[R] help with LDA topic modelling..
dear members, I am using LDA for topic modelling of news articles (topicmodels package). I am visualizing the accuracy with the LDAvis package. The visualization shows clusters as circles, probably intersecting. My question is, if a find the optimal number of topics, k, and if the circles representing the topics doesn't intersect, then I have achieved perfect segregation. AM I right? Thanking You, Yours sincerely, AKSHAY M KULKARNI [[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.
Re: [R] Sum every n (4) observations by group
Milu,
Your data seems to be very consistent in that each value of ID has eight
rows. You seem to want to just sum every four so that fits:
ID Date Value
1 A 4140 0.000207232
2 A 4141 0.000240141
3 A 4142 0.000271414
4 A 4143 0.000258384
5 A 4144 0.000243640
6 A 4145 0.000271480
7 A 4146 0.000280585
8 A 4147 0.000289691
9 B 4140 0.000298797
10 B 4141 0.000307903
11 B 4142 0.000317008
12 B 4143 0.000326114
13 B 4144 0.000335220
14 B 4145 0.000344326
15 B 4146 0.000353431
16 B 4147 0.000362537
17 C 4140 0.000371643
18 C 4141 0.000380749
19 C 4142 0.000389854
20 C 4143 0.000398960
21 C 4144 0.000408066
22 C 4145 0.000417172
23 C 4146 0.000426277
24 C 4147 0.000435383
There are many ways to do what you want, some more general than others, but
one trivial way is to add a column that contains 24 numbers ranging from 1
to 6 like this assuming mydf holds the above:
Here is an example of such a vector:
rep(1:(nrow(mydf)/4), each=4)
[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6
So you can add a column like:
> mydf$fours <- rep(1:(nrow(mydf)/4), each=4)
> mydf
ID Date Value fours
1 A 4140 0.000207232 1
2 A 4141 0.000240141 1
3 A 4142 0.000271414 1
4 A 4143 0.000258384 1
5 A 4144 0.000243640 2
6 A 4145 0.000271480 2
7 A 4146 0.000280585 2
8 A 4147 0.000289691 2
9 B 4140 0.000298797 3
10 B 4141 0.000307903 3
11 B 4142 0.000317008 3
12 B 4143 0.000326114 3
13 B 4144 0.000335220 4
14 B 4145 0.000344326 4
15 B 4146 0.000353431 4
16 B 4147 0.000362537 4
17 C 4140 0.000371643 5
18 C 4141 0.000380749 5
19 C 4142 0.000389854 5
20 C 4143 0.000398960 5
21 C 4144 0.000408066 6
22 C 4145 0.000417172 6
23 C 4146 0.000426277 6
24 C 4147 0.000435383 6
You now use grouping any way you want to apply a function and in this case
you want a sum. I like to use the tidyverse functions so will show that as
in:
mydf %>%
group_by(ID, fours) %>%
summarize(sums=sum(Value), n=n())
I threw in the extra column in case your data sometimes does not have 4 at
the end of a group or beginning of next. Here is the output:
# A tibble: 6 x 4
# Groups: ID [3]
IDfours sums n
1 A 1 0.000977 4
2 A 2 0.00109 4
3 B 3 0.00125 4
4 B 4 0.00140 4
5 C 5 0.00154 4
6 C 6 0.00169 4
Of course there are all kinds of ways to do this in standard R, including
trivial ones like looping over indices starting at 1 and taking four at a
time and getting the Value data for mydf$Value[N] + mydf$Value[N+1] ...
-Original Message-
From: R-help On Behalf Of Miluji Sb
Sent: Sunday, December 19, 2021 1:32 PM
To: r-help mailing list
Subject: [R] Sum every n (4) observations by group
Dear all,
I have a dataset (below) by ID and time sequence. I would like to sum every
four observations by ID.
I am confused how to combine the two conditions. Any help will be highly
appreciated. Thank you!
Best.
Milu
## Dataset
structure(list(ID = c("A", "A", "A", "A", "A", "A", "A", "A", "B", "B", "B",
"B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C", "C", "C"), Date =
c(4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L, 4147L, 4140L, 4141L,
4142L, 4143L, 4144L, 4145L, 4146L, 4147L, 4140L, 4141L, 4142L, 4143L, 4144L,
4145L, 4146L, 4147L ), Value = c(0.000207232, 0.000240141, 0.000271414,
0.000258384, 0.00024364, 0.00027148, 0.000280585, 0.000289691, 0.000298797,
0.000307903, 0.000317008, 0.000326114, 0.00033522, 0.000344326, 0.000353431,
0.000362537, 0.000371643, 0.000380749, 0.000389854, 0.00039896, 0.000408066,
0.000417172, 0.000426277, 0.000435383 )), class = "data.frame", row.names =
c(NA, -24L))
[[alternative HTML version deleted]]
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Re: [R] Sum every n (4) observations by group
Dear Peter,
Thanks so much for your reply and the code! This is helpful.
