I have already checked the forum for the solution but could not find the solution, hence, posting the problem here.
I am trying to change a name of a column using Rename, although the command is executed without any error but the names of the columns does not change.
I use this coderename(red,parts = category) Thanks
You need to pass a named vector as the second argument, and the names need to be quoted. Also, don't forget that you need to overwrite red with the result.
red <- reshape::rename(red, c(parts = "category"))
If you want to change multiple columns, add the items to the named vector you supply to the second argument.
Related
This is the code I am trying to run:
data_table<-data_table%>%
merge(new_table, by = 'Sample ID')%>%
mutate(Normalized_value = ((1.8^(data_table$Ubb - data_table$Ct_adj))*10000))
I want to first add the new column ("Ubb") from "new_table" and then add a calculated column using that new column. However, I get an error saying that Ubb column does not exist. So it's not performing merge before running mutate? When I separate the functions everything works fine:
data_table<-data_table%>%
merge(new_table, by = 'Sample ID')
data_table<-data_table%>%
mutate(Normalized_value = ((1.8^(data_table$Ubb - data_table$Ct_adj))*10000))
I would like to keep everything together just for style, but I'm also just curious, shouldn't R perform merge first and then mutate? How does order of operation during piping work?
Thank you!
you dont need to refer to column name with $ sign. i.e. use Normalized_value = ((1.8^(Ubb - Ct_adj))*10000)
because it is merged now. with $ sign I believe R, even though does the merge, has original data_table still in memory. because the assignment operator did not work yet. the assignment will take place after all operations.
Try running the code like this:
data_table<-data_table%>%
merge(new_table, by = 'Sample ID')%>%
mutate(Normalized_value = ((1.8^(Ubb - Ct_adj))*10000))
Notice that I'm not using the table name with the $ within the pipe. Your way is telling the mutate column to look at a vector. Maybe it's having some trouble understanding the length of that vector when used with the merge. Just call the variable name within the pipe. It's easiest to think of the %>% as 'and then': data_table() and then merge() and then mutate(). You might also want to think about a left_join instead of a merge.
I'm hoping to produce a pipe-friendly function where a user specifies the "name of choice" for a new column produced by the function as one of the function arguments.
In the function below, I'd like name_for_elective to be something that the user can set at will, and afterwards, the user could expect that there will be a new column in their data with the name that they provided here.
I've looked at https://dplyr.tidyverse.org/reference/dplyr_data_masking.html, the mutate() function documentation, searched here, and tried working with https://dplyr.tidyverse.org/reference/rename.html, but to no avail.
elective_open<-function(.data,name_for_elective,course,tiebreaker){
name_for_elective<-rlang::ensym(name_for_elective)
course<-rlang::ensym(course)
tiebreaker<-rlang::ensym(tiebreaker)
.data%>%
mutate(!!name_for_elective =ifelse(!!tiebreaker==max(!!tiebreaker),1,0))%>%mutate(!!name_for_elective=ifelse(!!name_for_elective==0,!!course[!!name_for_elective==1],""))%>%
filter(!(!!course %in% !!name_for_elective))
}
I've included this example function because there are several references to the desired new column name, and I'm unsure if the context in which the reference is made changes syntax.
As you can see, I was hoping !!name_for_elective would let me name our new column, but no. I've played with {{}}, not using rlang::ensym, and still haven't got this figured out.
Any solution would be greatly appreciated.
This: Use dynamic variable names in `dplyr` may be helpful, but I can't seem to figure out how to extend this to work in the case where multiple references are made to the name argument.
Example data, per a good suggestion by #MrFlick, takes the form below:
dat<-tibble(ID=c("AA","BB","AA","BB","AA","BB"),Class=c("A_Class","B_Class","C_Class","D_Class","E_Class","F_Class"),
randomNo=c(.75,.43,.97,.41,.27,.38))
The user could then run something like:
dat2<-dat%>%
elective_open(MyChosenName,Class,randomNo)
A desired result, if using the function a single time, would be:
desired_result_1<-tibble(ID=c("AA","BB","AA","BB"),
Class=c("A_Class","D_Class","E_Class","F_Class"),
randomNo=c(.75,.41,.27,.38),
MyChosenName=c("C_Class","B_Class"))
The goal would be to allow this function to be used again if so desired, with a different name specified.
In the case where a user runs:
dat3<-dat%>%
elective_open(MyChosenName,Class,randomNo)%>%
mutate(Just_Another_One=1)%>%
elective_open(SecondName,Class,randomNo)
The output would be:
desired_result_2<-tibble(ID=c("AA","BB"),
Class=c("E_Class","F_Class"),
randomNo=c(.27,.38),
MyChosenName=c("C_Class","B_Class"),
Just_Another_One=c(1,1),
SecondName=c("A_Class","D_Class"))
In reality, there may be any number of records with the same ID, and any number of Class-es.
