Got it, thank you!
On Tue, 10 Aug 2021, 00:12 David Winsemius, wrote:
>
> On 8/9/21 12:22 PM, Luigi Marongiu wrote:
> > Thank you! it worked fine! The only pitfall is that `NA` became
> > ``. This is essentially the same thing anyway...
>
>
> It's not "essentially the same thing". It IS the same
Here is my hasty attempt last night checked in the light of morning.
It seems to return the correct extreme values and contains an example.
Jim
On Mon, Aug 9, 2021 at 10:50 PM Md. Moyazzem Hossain
wrote:
>
> Dear Jim,
>
> Thank you very much for your kind help.
>
> Take care.
>
> Md
>
> On Mon,
Rolf,
Sorry for only briefly chiming in, and late, but I don't usually follow
r-help that much these days.
I am writing this from an Ubuntu machine running R as well as RStudio from
pre-made binary .deb packages. R comes via apt from CRAN (using Michael's
binaries), RStudio from them via helper
I thought that I should let everyone know that I have, in some sense at
least, resolved my problem with 'no "doc" directory' and Rstudio. I
got a useful reply off-list from Duncan Murdoch (thanks Duncan) to the
effect that Rstudio requires its own purpose-specific binaries.
I was always under t
On 8/9/21 12:22 PM, Luigi Marongiu wrote:
Thank you! it worked fine! The only pitfall is that `NA` became
``. This is essentially the same thing anyway...
It's not "essentially the same thing". It IS the same thing. The print
function displays those '<>' characters flanking NA's when the cl
Thank you! it worked fine! The only pitfall is that `NA` became
``. This is essentially the same thing anyway...
On Mon, Aug 9, 2021 at 5:18 PM Ivan Krylov wrote:
>
> Thanks for providing a reproducible example!
>
> On Mon, 9 Aug 2021 15:33:53 +0200
> Luigi Marongiu wrote:
>
> > df[df[['vect[2]'
Thanks for providing a reproducible example!
On Mon, 9 Aug 2021 15:33:53 +0200
Luigi Marongiu wrote:
> df[df[['vect[2]']] == 2, 'vect[2]'] <- "No"
Please don't quote R expressions that you want to evaluate. 'vect[2]'
is just a string, like 'hello world' or 'I want to create a new column
named "
Hi,
You are going to need to provide more information than what you have
below and I may be mis-interpreting what you have provided.
Presuming you are designing a prospective, three-group, randomized
allocation study, there is typically an a priori specification of the
ratios of the sample s
FWIW:
Yes, thanks for noting that.
My own preference is to always propagate NA's and manually decide how
to deal with them, but others may disagree.
Best,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley
Dear All: good morning
*Re:* Sample Size Determination to Compare Three Independent Proportions
*Situation:*
Three Binary variables (Yes, No)
Three independent populations with fixed sizes (*say:* N1 = 1500, N2 = 900,
N3 = 1350).
Power = 0.80
How to choose the sample sizes to compare th
Thank you, it works!
On Mon, Aug 9, 2021 at 3:26 PM Andrew Simmons wrote:
>
> Hello,
>
>
> There are two convenient ways to access a column in a data.frame using `$`
> and `[[`. Using `df` from your first email, we would do something like
>
> df <- data.frame(VAR = 1:3, VAL = c("value is blue",
You are right, vect will contain the names of the columns of the real
dataframe buyt the actual simulation of the real case is more like
this:
```
> df = data.frame(A = 1:5, B = c(1, 2, NA, 2, NA), C = c("value is blue",
> "Value is red", "empty", " value is blue", " Value is green"), D = 9:13, E
Hello,
There are two convenient ways to access a column in a data.frame using `$`
and `[[`. Using `df` from your first email, we would do something like
df <- data.frame(VAR = 1:3, VAL = c("value is blue", "Value is red",
"empty"))
df$VAL
df[["VAL"]]
The two convenient ways to update / / replac
I wanted to remove possible white spaces before or after the string.
Actually, it worked, I used `gsub("[:blank:]*val[:blank:]*", "",
df$VAL, ignore.case=TRUE)`. I don't know why in the example there were
extra columns -- they did not came out in the real case.
Thank you, I think the case is closed
Dear Jim,
Thank you very much for your kind help.
Take care.
Md
On Mon, Aug 9, 2021 at 1:17 PM Jim Lemon wrote:
> And if you really don't like programming:
>
> whipple_index<-function(x,td=c(0,5)) {
> wi<-rep(NA,11)
> names(wi)<-c(paste0("wi",0:9),"O/all")
> for(i in 0:9) {
> ttd<-which(
And if you really don't like programming:
whipple_index<-function(x,td=c(0,5)) {
wi<-rep(NA,11)
names(wi)<-c(paste0("wi",0:9),"O/all")
for(i in 0:9) {
ttd<-which((x %% 10) %in% i)
wi[i+1]<-length(ttd) * 100/length(x)
}
ttd<-which((x %% 10) %in% td)
wi[11]<-length(ttd) * 100/(length(x)/le
According to Wikipedia, this is the definition of Whipple's index:
"The index score is obtained by summing the number of persons in the
age range 23 and 62 inclusive, who report ages ending in 0 and 5,
dividing that sum by the total population between ages 23 and 62 years
inclusive, and multiplyin
Dear Greg,
Thank you very much for your suggestion. I will try it and follow your
advice.