What I would like is the data.frame below - sum values for *4140, 4141,
4142, 4143 *and then for *4144, 4145, 4146, 4147 *for IDs A, B, and C. Does
that make sense? Thanks again!
Best.
Milu
results <- structure(list(ID = c("A", "A", "A", "A", "A", "A", "A", "A",
"B", "B", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C",
"C", "C", "C"), Date = c(4140L, 4141L, 4142L, 4143L, 4144L, 4145L,
4146L, 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L,
4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L, 4147L
), Value = c(0.000207232, 0.000240141, 0.000271414, 0.000258384,
0.00024364, 0.00027148, 0.000280585, 0.000289691, 0.000298797,
0.000307903, 0.000317008, 0.000326114, 0.00033522, 0.000344326,
0.000353431, 0.000362537, 0.000371643, 0.000380749, 0.000389854,
0.00039896, 0.000408066, 0.000417172, 0.000426277, 0.000435383
), sum = c(NA, NA, NA, 0.000977171, NA, NA, NA, 0.001054089,
NA, NA, NA, 0.001213399, NA, NA, NA, 0.001395514, NA, NA, NA,
0.001541206, NA, NA, NA, 0.001686898)), class = "data.frame", row.names =
c(NA,
-24L))
On Sun, Dec 19, 2021 at 7:50 PM Peter Langfelder
wrote:
> I'm not sure I understand the task, but if I do, assuming your data
> frame is assigned to a variable named df, I would do something like
>
> sumNs = function(x, n)
> {
>if (length(x) %%n !=0) stop("Length of 'x' must be a multiple of 'n'.")
>n1 = length(x)/n
>ind = rep(1:n1, each = n)
>tapply(x, ind, sum)
> }
> sums = tapply(df$Value, df$ID, sumNs, 4)
>
> Peter
>
> On Sun, Dec 19, 2021 at 10:32 AM Miluji Sb wrote:
> >
> > Dear all,
> >
> > I have a dataset (below) by ID and time sequence. I would like to sum
> every
> > four observations by ID.
> >
> > I am confused how to combine the two conditions. Any help will be highly
> > appreciated. Thank you!
> >
> > Best.
> >
> > Milu
> >
> > ## Dataset
> > structure(list(ID = c("A", "A", "A", "A", "A", "A", "A", "A",
> > "B", "B", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C",
> > "C", "C", "C"), Date = c(4140L, 4141L, 4142L, 4143L, 4144L, 4145L,
> > 4146L, 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L,
> > 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L, 4147L
> > ), Value = c(0.000207232, 0.000240141, 0.000271414, 0.000258384,
> > 0.00024364, 0.00027148, 0.000280585, 0.000289691, 0.000298797,
> > 0.000307903, 0.000317008, 0.000326114, 0.00033522, 0.000344326,
> > 0.000353431, 0.000362537, 0.000371643, 0.000380749, 0.000389854,
> > 0.00039896, 0.000408066, 0.000417172, 0.000426277, 0.000435383
> > )), class = "data.frame", row.names = c(NA, -24L))
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] Sum every n (4) observations by group
I'm not sure I understand the task, but if I do, assuming your data
frame is assigned to a variable named df, I would do something like
sumNs = function(x, n)
{
if (length(x) %%n !=0) stop("Length of 'x' must be a multiple of 'n'.")
n1 = length(x)/n
ind = rep(1:n1, each = n)
tapply(x, ind, sum)
}
sums = tapply(df$Value, df$ID, sumNs, 4)
Peter
On Sun, Dec 19, 2021 at 10:32 AM Miluji Sb wrote:
>
> Dear all,
>
> I have a dataset (below) by ID and time sequence. I would like to sum every
> four observations by ID.
>
> I am confused how to combine the two conditions. Any help will be highly
> appreciated. Thank you!
>
> Best.
>
> Milu
>
> ## Dataset
> structure(list(ID = c("A", "A", "A", "A", "A", "A", "A", "A",
> "B", "B", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C",
> "C", "C", "C"), Date = c(4140L, 4141L, 4142L, 4143L, 4144L, 4145L,
> 4146L, 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L,
> 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L, 4147L
> ), Value = c(0.000207232, 0.000240141, 0.000271414, 0.000258384,
> 0.00024364, 0.00027148, 0.000280585, 0.000289691, 0.000298797,
> 0.000307903, 0.000317008, 0.000326114, 0.00033522, 0.000344326,
> 0.000353431, 0.000362537, 0.000371643, 0.000380749, 0.000389854,
> 0.00039896, 0.000408066, 0.000417172, 0.000426277, 0.000435383
> )), class = "data.frame", row.names = c(NA, -24L))
>
> [[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] Sum every n (4) observations by group
Dear all,
I have a dataset (below) by ID and time sequence. I would like to sum every
four observations by ID.
