How can I access data in a nested R list? - r

I want to learn how to access data from a nested list in R. I am relatively new to the R programming language, so I am unsure how to proceed.
The data is a 'large list(947 elements, 654.9mb) and takes the form:
The numbers within the datalist refer to station numbers and when I click on one (in Rstudio) it looks like this:
I want to kow how I can access the data within 'doy' for example. I have tried:
data[[1]]
which returns all the data for the first element of the list (site, location, doy,ltm etc). So clearly the number used within the square brackets is interpreted as an index for the list, as opposed to an identifier for the elements/station in the list.
Then I tried:
data$1
but it returned the error:
Error: unexpected numeric constant in "data$1"
Then I tried:
data[data$1==doy]
But was returned this:
Error: unexpected numeric constant in "data[data$1"
So at this point, I realise that it is not construing the number of the station as a category/factor within the list. It's just reading it as a number. So I thought I'd put some quotes around it to see if that changed what happened:
data[data$"1"=="doy"]
This returned
named list()
But when I looked at it in the environment, it was a list of 0.
I looked at some of the similar question here on Stack (like: accessing nested lists in R) and tried:
data[data$"1"=="doy",][[1]]
But just got:
Error in data[data$"1" == "doy", ] : incorrect number of dimensions
How can I access this data? It reminds me of a structure in Matlab, but it doesn't seem to be indexed in a similar fashion in R.

Let's look at some ways to do what you want:
data[[1]]
This returns the first element of the list, which is itself a list. You can use the $ subsetting shorthand, but the name of the first element is nonstandard. R prefers names that start with letters and include only alphanumeric characters, periods and underscores. You can escape this behavior with backticks:
data$`1`
If you want to access one of the elements of list 1 in your list of lists, you need to further subset. To get to doy, which is the third element of 1. You can do that four ways.
data[[1]][[3]]
data$`1`[[3]]
data[[1]]$doy
data$`1`$doy

One way (in addition to what Ben Norris has shown):
our_list[[c("1", "doy")]]
Reproducible example data (please provide next time)
our_list <- list(`1` = list(site = "x", doy = 3))

Related

Logical comparison of elements from named list vs named vector in R

Sorry if this is a duplicate, I read through a number of threads but couldn't really find a good explanation.
I have a dataset (dataframe) where I calculated the mean value of each column. I now want to do some logical comparisons between these values. I used lapply to get the means
means_list <- lapply(dataset_df, mean)
which outputs a named list. But when I try to compare two elements of this list, e.g.
means_list["condition1"] > means_list["condition2"]
I get an error ("comparison of these types is not implemented").
I don't get that error if I use sapply instead so that I'm working with a named vector. I can also get around the error by converting the list to a dataframe with as.data.frame first.
So, I feel like I'm doing something wrong when subsetting a named list here but I don't quite understand how. Is there a correct way to subset the list so that I can do the logical comparison? Or is this not possible with named lists?
Thanks!
To access to the element of a list by its name, you have to use double brackets:
means_list[["condition1"]] > means_list[["condition2"]]

Filtering data, comma vs not comma

I have the following code
#abnormal return
exp.ret <- lm((RET-rf)~mkt.rf+smb+hml, data=tesla[tesla$period=="estimation.period",])
tesla$abn.ret <- (tesla$RET-tesla$rf)-predict(exp.ret,tesla)
#CAR during event window
CAR <- sum(tesla$abn.ret[tesla$period=="event.period",])
First section runs fine, but second gets this error:
"Error in tesla$abn.ret[tesla$period == "event.period", ] :
incorrect number of dimensions
I know that the solution is to remove the last comma:
#CAR during event window
CAR <- sum(tesla$abn.ret[tesla$period=="event.period"])
Just wondering what is the right pedagogical way of understanding it, why do I need a comma in the end in some cases, but some not, when I'm filtering for only parts of the data frame.
$ sign, [[]] and [] have different meanings.
In short:
$ sign and [[]] subsets one column of a dataframe or one item of a list.
The output of a subsetted dataframe will be a vector, while the output of a subsetted list will be a variable the same class as the original item, which can be a dataframe, another list, etc...
It's important to note that $ doesn't accept a column index (only a column name) and that you cannot insert two column names/index after $ or inside [[]].
[] slices a dataframe or a list sorting out one or more elements.
the class of the output variable will be the same as the original variable.
if you slice a dataframe using [], the output will be a dataframe, the same applies for lists, etc...
In your specific case, you used $ sign to subset your variable. Then, you tried to slice this output from the subset action using [ , ], but it turned out that the output is a vector, and a vector has always only one dimension and an error was fired. You should slice your vector using [] (the output will be a vector) or [[]] (the output will be a vector with length = 1).
Possible ways to subset tesla as you wish:
tesla$abn.ret[tesla$period == "event.period"]
tesla[["abn.ret"]][tesla$period == "event.period"]
tesla[tesla$period == "event.period", "abn.ret"]
You would achieve the same result using tesla[["period"]] instead of tesla$period.
For some extra details/examples, refer to An introduction to R, published by CRAN.
I hope it helped you somehow..!
tesla$abn.ret is one-dimensional. Each comma separates a dimension, so yours implies 2 dimensions.
Alternatively you could run
tesla[tesla$period=="event.period", "abn.ret"]
And get the same results, since tesla is 2-d.
If you look at the documentation with command ?'[', you find that the default behaviour of syntax x[i] is to drop one dimension away.
If you want to disable the dropping of the dimension, you have explicitly to write x[i,drop=False].

