As a follow-up to this, thanks to Bill Dunlap I was able to resolve what was
causing this problem (I still had problems with covr::package_coverage() - but
of a different sort and not directly related to this report, I had an existing
.R/Makevars file, created in Nov. 2014 related to the inst
> On Sep 9, 2017, at 11:45 AM, Andre Mikulec wrote:
>
> list subselect by name ?
> -
>
> I have this 'list of two elements of named elements.'
>
>> list(letters=letters, LETTERS=LETTERS)[c("letters","LETTERS")]
>
> $letters
> [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"
list subselect by name ?
-
I have this 'list of two elements of named elements.'
> list(letters=letters, LETTERS=LETTERS)[c("letters","LETTERS")]
$letters
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"
$L
I would try fininterval as well. It should do what you have asked provided
that you take care of the issue Ulrik pointed out.
Best of luck--EK
On Fri, Sep 8, 2017 at 6:15 AM, Hemant Sain wrote:
> i have a vector containing values ranging from 0 to 24
> i want to create another variable which can
You appear to already have something named subtree in your environment:
The following object is masked _by_ ‘.GlobalEnv’:
subtree
You could get rid of it, or you could specify that you want to use
subtree from extracat
hcs <- extracat::subtree(hc, k = 7)
Sarah
On Sat, Sep 9, 2017 at 8:58
Dear R community,
I would like to plot a partial hclust output, so I am looking for a subtree
function that would return an tree structure I can plot.
I ran the test code of subtree following the instruction on
http://finzi.psych.upenn.edu/library/extracat/html/subtree.html
However, an error mess
Exactly what I was looking for Eric, thanks!
I agree on your second point.
Best,
Lars.
On Sat, Sep 9, 2017 at 9:02 AM, Eric Berger wrote:
> Hi Lars,
> Two comments:
> 1. You can achieve what you want with a slight modification of your
> definition of s(), using the hint from the error message
Hi Lars,
Two comments:
1. You can achieve what you want with a slight modification of your
definition of s(), using the hint from the error message that you need an
argument '.':
s <- function(.) {
dplyr::summarise(., x1m = mean(X1),
x2m = mean(X2),
x3m = mea
Hi Lars
I am not very sure what you really want. However, I am suggesting the
following code that enables (1) to obtain the full summary of your data and
(2) retrieve only mean of X values as function of factors f1 and f2.
library(tidyverse)
library(psych)
df <- data.frame(matrix(rnorm(40), 10, 4
Dear group,
Is there a way I could avoid the sort of duplication illustrated below?
i.e., I have the same dplyr::summarise function on different group_by
arguments. So I'd like to create a single summarise function that could be
applied to both. My attempt below fails.
df <- data.frame(matrix(rno
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