I am confused how to combine the two conditions. Any help will be highly
appreciated. Thank you!
Best.
Milu
## Dataset
structure(list(ID = c("A", "A", "A", "A", "A", "A", "A", "A",
"B", "B", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C",
"C", "C", "C"), Date = c(4140L, 4141L, 4142L, 4143L, 4144L, 4145L,
4146L, 4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L,
4147L, 4140L, 4141L, 4142L, 4143L, 4144L, 4145L, 4146L, 4147L
), Value = c(0.000207232, 0.000240141, 0.000271414, 0.000258384,
0.00024364, 0.00027148, 0.000280585, 0.000289691, 0.000298797,
0.000307903, 0.000317008, 0.000326114, 0.00033522, 0.000344326,
0.000353431, 0.000362537, 0.000371643, 0.000380749, 0.000389854,
0.00039896, 0.000408066, 0.000417172, 0.000426277, 0.000435383
)), class = "data.frame", row.names = c(NA, -24L))
[[alternative HTML version deleted]]
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Re: [R] Bug in list.files(full.names=T)
I don't know the answer to your question, but I see the same behaviour
on MacOS, e.g. list.files("./") includes ".//R" in the results on my
system. Both "./R" and ".//R" are legal ways to express that path on
MacOS, so it's not a serious bug, but it does look ugly.
Duncan Murdoch
On 18/12/2021 9:55 a.m., Mario Reutter wrote:
Dear everybody,
I'm a researcher in the field of psychology and a passionate R user. After
having updated to the newest version, I experienced a problem with
list.files() if the parameter full.names is set to TRUE.
A path separator "/" is now always appended to path in the output even if
path %>% endsWith("/"). This breaks backwards compatibility in case path
ends with a path separator. The problem occurred somewhere between R
version 3.6.1 (2019-07-05) and 4.1.2 (2021-11-01).
Example:
list.files("C:/Data/", full.names=T)
C:/Data//file.csv
Expected behavior:
Either a path separator should never be appended in accordance with
the documentation: "full.names
a logical value. If TRUE, the directory path is prepended to the file names
to give a relative file path."
Or it could only be appended if path doesn't already end with a path
separator.
My question would now be if this warrants a bug report? And if you agree,
could someone issue the report since I'm not a member on Bugzilla?
Thank you and best regards,
Mario Reutter
[[alternative HTML version deleted]]
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[R] Speed up studentized confidence intervals ?
Dear R-experts,
Here below my R code working but really really slowly ! I need 2 hours with my
computer to finally get an answer ! Is there a way to improve my R code to
speed it up ? At least to win 1 hour ;=)
Many thanks
library(boot)
s<- sample(178:798, 10, replace=TRUE)
mean(s)
N <- 1000
out <- replicate(N, {
a<- sample(s,size=5)
mean(a)
dat<-data.frame(a)
med<-function(d,i) {
temp<-d[i,]
f<-mean(temp)
g<-var(replicate(50,mean(sample(temp,replace=T
return(c(f,g))
}
boot.out <- boot(data = dat, statistic = med, R = 1)
boot.ci(boot.out, type = "stud")$stud[, 4:5]
})
mean(out[1,] < mean(s) & mean(s) < out[2,])
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Bug in list.files(full.names=T)
Dear everybody,
I'm a researcher in the field of psychology and a passionate R user. After
having updated to the newest version, I experienced a problem with
list.files() if the parameter full.names is set to TRUE.
A path separator "/" is now always appended to path in the output even if
path %>% endsWith("/"). This breaks backwards compatibility in case path
ends with a path separator. The problem occurred somewhere between R
version 3.6.1 (2019-07-05) and 4.1.2 (2021-11-01).
Example:
>> list.files("C:/Data/", full.names=T)
C:/Data//file.csv
Expected behavior:
Either a path separator should never be appended in accordance with
the documentation: "full.names
a logical value. If TRUE, the directory path is prepended to the file names
to give a relative file path."
Or it could only be appended if path doesn't already end with a path
separator.
My question would now be if this warrants a bug report? And if you agree,
could someone issue the report since I'm not a member on Bugzilla?
Thank you and best regards,
Mario Reutter
[[alternative HTML version deleted]]
__
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
and provide commented, minimal, self-contained, reproducible code.
