Note: I'm aware that there are tons of post about basic deletion on R, but the conventional method doesnt work for this case:
Using R, I've built a key-value pair list, in which both key and value are the same string. (Might seem strange but that's a different discussion ..)
name <- "foo"
list[[name]] <- name
Now I want to delete an element from the list again. Using the conventional method for deletion I tried:
list[[name]] <- NULL
However, this throws the following error:
Warning: Error in <-: more elements supplied than there are to replace
Does anyone know how to delete an element in a key-value list?
Cheers
Edit: Here are the relevant parts taken directly from my code
# saving the list element
networkName <- "Simulation_1_Run_1"
.GlobalEnv$netDynListName[[networkName]] <- (networkName)
# retrieval of the element
# ...
# deletion of list element which causes an error
.GlobalEnv$netDynListName[[networkName]] <- NULL
Related
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 just learning R and having a hard time wrapping my head around how to extract elements from objects in a list. I've parsed a json file into R giving me list object. But I can't figure out how, from there, to extract various json elements from the list. here's a truncated look at how my data appears after parsing the json:
> #Parse data into R objects#
> list.Json= fromJSON(,final.name, method = "C")
> head(listJson,6)
[[1]]
[[1]]$contributors
NULL
[[1]]$favorited
[1] FALSE
...[truncating]...
[[5]]
[[5]]$contributors
NULL
[[5]]$favorited
[1] FALSE
I can figure out how to extract the favorites data for one of the objects in the list
> first.object=listJson[1]
> ff=first.object[[1]]$favorited
> ff
[1] FALSE
But I'm very confused about how to extract favorited for all objects in the list. I've looked into sappily, is that the correct approach? Do I need to put the above code into a for...next loop?
sapply is going to apply some function to every element in your list. In your case, you want to access each element in a (nested) list. sapply is certainly capable of this. For instance, if you want to access the first child of every element in your list:
sapply(listJson, "[[", 1)
Or if you wanted to access the item named "favorited", you could use:
sapply(listJson, "[[", "favorited")
Note that the [ operator will take a subset of the list you're working with. So when you access myList[1], you still have a list, it's just of length 1. However, if you reference myList[[1]], you'll get the contents of the first space in your list (which may or may not be another list). Thus, you'll use the [[ operator in sapply, because you want to get down to the contents of the list.
I've got a large function in R and the users have the ability to not include/specify an object. If they DO, the code checks to make sure the names in that object match the names in another. If they DON'T, there's no need to do that checking. The code line is:
if(exists("grids")) if(!all(expvarnames %in% names(grids))) {stop("Not all expvar column names found as column names in grids")}
But I'm getting the following error:
Error in match(x, table, nomatch = 0L) : argument "grids" is missing, with no default
Well in this trial run, grids is SUPPOSED to be missing. If I try
if(exists("grids")) print("yay")
Then nothing prints, i.e. the absence of grids means the expression isn't evaluated, which is as I'd expect. So can anyone think why R seems to be evaluating the subsequent IF statement in the main example? Should I slap another set of curly brackets around the second one??
Thanks!
Edit: more problems. Removing "grids," from the functions list of variables means it works if there's no object called grids and you don't specify it in the call (i.e. function(x,grids=whatever)). And keeping "grids," IN the functions list of variables means it works if there IS an object called grids and you do specify it in the call.
Please see this: http://i.imgur.com/9mr1Lwi.png
using exists(grids) is out because exists wants "quotes" and without em everything fails. WITH them ("grids"), I need to decide whether to keep "grids," in the functions list. If I don't, but I specify it in the call (function(x,grids=whatever)) then I get unused argument fail. If I DO, but don't specify it in the call because grids doesn't exist and I don't want to use it, I get match error, grids missing no default.
