
newname
11
22
33
44
55
66
77
88
99
10 10
11 11
12 12
The number of ways to do what you want is huge. You can pick a way that
makes more sense to you, especially the ones others have supplied, or one
that seems more efficient. As noted, all methods may also need to deal with
your NA issue at some stage.
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Re: [R] How to Reformat a dataframe
Às 04:13 de 28/10/2023, Paul Bernal escreveu: Dear friends, I have the following dataframe: dim(alajuela_df) [1] 126 12 dput(alajuela_df) structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465, 28.00536465, 120.0068103, 31.9980405, 85.0071837, 40.1532933, 19.3120917, 113.12581575, 28.45843425, 114.400074, 143.925, 46.439634, 20.7845679, 50.82874575, 36.9818061, 44.6273556, 40.57804605, 30.38398005, 47.94042705, 36.38715225, 28.06199835, 28.4867511, 122.86681215, 56.4071652, 35.9057658, 52.669341, 24.94714485, 54.4249857, 61.164396, 47.88379335, 30.582198, 26.051502, 43.041612, 64.59073485, 51.6499344, 78.8202902886201, 35.2390173175627, 82.2394568898745, 47.760850180466, 54.3654763342294, 49.4878058854839, 32.8813266149642, 38.9301880693548, 51.9506275455197, 55.4404001992832, 50.7979761262545, 37.1198211413082, 36.9144309425627, 33.7829493281362, 32.8647492475806, 42.892686344, 63.9814428257048, 39.219040238172, 88.7557324417563, 42.0964144925627, 129.15973304991, 117.872998635484, 35.4004098300179, 83.4102757505377, 38.6443638074373, 100.491764259319, 40.219162961828, 35.901029409319, 85.2814714674731, 26.5821299974014, 33.0141992892473, 52.4006692386201, 62.7450310643369, 33.3732853655914, 36.4616827405018, 146.501706130466, 59.9869292767025, 132.659282967294, 29.9692075517921, 45.8436438112007, 29.3028312143369, 49.9042699517921, 27.467868886, 29.8483957580645, 26.1192323867384, 62.1371491517921, 53.92489050681, 64.5840869873656, 28.4476420455197, 52.9893180218638, 36.4730500781362, 40.8595060662186, 33.4571194806452, 35.6201445262545, 47.5373940344086, 62.5177012273297, 31.970248036, 29.6637272685484, 49.9578249978495, 26.1936613831541, 30.8051128442652, 153.165249601165, 36.7652008047491, 14.3854334390681, 55.9930862447133, 87.5882628859319, 88.414763059767, 58.6046644034946, 20.5301898890681, 39.0037205346774, 67.5971386298632, 25.1295804696237, 32.7864973072581, 59.5662701215054, 56.3503435331631, 25.2768483884409, 120.187620533964, 51.3699688189953, 42.6889983487868, 34.5183987020052, 32.9513272936733, 20.6141437461215, 43.4197954932283, 107.258904254306, 23.7548470509806, 48.4457090643702, 72.3249287581108, 32.7272590911596, 48.6115621286089), ...2 = c(34.9996266, 22.00219245, 22.9932822, 19.00060635, 35.99071635, 23.9560551, 44.99547465, 22.00219245, 17.35822905, 38.9639856, 16.4804067, 93.87035775, 81.2693595, 65.4119235, 14.6114946, 33.32893245, 23.3330844, 66.68618175, 29.1097218, 18.77407155, 28.1469489, 20.64298365, 15.31941585, 25.79665035, 39.5869563, 35.9057658, 28.11863205, 27.09922545, 18.03783345, 52.3295388, 47.99706075, 21.88892505, 18.8590221, 17.64139755, 29.2796229, 27.750513, 26.9576412, 36.8932410314484, 19.1194611744253, 37.5166237491071, 37.9625421619439, 23.444984650496, 37.4408609345785, 43.1033210084325, 28.1252430606151, 38.4216392645833, 41.0532302780651, 34.4822260610119, 27.8096188059524, 25.9720387760913, 19.5989670673372, 21.1366557174603, 34.0091174203373, 111.149568174301, 24.5438738482759, 62.923693488747, 