Actually, I want to find out the index for each digit like 0, 1, ..., 9.
Thanks in advance. Take care.
Md
On Mon, Aug 9, 2021 at 12:05 PM Greg Minshall wrote:
> Md,
>
> if this is what you are looking
Hi Luigi,
You want to get rid of certain strings in the "VAL" column. You are
assigning to:
df[df$VAL]
Error in `[.data.frame`(df, df$VAL) : undefined columns selected
when I think you should be assigning to:
df$VAL
What do you want to remove other than "[V|v]alue is" ?
JIim
On Mon, Aug 9, 20
On Mon, 9 Aug 2021 13:16:02 +0200
Luigi Marongiu wrote:
> df = data.frame(VAR = ..., VAL = ...)
> vect = letters[1:5]
What is the relation between vect and the column names of the data
frame? Is it your intention to choose rows or columns using `vect`?
> df[df[['vect[2]']] == 2, 'vect[2]']
'..
Thank you but I think I got it wrong:
```
> df = data.frame(VAR = letters[1:5], VAL = c(1, 2, NA, 2, NA)); df
VAR VAL
1 a 1
2 b 2
3 c NA
4 d 2
5 e NA
> vect = letters[1:5]
> df[df[['vect[2]']] == 2, 'vect[2]'] <- "No"; df
VAR VAL vect[2]
1 a 1
2 b 2
3 c NA
Md,
if this is what you are looking for:
https://en.wikipedia.org/wiki/Whipple%27s_index
then, the article says the algorithm is
The index score is obtained by summing the number of persons in the age
range 23 and 62 inclusive, who report ages ending in 0 and 5, dividing
that sum b
Sorry, silly question, gsub works already with regex. But still, if I
add `[[:blank:]]` still I don't get rid of all instances. And I am
keeping obtaining extra columns
```
> df[df$VAL] = gsub("[[:blank:]Value is]", "", df$VAL, ignore.case=TRUE)
> df[df$VAL] = gsub("[[:blank:]Value is]", "", df$VAL
Thank you, that is much appreciated. But on the real data, the
substitution works only on few instances. Is there a way to introduce
regex into this?
Cheers
Luigi
On Mon, Aug 9, 2021 at 11:01 AM Jim Lemon wrote:
>
> Hi Luigi,
> Ah, now I see:
>
> df$VAL<-gsub("Value is","",df$VAL,ignore.case=TRU
Dear Avi Gross,
Thank you very much for your email. Actually, I have a little knowledge of
R programming.
I have a dataset of ages ranging from 10 to 90. Now, I want to find out the
Whipple’s index for age heaping among individuals for each digit like
0,1,...,9.
I have searched in google I got t
On Mon, 9 Aug 2021 10:26:03 +0200
Luigi Marongiu wrote:
> vect = names(df)
> sub_df[vect[1]]
> df$column[df$column == value] <- new.value
Let's see, an equivalent expression without the $ syntax is
`df[['column']][df[['column']] == value] <- new.value`. Slightly
shorter, matrix-like syntax woul
Hi Luigi,
Ah, now I see:
df$VAL<-gsub("Value is","",df$VAL,ignore.case=TRUE)
df
VAR VAL
1 1 blue
2 2 red
3 3 empty
Jim
On Mon, Aug 9, 2021 at 6:43 PM Luigi Marongiu wrote:
>
> Hello,
> I have a dataframe where I would like to change the string of certain
> rows, essentially I am lo
Thank you very much, but that would make even more work due to the
duplication...
On Mon, Aug 9, 2021 at 10:53 AM Jim Lemon wrote:
>
> Hi Luigi,
> It looks to me as though you will have to copy the data frame or store
> the output in a new data frame.
>
> Jim
>
> On Mon, Aug 9, 2021 at 6:26 PM Lu
Hi Luigi,
It looks to me as though you will have to copy the data frame or store
the output in a new data frame.
Jim
On Mon, Aug 9, 2021 at 6:26 PM Luigi Marongiu wrote:
>
> Hello,
> I would like to recursively select the columns of a dataframe by
> strong the names of the dataframe in a vector
Hello,
I have a dataframe where I would like to change the string of certain
rows, essentially I am looking to remove some useless text from the
variables.
I tried with:
```
> df = data.frame(VAR = 1:3, VAL = c("value is blue", "Value is red", "empty"))
> df[df$VAL] = gsub("value is ", "", df$VAL,
Hello,
I would like to recursively select the columns of a dataframe by
strong the names of the dataframe in a vector and extracting one
element of the vector at a time. This I can do with, for instance:
```
vect = names(df)
sub_df[vect[1]]
```
The problem is that I would like also to change the v
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