I have several things I considered about this topic.
It is, in general, not possible to do some things in one language or another
even if you find a bridge. Python lets you place all kinds of things into a
dictionary including many static objects like tuples or even other
dictionaries. What is all
cat in R behaves similarly to cat in unix-alikes, sends text to a stdout.
Usually, that stdout would be a file, but usually in R it is the R Console.
I think it might also help to note the difference between cat and print:
x <- "test\n"
cat(x)
print(x)
produces
> cat(x)
test
> print(x)
[1] "t
On Tue, 2 Nov 2021 10:18:07 -0700 (PDT)
Rich Shepard wrote:
> 'corvalis discharge summary\n'
> summary(cor_disc)
> sd(cor_disc$cfs)
> '-\n'
In the interactive mode, on the top level of execution, these commands
behave as if you had written
print(sink('data-summaries.txt'))
print
On Tue, 2 Nov 2021, Bert Gunter wrote:
What do you think these 2 lines are doing?
cat ('corvalis discharge summary\n')
print(cat)
Bert,
If I used them in linux cat would display the file (as do more and less) and
print() would be replaced with lpr
Please consult ?cat . You might also spend
What do you think these 2 lines are doing?
cat ('corvalis discharge summary\n')
print(cat)
Please consult ?cat . You might also spend a bit of (more?) time with an R
tutorial or two as you seem confused about how assignment (<-) works. Or
maybe I'm confused about what is confusing you
Bert G
On Tue, 2 Nov 2021, Andrew Simmons wrote:
You probably want to use cat and print for these lines. These things won't
print when not run at the top level, so if you want them to print, you must
specify that.
Andrew,
I modified the file to this:
sink('data-summaries.txt')
cat ('corvalis dischar
You probably want to use cat and print for these lines. These things won't
print when not run at the top level, so if you want them to print, you must
specify that.
On Tue, Nov 2, 2021, 13:18 Rich Shepard wrote:
> I've read ?sink and several web pages about it but it's not working
> properly
> w
Note that an environment carries a hash table with it, while a named list
does not. I think that looking up an entry in a list causes a hash table
to be created and thrown away. Here are some timings involving setting and
getting various numbers of entries in environments and lists. The times
ar
I've read ?sink and several web pages about it but it's not working properly
when I have the commands in a script and source() them.
The file:
library(tidyverse)
library(lubridate)
sink('data-summaries.txt')
'corvalis discharge summary\n'
summary(cor_disc)
sd(cor_disc$cfs)
'-\n'
If you're thinking about using environments, I would suggest you initialize
them like
x <- new.env(parent = emptyenv())
Since environments have parent environments, it means that requesting a
value from that environment can actually return the value stored in a
parent environment (this isn't an
On Tue, 2 Nov 2021 at 10:48, Yonghua Peng wrote:
>
> I know this is a newbie question. But how do I implement the hash
structure
> which is available in other languages (in python it's dict)?
>
As other posters wrote then environments are the solution.
data.frames, vectors and lists are much slow
On Tue, 2 Nov 2021, Ivan Krylov wrote:
That's because mutate() doesn't, well, mutate its argument. It _returns_
its changes, but it doesn't save them in the original variable. It's your
responsibility to assign the result somewhere:
Ivan,
I realized this after thinking more about the issue.
On Mon, 1 Nov 2021, jim holtman wrote:
drop the select, or put tz in the select
Jim,
Thinking more about the process after logging out for the evening it
occurred to me that I need to select all columns to retain them in the
tibble. I just tried that and, sure enough, that did the job.
Thank
Yes. A data.frame is basically a list where all elements are vectors of
the same length. So this issue also exists in a data.frame. However, the
data.frame construction method will detect this and generate unique
names (which also might not be what you want):
> data.frame(a=1:3, a=1:3)
On 02/11/2021 4:13 a.m., Yonghua Peng wrote:
I know this is a newbie question. But how do I implement the hash structure
which is available in other languages (in python it's dict)?
I know there is the list, but list's names can be duplicated here.
As Eric said, the environment comes pretty cl
Hi
Although you got several answers, simple aggregate was omitted.
> with(dat, aggregate(wt, list(Year=Year, Sex=Sex), mean))
Year Sexx
1 2001 F 12.0
2 2002 F 13.3
3 2003 F 12.0
4 2001 M 15.0
5 2002 M 16.3
6 2003 M 15.0
you can reshape the result
>
But for data.frame the colnames can be duplicated. Am I right?
Regards.
On Tue, Nov 2, 2021 at 6:29 PM Jan van der Laan wrote:
>
> True, but in a lot of cases where a python user might use a dict an R
> user will probably use a list; or when we are talking about arrays of
> dicts in python, the
True, but in a lot of cases where a python user might use a dict an R
user will probably use a list; or when we are talking about arrays of
dicts in python, the R solution will probably be a data.frame (with each
dict field in a separate column).
Jan
On 02-11-2021 11:18, Eric Berger wro
hi, Heinz,
> You are right - match seems obviously better, but why not do
>
> x <- c("a","b","c")
> match(x, letters[])
no, really, i wasn't sure of the requirements!
cheers, Greg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
h
One choice is
new.env(hash=TRUE)
in the base package
On Tue, Nov 2, 2021 at 11:48 AM Yonghua Peng wrote:
> I know this is a newbie question. But how do I implement the hash structure
> which is available in other languages (in python it's dict)?
>
> I know there is the list, but list's names c
I know this is a newbie question. But how do I implement the hash structure
which is available in other languages (in python it's dict)?
I know there is the list, but list's names can be duplicated here.
> x <- list(x=1:5,y=month.name,x=3:7)
> x
$x
[1] 1 2 3 4 5
$y
[1] "January" "Februar
Greg,
Greg Minshall wrote/hat geschrieben on/am 02.11.2021 08:57:
Heinz,
x <- c("a","b","c")
lettersnum <- 1:length(letters[])
names(lettersnum) <- letters[]
lettersnum[x]
lettersnum[x]
a b c
1 2 3
i'm not sure if the following is obviously better, but one might do
b <- match(a, a)
n
Heinz,
> x <- c("a","b","c")
> lettersnum <- 1:length(letters[])
> names(lettersnum) <- letters[]
> lettersnum[x]
> > lettersnum[x]
> a b c
> 1 2 3
i'm not sure if the following is obviously better, but one might do
> b <- match(a, a)
> names(b) <- a
> b
a b c
1 2 3
cheers, Greg
_
On Mon, 1 Nov 2021 15:24:47 -0700 (PDT)
Rich Shepard wrote:
> + mutate(
> + sampdt = make_datetime(year, mon, day, hr, min)
> + )
<...>
> produces the sampdt column, but it, and the timezone, are not present
> in the cor_disc tibble
That's because mutate() doesn't, well, mutate its argument. I
24 matches
Mail list logo
