Dear Jim, Good morning, hope you are doing very well, Here I want to
calculate the specific parameters based on the previous data. I have
attached below.
can you help me with r functions or code to solve the above equations
Kind regards,
Hana
On 6/13/22, Jim Lemon wrote:
> Hi Hana,
> This is a
Hi, Dear Respected Professor! I hope that you are doing well. Kindly help
me with the following:
I have the R-codes for the "Quantile Augmented Mean Group" method. The
relevant codes and data are attached herewith.
Note: The link to the reference paper is:
https://www.econ.cam.ac.uk/people-files/e
Dear, Jim I try the following codes but I got an error
Dfm1<-data.table(Dfm1)
# the names of the columns to be multiplied by BETA
ID_names <- paste0("ID",4:8)
# the names of the new columns that have been multiplied by beta
new_ID_names <- paste0("new_",ID_names)
# Now with a simple command we
Dear Tim it works, But i want to treat them as a column name. Is it
possible to treat as a column name ?
Best,
Hana
On 6/13/22, Ebert,Timothy Aaron wrote:
> ID5 <- as.data.frame(ID_names1)
> colnames(ID5)<-"val" #rename the columns so they have the same name.
> ID6 <- as.data.frame(ID_names2)
> c
Hello,
Are you looking for this?
ID_names <- sprintf("id%05d", 1:5)
head(ID_names)
#> [1] "id1" "id2" "id3" "id4" "id5" "id6"
tail(ID_names)
#> [1] "id49995" "id49996" "id49997" "id49998" "id4" "id5"
sprintf with the format %05d pads the integers with zeros
Yes, It just takes another step. Here is the new pattern:
ID_names1 <- paste0("id",1:9)
ID_names2 <- paste0("id000",10:99)
ID5 <-as.data.frame(ID_names1) #convert to a dataframe.
ID5$name<-colnames(ID5) #make a new variable from column name.
colnames(ID5)<-c("name","val") #rename columns to be
Sorry, the last post had a small mistake. I got name and val switched. Here
they are in the correct order.
ID_names1 <- paste0("id",1:9)
ID_names2 <- paste0("id000",10:99)
ID5 <-as.data.frame(ID_names1)
ID5$name<-colnames(ID5)
colnames(ID5)<-c("val","name")
ID6 <-as.data.frame(ID_names2)
ID6$
Can anyone explain why Inf appears in the following results?
> sapply(-1:-10, \(ncp) qt(1-1*(10^(-4+ncp)), 35, ncp))
[1] 3.6527153 3.0627759 2.4158355 1.7380812 1.0506904 0.3700821
[7]Inf -0.9279783 -1.5341759 -2.1085213
> sapply(seq(-6.9, -7.9, -0.1), \(ncp) qt(1-1*(10^(-4+ncp)),
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