How do I get around this? Maybe list it in the function variables list as grids="NULL", then rather than if(exists("grids")) do if(grids!="NULL")
I still don't know why the original match problem is happening though. Match is from the expvarnames/grids names checker, which is AFTER if(exists("grids")) which evaluates to FALSE. WAaaaaaaiiiiittttt..... If I specify grids in the function variables list, i.e. simply putting function(x,grids,etc){do stuff}, does that mean the function CREATES an object called grids, within its environment?
Man this is so f'd up....
testfun <- function(x,grids)
{if(exists("grids")) globalgrids<<-grids
print(x+1)}
testfun(1) # Error in testfun(1) : argument "grids" is missing, with no default
testfun <- function(x,grids)
{if(exists("grids")) a<<-c(1,2,3)
print(x+1)}
testfun(1) #2 (and globally assigns a)
So in the first example, the function seems to have created an object called "grids" because exists("grids") evaluates to true. But THEN, ON THE SAME LINE, when asked to do something with grids, it says it doesn't exist! Schroedinger's object?!
This is proven in example 2: grids evaluates true and a is globally assigned then the function does its thing. Madness. Complete madness. Does anyone know WHY this ridiculousness is going on? And is the best solution to use my grids="NULL" default in the functions variables list?
Thanks.
Reproducible example, if you want to but I've already done it for every permutation:
testfun <- function(x,grids)
{if(exists("grids")) if(!all(expvarnames %in% names(grids))) {stop("Not all expvar column names found as column names in grids")}
print(x+1)}
testfun(1)
testfun(x=1,grids=grids)
grids<-data.frame(c(1,2,3),c(1,2,3),c(1,2,3))
expvarnames <- c("a","b","c")
colnames(grids) <- c("a","b","c")
Solution
Adapting your example use:
testfun <- function(x,grids = NULL)
{
if(!is.null(grids)){
if(!all(expvarnames %in% names(grids))){
stop("Not all expvar column names found as column names in grids")
}
print(x+1)
}
}
Using this testfun(1) will return nothing. By specifying a default argument in the function as NULL the function then checks for this (i.e. no argument specified) and then doesn't continue the function if so.
The Reason the Problem Occurs
We go through each of the examples:
testfun <- function(x,grids)
{if(exists("grids")) globalgrids<<-grids
print(x+1)}
testfun(1) # Error in testfun(1) : argument "grids" is missing, with no default
Here we call the function testfun, giving only the x argument. testfun knows it needs two arguments, and so creates local variables x and grids. We have then given an argument to x and so it assigns the value to x. There is no argument to grids, however the variable has still been created, even though no value has been assigned to it. So grids exists, but has no value.
From this exists("grids") will be TRUE, but when we try to do globalgrids<<-grids we will get an error as grids has not been assigned a value, and so we can't assign anything to globalgrids.
testfun <- function(x,grids)
{if(exists("grids")) a<<-c(1,2,3)
print(x+1)}
testfun(1) #2 (and globally assigns a)
This, however is fine. grids exists as in the previous case, and we never actually try and access the value stored in grids, which would cause an error as we have not assigned one.
In the solution, we simply set a default value for grids, which means we can always get something whenever we try and access the variable. Unlike in the previous cases, we will get NULL, not that nothing is stored there.
The main point of this is that when you declare arguments in your function, they are created each time you use the function. They exist. However, if you don't assign them values in your function call then they will exist, but have no value. Then when you try and use them, their lack of values will throw an error.
> a <- c(1,2,3,4)
> b <- c(2,4,6,8)
> if(exists("a")) if(!all(a %in% b)) {stop("Not all a in b")}
Error: Not all a in b
> rm(a)
> if(exists("a")) if(!all(a %in% b)) {stop("Not all a in b")}
>
When a does not exist, the expression does not evaluate, as expected. Before testing your first expression, make sure that grids does not exist by running rm(grids) in the console.
Richard Scriven's comment got me thinking: grids was an argument in my function but was optional, so maybe shouldn't be specified (like anything in "..." optional functions). I commented it out and it worked. Hooray, cheers everyone.
