I thought that R did not do partial matching on named lists, so I'm confused by the example below. I tried reading the Argument matching document but I'm still not sure of what's going on. Any help understanding this example would be appreciated.
ll <- list("dir_session" = "some_directory")
print(ll$dir_session) # prints contents of ll$dir_session as expected
print(ll$dir) # prints contents of ll$dir_session, but I expected to print NULL
print(ll[["dir"]]) # prints NULL as expected
Not sure if it makes a difference but I'm using R version 3.3.3 (2017-03-06).
I'm afraid the answer is you thought wrong. It has less to do with the class of object (a named list) and more to do with the "$" operator which does partial matching. See the ?Extract help page. This is different than argument matching when calling a function.
Related
My question refers to redundant code and a problem that I've been having with a lot of my R-Code.
Consider the following:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
combined_df_putnam$fu_time<-combined_df_putnam$age*365.25
combined_df_einstein$fu_time<-combined_einstein$age*365.25
combined_df_newton$fu_time<-combined_newton$age*365.25
...
combined_leibniz$fu_time<-combined_leibniz$age*365.25
I am trying to slim-down my code to do something like this:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1)
paste0("combined_df_",list_names[0:7]) <- paste0("combined_df_",list_names[0:7])$age*365.25
When I try to do that, I get "target of assignment expands to non-language object".
Basically, I want to create a list that contains descriptors, use that list to create a list of dataframes/lists and use these shortcuts again to do calculations. Right now, I am copy-pasting these assignments and this has led to various mistakes because I failed to replace the "name" from the previous line in some cases.
Any ideas for a solution to my problem would be greatly appreciated!
The central problem is that you are trying to assign a value (or data.frame) to the result of a function.
In paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1), the left-hand-side returns a character vector:
> paste0("combined_df_",list_names[0:7])
[1] "combined_df_putnam" "combined_df_einstein" "combined_df_newton"
[4] "combined_df_kant" "combined_df_hume" "combined_df_locke"
[7] "combined_df_leibniz"
R will not just interpret these strings as variables that should be created and be referenced to. For that, you should look at the function assign.
Similarily, in the code paste0("combined_df_",list_names[0:7])$age*365.25, the paste0 function does not refer to variables, but simply returns a character vector -- for which the $ operator is not accepted.
There are many ways to solve your problem, but I will recommend that you create a function that performs the necessary operations of each data frame. The function should then return the data frame. You can then re-use the function for all 7 philosophers/scientists.
I'm trying to remove all objects from my RStudio environment where the object names are NOT equal to a pattern.
rm(list=ls(pattern!="may19"))
However this gives me an error
Error in as.environment(pos) : no item called "pattern != "may19""
on the search list
Is there another way to approach this?
Thanks in advance
We could use one of the following(other variants might exist, you can add all=TRUE or all.names=TRUE) for completeness:
rm(list=setdiff(ls(),"may19"))
rm(list=ls(pattern = "[^may19]"))
I am new to R and I have troubles understanding how displaying an index works.
# Find indices of NAs in Max.Gust.SpeedMPH
ind <- which(is.na(weather6$Max.Gust.SpeedMPH))
# Look at the full rows for records missing Max.Gust.SpeedMPH
weather6[ind, ]
My code here works, no problem but I don't understand why weather6[ind] won't display the same thing as weather6[ind, ] . I got very lucky and mistyped the first time.
I apologize in advance that the question might have been posted somewhere else, I searched and couldn't find a proper answer.
So [ is a function just like any other function in R, but we call it strangely. Another way to write it in this case would be:
'[.data.frame'(weather6,ind,)
or the other way:
'[.data.frame'(weather6,ind)
The first three arguments to the function are named x, i and j. If you look at the code, early on it branches with the line:
if (Narg < 3L)
Putting the extra comma tells R that you've called the function with 3 arguments, but that the j argument is "missing". Otherwise, without the comma, you have only 2 arguments, and the function code moves on the the next [ method for lists, in which it will extract the first column instead.
I have a list of identifiers as follows:
url_num <- c('85054655', '85023543', '85001177', '84988480', '84978776', '84952756', '84940316', '84916976', '84901819', '84884081', '84862066', '84848942', '84820189', '84814935', '84808144')
And from each of these I'm creating a unique variable:
for (id in url_num){
assign(paste('test_', id, sep = ""), FUNCTION GOES HERE)
}
This leaves me with my variables which are:
test_8505465, test_85023543, etc, etc
Each of them hold the correct output from the function (I've checked), however my next step is to combine them into one big vector which holds all of these created variables as a seperate element in the vector. This is easy enough via:
c(test_85054655,test_85023543,test_85001177,test_84988480,test_84978776,test_84952756,test_84940316,test_84916976,test_84901819,test_84884081,test_84862066,test_84848942,test_84820189,test_84814935,test_84808144)
However, as I update the original 'url_num' vector with new identifiers, I'd also have to come down to the above chunk and update this too!
Surely there's a more automated way I can setup the above chunk?
Maybe some sort of concat() function in the original for-loop which just adds each created variable straight into an empty vector right then and there?
So far I've just been trying to list all the variable names and somehow get the output to be in an acceptable format to get thrown straight into the c() function.
for (id in url_num){
cat(as.name(paste('test_', id, ",", sep = "")))
}
...which results in:
test_85054655,test_85023543,test_85001177,test_84988480,test_84978776,test_84952756,test_84940316,test_84916976,test_84901819,test_84884081,test_84862066,test_84848942,test_84820189,test_84814935,test_84808144,
This is close to the output I'm looking for but because it's using the cat() function it's essentially a print statement and its output can't really get put anywhere. Not to mention I feel like this method I've attempted is wrong to begin with and there must be something simpler I'm missing.
Thanks in advance for any help you guys can give me!
Troy
I'm having trouble working with a data table in R. This is probably something really simple but I can't find the solution anywhere.
Here is what I have:
Let's say t is the data table
colNames <- names(t)
for (col in colNames) {
print (t$col)
}
When I do this, it prints NULL. However, if I do it manually, it works fine -- say a column name is "sample". If I type t$"sample" into the R prompt, it works fine. What am I doing wrong here?
You need t[[col]]; t$col does an odd form of evaluation.
edit: incorporating #joran's explanation:
t$col tries to find an element literally named 'col' in list t, not what you happen to have stored as a value in a variable named col.
$ is convenient for interactive use, because it is shorter and one can skip quotation marks (i.e. t$foo vs. t[["foo"]]. It also does partial matching, which is very convenient but can under unusual circumstances be dangerous or confusing: i.e. if a list contains an element foolicious, then t$foo will retrieve it. For this reason it is not generally recommended for programming.
[[ can take either a literal string ("foo") or a string stored in a variable (col), and does not do partial matching. It is generally recommended for programming (although there's no harm in using it interactively).