36.056976280754, 36.934798630754, 47.0895472967433, 23.0987961767857, 45.1991799260913, 23.0274422095238, 26.8716961246169, 21.5009916962302, 25.315076619246, 41.2098751296627, 15.6537752838123, 20.1723471152778, 31.7437694203373, 34.6486999232143, 27.6318028338123, 23.9637897951389, 50.6646878095238, 33.3208231933532, 38.0673028746602, 19.9469359998016, 27.0503264451389, 19.838575356, 29.0062549955939, 17.231889453373, 25.332044256, 17.651735447619, 37.12988188841, 37.7982565469246, 33.1890524548611, 15.4283008803571, 33.5040593288314, 24.9726150325397, 29.0612348127976, 30.2629592267857, 32.8065534719349, 43.3676303365079, 39.19334780424, 23.9252474159722, 15.8641806903257, 27.51501019685, 21.5861832758362, 18.5136698379577, 54.9166790494253, 29.2865335598214, 12.8130786005824, 46.2078437446569, 37.9351942968391, 31.8758398080357, 25.483024048, 17.4062173291544, 19.1528843310516, 37.2160194109127, 25.3577400835684, 21.3381885368056, 37.5842732179795, 62.4856267757438, 19.2393143990079, 49.4806272500855, 29.5274325254668, 18.7819282827049, 19.5089694997422, 25.524181172185, 15.9275217745507, 25.461470312583, 37.4026236465872, 14.6120197959536, 26.5773978462589, 27.6541154512847, 22.5193597452548, 27.9919830179818), ...3 = c(24.04100565, 22.00219245, 15.0079305, 12.99743415, 22.9932822, 13.9885239, 27.0142749, 16.3671393, 12.2895129, 28.68496905, 16.56535725, 32.70596175, 46.01488125, 24.94714485, 10.8170367, 19.68021075, 15.0079305, 26.41962105, 21.9738756, 15.71585175, 21.26595435, 13.28060265, 11.6099085, 15.5742675, 25.96655145, 20.5013994, 15.99902025, 18.03783345, 11.07188835, 32.59269435, 29.7326925, 15.82911915, 13.4788206, 22.11545985, 17.6130807, 22.3136778, 17.01842685, 24.8211230265233, 13.237941302509, 22.3448026530466, 26.816253605, 19.755485530466, 28.9524567137097, 35.9159180773958, 29.0510074712366, 29.6768770591437, 19.74645255681, 21.3046743438172, 18.2600040043011, 14.1888055630824, 13.2245440831541,
Re: [R] How to Reformat a dataframe
The tidyverse idiom looks very different but does what you want and I have come
to like it.
What idiom of R one likes, for the mostly small datasets I handle, is largely a
matter
of preferenceds for "readability", itself very personal. Here's my tidyverse
way of doing
what you wanted:
### start of code
library(tidyverse)
# tmpDF <- structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465,
### I omitted the rest of reconstructing the dataframe for brevity, easy to
reconstruct
tmpDF %>%
pivot_longer(cols = everything()) -> tmpTibLong
tmpTibLong
### showed:
# # A tibble: 1,512 × 2
# name value
#
# 1 ...1 93.0
# 2 ...2 35.0
# 3 ...3 24.0
# 4 ...4 43.0
# 5 ...5 53.0
# 6 ...6 62.0
# 7 ...7 91.0
# 8 ...8 89.0
# 9 ...9 54.0
# 10 ...10 75.0
# # ℹ 1,502 more rows
# # ℹ Use `print(n = ...)` to see more rows
### as you don't want the missing values
tmpTibLong %>%
drop_na()
### gave:
# # A tibble: 1,509 × 2
# name value
#
# 1 ...1 93.0
# 2 ...2 35.0
# 3 ...3 24.0
# 4 ...4 43.0
# 5 ...5 53.0
# 6 ...6 62.0
# 7 ...7 91.0
# 8 ...8 89.0
# 9 ...9 54.0
# 10 ...10 75.0
# # ℹ 1,499 more rows
# # ℹ Use `print(n = ...)` to see more rows
### end
Very best all,
Chris
On 28/10/2023 07:41, Paul Bernal wrote:
Hi Iris,
Thank you so much for your valuable feedback. I wonder why your code gives
you 1512 rows, given that the original structure has 12 columns and 126
rows, so I would expect (125*12)+ 9=1,509 total rows.