I find the following behaviour of the R person object rather unexpected:
Let's create a multi-person object:
a = c(person("Huck", "Finn"), person("Tom", "Sawyer"))
Imagine we want to update the given name of one person in the object:
a[[1]]$given <- 'Huckleberry'
Then if we inspect our object, to my surprise we have:
> a
[1] " <> [] ()" "Tom Sawyer"
Where'd Huckleberry Finn go?! (Note that if we try this with just a single person object, it works fine.) Why does this happen?
How can we do the above so that we get the more logical behavior of correcting just the first name?
The syntax you want here is
a <- c(person("Huck", "Finn"), person("Tom", "Sawyer"))
a[1]$given<-"Huckleberry"
a
#[1] "Huckleberry Finn" "Tom Sawyer"
A group of people is still a "person" and it has it's own special indexing function [.person and concat function c.person so it has perhaps different behavior than you were expecting. The problem was that [[ ]] was messing with the underlying hidden list.
Actually, it's interesting because they've overloaded nearly all the indexing methods for person but not the [<- or [[<- and that's really what's causing the error. Because up to here, we're the same
`$<-`(`[`(a,1), "given", "Huckleberry") #works
`$<-`(`[[`(a,1), "given", "Huckleberry") #works
but when we get to
`[<-`(a, 1, `$<-`(`[`(a,1), "given", "Huckleberry")) #works
`[[<-`(a, 1, `$<-`(`[[`(a,1), "given", "Huckleberry")) #no work
we see a difference. The special wrapping/unwrapping that happens during retrieval does not happen during assignment.
So what's going on is that a "person" is always a list of lists. The outer list holds all the people and the inner lists hold the data. You can think of the data like this
x<-list(
list(name="a"),list(name="b")
)
y<-list(
list(name="c")
)
where x is a collection of two people and y is a "single" person. When you do
x[1]<-y
x
you end up with
list(
list(name="c"),list(name="b")
)
since you're replacing a list with a list which is how [ indexing works with lists. But if you try to replace the element at [[1]] with a list of lists, that list will get nested. For example
x[[1]]<-y
x
becomes
x<-list(
list(list(name="c")),list(name="b")
)
And that extra level of nesting is what's confusing R when it goes to print the person in the first position. That first person won't have any named elements at the second level, so when it goes to print, it will return
emptyp <- structure(list(structure(list(), class="person")), class="person")
utils:::format.person(emptyp)
# " <> [] ()"
which gives is the symbols where it's trying to place the name, e-mail address, role, and comment.
I'm just learning R and having a hard time wrapping my head around how to extract elements from objects in a list. I've parsed a json file into R giving me list object. But I can't figure out how, from there, to extract various json elements from the list. here's a truncated look at how my data appears after parsing the json:
> #Parse data into R objects#
> list.Json= fromJSON(,final.name, method = "C")
> head(listJson,6)
[[1]]
[[1]]$contributors
NULL
[[1]]$favorited
[1] FALSE
...[truncating]...
[[5]]
[[5]]$contributors
NULL
[[5]]$favorited
[1] FALSE
I can figure out how to extract the favorites data for one of the objects in the list
> first.object=listJson[1]
> ff=first.object[[1]]$favorited
> ff
[1] FALSE
But I'm very confused about how to extract favorited for all objects in the list. I've looked into sappily, is that the correct approach? Do I need to put the above code into a for...next loop?
sapply is going to apply some function to every element in your list. In your case, you want to access each element in a (nested) list. sapply is certainly capable of this. For instance, if you want to access the first child of every element in your list:
sapply(listJson, "[[", 1)
Or if you wanted to access the item named "favorited", you could use:
sapply(listJson, "[[", "favorited")
Note that the [ operator will take a subset of the list you're working with. So when you access myList[1], you still have a list, it's just of length 1. However, if you reference myList[[1]], you'll get the contents of the first space in your list (which may or may not be another list). Thus, you'll use the [[ operator in sapply, because you want to get down to the contents of the list.