In this case you can just stick to using the embrace {{}} option for your variables. If you want to dynamically create column names, you're going to still need to use :=. The difference here is that you can use the glue-style syntax with the embrace operator to get the name of the symbol. This works with the data provided.
elective_open <- function(.data, name_for_elective, course, tiebreaker){
.data%>%
mutate("{{name_for_elective}}" := ifelse({{tiebreaker}}==max({{tiebreaker}}),1,0)) %>%
mutate("{{name_for_elective}}" := ifelse({{name_for_elective}}==0,{{course}}[{{name_for_elective}}==1],"")) %>%
filter(!({{course}} %in% {{name_for_elective}}))
}
In R, I am trying to dynamically select columns of a data.frame called DF. If
cutOffYear=2014
and
forecast_years=3
Then this piece of code
paste0("DF$X",cutOffYear+1:forecast_years)
yields:
[1] "DF$X2015" "DF$X2016" "DF$X2017"
Assuming all three columns exist in DF how do I assign the column variables to the characters?
I have tried a lot of combinations of get, assign and paste0 but I am failing.
We can try with [ to select the columns. It is often error prone when using $. If we need to get the output as a data.frame with columns specified in the pasted combination of 'cutOffYear', 'forecast_years', then the below code should work fine
DF[paste0("X", cutOffYear+1:forecast_years)]
I've got csv, that has cols' heads like "2015-10-10", etc.
It's imported to R as strange X2015.10.10
I use Date class objects like 2014-12-31 and I want to get entire column, with the same date in top.
I've checked reading csv with ColClasses, etc. I need good hint.
Best,
JS
We can use check.names=FALSE in the read.csv/read.table. If you need to have a different column names, use header=FALSE which will provide default column names like V1, V2, etc.
While testing a simulation in R using randomly generated input data, I have found and fixed a few bugs and would now like to re-run the simulation with the same data, but with all intermediate variables removed to ensure it's a clean test.
Is there a way to remove several dozen manually selected variables from the workspace without having to:
a) clobber the entire workspace, e.g. rm(list=ls()), or b) type each variable name, e.g. remove(name1, name2, ...)?
Ideal solution would be to use ls() to inspect the definitions and then pick out the indices of the ones I want to remove, e.g.
ls() # inspect definitions
delme <- c(3,5,7:9,11,13) # names selected for removal
remove(ls()[delme]) # DESIRED SOLUTION -- doesn't quite work this way
(In hindsight, I should have used a fixed seed to generate the random input data, which allow clearing everything and then re-running the test...)
There is a much simpler and more direct solution:
vars.to.remove <- ls()
vars.to.remove <- temp[c(1,2,14:15)]
rm(list = vars.to.remove)
Or, better yet, if you are good about variable naming schemes, you can use the following pattern matching strategy:
E.g. I name all temporary variables with the starting string "Temp."
... so, you can have Temp.Names, Temp.Values, Temp.Whatever
The following produces the list of variables that match this pattern
ls(pattern = "^Temp\\.")
So, you can remove all unneeded variables using ONE line of code, as follows:
rm(list = ls(pattern = "^Temp\\."))
Hope this helps.
Assad, while I think the actual answer to the question is in the comments, let me suggest this pattern as a broader solution:
rm(list=
Filter(
Negate(is.na), # filter entries corresponding to objects that don't meet function criteria
sapply(
ls(pattern="^a"), # only objects that start with "a"
function(x) if(is.matrix(get(x))) x else NA # return names of matrix objects
) ) )
In this case, I'm removing all matrix object that start with "a". By modifying the pattern argument and the function used by sapply here, you can get pretty fine control over what you delete, without having to specify many names.
If you are concerned that this could delete something you don't want to delete, you can store the result of the Filter(... operation in a variable, review the contents, and then execute the rm(list=...) command.
Try
eval(parse(text=paste("rm(",paste(ls()[delme],sep=","),")")))
I had a similar requirement. I pulled all the elements I needed to a list:
varsToPurge = as.list(ls())
I then reassign the few values I wish to keep with new variable names which will not be in the variable varsToPurge. After that I looped through the elements
for (j in 1:length(varsToPurge)){
rm(list = as.character(varsToPurge[j]))
}
Do a little garbage collecting, and you maintain a clean environment as you go through your code.
gc()
You can also use a vector of row numbers you wish to keep instead and run through the vector in the loop but it won't be as dynamic if you add rough work you wish to remove.