Cheers,
Paul
El El vie, 27 de oct. de 2023 a la(s) 10:40 p. m., Iris Simmons <
ikwsi...@gmail.com> escribió:
You are not getting the structure you want because the indexes are
wrong. They should be something more like this:
i <- 0
for (row in 1:nrow(alajuela_df)){
for (col in 1:ncol(alajuela_df)){
i <- i + 1
df[i,1]=alajuela_df[row,col]
}
}
but I think what you are doing can be written much shorter and will run
faster:
## transpose here matches your original code
df <- data.frame(aportes_alajuela = c(t(alajuela_df)))
## but if you do not want to transpose, then do this
df <- data.frame(aportes_alajuela = unlist(alajuela_df, use.names = FALSE))
However, you said you expected 1509 observations, but this gives you
1512 observations. If you want to exclude the 3 NA observations, do
something like:
df <- df[!is.na(df$aportes_alajuela), , drop = FALSE]
On Fri, Oct 27, 2023 at 11:14 PM Paul Bernal
wrote:
Dear friends,
I have the following dataframe:
dim(alajuela_df)
[1] 126 12
[dput snipped]
What I want to do is, instead of having 12 observations by row, I want
to
have one observation by row. I want to have a single column with 1509
observations instead of 126 rows with 12 columns per row.
I tried the following:
df = data.frame(matrix(nrow = Length, ncol = 1))
colnames(df) = c("aportes_alajuela")
for (row in 1:nrow(alajuela_df)){
for (col in 1:ncol(alajuela_df)){
df[i,1]=alajuela_df[i,j]
}
}
But I am not getting the data in the structure I want.
Any help will be greatly appreciated.
Best regards,
Paul
[[alternative HTML version deleted]]
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--
Chris Evans (he/him)
Visiting Professor, UDLA, Quito, Ecuador & Honorary Professor,
University of Roehampton, London, UK.
Work web site: https://www.psyctc.org/psyctc/
CORE site: http://www.coresystemtrust.org.uk/
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(Beware: French time, generally an hour ahead of UK)
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[R] weights vs. offset (negative binomial regression)
Colleagues, I have a dataset that includes five variables. - Catch: the catch number counted in some species (ind.) - Effort: fishing effort (the number of fishing vessels) - xx1, xx2, xx3: some environmental factors As an overdispersion test on the “Catch” variable, I modeled with negative binomial distribution using a GLM. The “Effort” variable showed a gradually decreasing trend during the study period. I was able to get the results I wanted when considered “Effort” function as a weights function in the negative binomial regression as follows: library(qcc) Catch=c(25,2,7,6,75,5,1,4,66,15,9,25,40,8,7,4,36,11,1,14,141,9,74,38,126,3) Effort=c(258,258,258,258,258,258,258,254,252,252,252,252,252,252,252,252,252,252,252,248,246,246,246,246,246,246) xx1=c(0.8,0.5,1.2,0.5,1.1,1.1,1.0,0.6,0.9,0.5,1.2,0.6,1.2,0.7,1.0,0.6,1.6,0.7,0.8,0.6,1.7,0.9,1.1,0.5,1.4,0.5) xx2=c(1.7,1.6,2.7,2.6,1.5,1.5,2.8,2.5,1.7,1.9,2.2,2.4,1.6,1.4,3.0,2.4,1.4,1.5,2.2,2.3,1.7,1.7,1.9,1.9,1.4,1.4) xx3=c(188,40,2,10,210,102,117,14,141,28,48,15,220,115,10,14,320,20,3,10,400,150,145,160,460,66) # edata <- data.frame(Catch, Effort, xx1, xx2, xx3) # qcc.overdispersion.test(edata$Catch, type="poisson") # summary(glm.nb(Catch~xx1+xx2+xx3, weights=Effort, data=edata)) summary(glm.nb(Catch~xx1+xx2+xx3+offset(log(Effort)), data=edata)) I am not sure the application of the weights function to the negative binomial regression is correct. Also I wonder if there is a better way doing this. Can anyone help? [[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.