How can I return a vector with a dataframe inside in R?

Here is a challenge for you: I was trying to make a tic tac toe based on R. First, the players have to configure putting in the name of the players, and the game should check if the name exists in a file called "Players.txt" (if not, the game will create one), if the name exists, the game will ask for a new one. The last part of the game is that the game should record all the punctuation of the players (each gambling chip used will subtract 5 points of 100 that the player has at the beginning of the game). The problem is when a player wins, the game shows the following error: "Error in table[location_name1, 3]: Incorrect number of dimension in R".
A vector can either be atomic or a list. Atomic vectors can only contain elements of one and the same data type. That means, you are "accidentally" creating a list with
vector=c(win,name1,name2,table)
with the result that each column of the data frame should become an entry.
You can solve it with
vector <- list(win, name1, name2, table)
vector is still a list but now it has the format I believe you want.
Having done that you still get errors. The reason is that these assignments fail.
location_name1=which(grepl(name1,table$gamers))
location_name2=which(grepl(name2,table$gamers))
They return an empty vector because earlier in the code you set win=vector[1]... table=vector[4]. Since vector is now a list, you have to subset it accordingly. That means you have to chance the statements to table=vector[[4]].
Now you are going to get another problem. The reason is that you treat the columns table$scores as text. When you read the data you need to make sure that this columns is not interpreted as text. You also have to eliminate all statements that coerce the column into text. Otherwise table[location_name1,3]=table[location_name1,3]+pointsx will obviously fail because you cannot add a number to a string.
For example, you coerce the column into a character column with this statement:
name1 <- data.frame(gamers=name1,games="1",scores="100")
games and scores are strings not numbers. Another example is the assigment after reading the table from the file. You can make sure that scoresare numeric by doing this.
scores <- as.numeric(table[,3])
Please get familiar with Rstudio debugging capabilities (https://support.rstudio.com/hc/en-us/articles/205612627-Debugging-with-RStudio). This way you can go through your code line by line and check consequences of each assignment to the data frame.

R varied length vector or list in variable

I am using R to prepare some data for a D3 visualization. The visualization was created using the following structure (this is a single row from a .csv file that is subsequently converted to JSON in javascript).
Joe.Schmoe, joe.schmoe#email.com, Sao Paulo, ["Community01", "Community02", "Community03"],
["workgroup01","workgroup02"]
This is a single row. The headers would be:
Person, Email, Location, Communities, Workgroups
You'll notice that the Communities and Workgroup columns contain lists. Furthermore, these lists will vary in length depending on what Communities and Workgroups each individual is associated with. I recognize that this is probably not best practice with regard to data "tidyness," but it is what this viz is expecting.
So ... in R (which I'm learning), I'm finding it impossible to recreate this structure because, when I try to populate the "communities" or "workgroups" variables, R seems to expect that each variable will be of equal length.
The code that I have is reading from a data.frame which is list of the members of a particular community, and adding the name of that community to a column in a master data.frame of all employees. I'm indexing by email address because it is unique. So this particular loop looks at each individual email address in a data.frame called "commTD" and finds it in a master data.frame called "testr." If it finds it, it looks at the communities variable and either replaces an NA value with the name of the community (in this case "Technical Design"), or if the vector already exists, appends Technical Design to it:
for(i in commTD$email){
if(i %in% testr$email){
tmpList <- testr[which(testr$email ==i) , 'communities']
if(is.na(tmpList)){
tmpList <- list(c("Technical Design"))
}
else{
tmpList <- append(tmpList[[1]][1], 'Technical Design')
}
testr[which(testr$email ==i) , 'communities'] <- list(tmpList)
}
}
This works fine for the initial replacement, but if I append a new community to the list, and then try to pass it back into the testr data.frame, I get an error:
Error in `[<-.data.frame`(`*tmp*`, which(testr$email == i), "communities",
: replacement has 2 rows, data has 1
You'll note that I'm trying to create a list of vectors, which is just one way I've tried to figure this out. I thought maybe I could force R to see the list as a single object, even though it contains multiple items -- or in this case a vector of multiple items.
Is this just impossible in R, to have varied length vectors or lists as a single variable in a data frame?
Data frames are by definition a list of vectors of equal length, so when you ask if this is possible as a class data.frame(), no its not.
You could either use as suggested another type of object like data.table, or another way would be to think of your desired output as a list of unequal vectors, to pass to your js.
That object would look like something like:
dataList <- list(name = c("Joe.Schmoe", "Joe.Bloe"),
email = c("joe.schmoe#email.com", "joe.bloe#email.com"),
location = c("Sao Paulo", "London"),
Communities = list(c("Community01", "Community02", "Community03"),
c("Community02", "Community05", "Community03")
),
Workgroups = list(c("workgroup01","workgroup02"),
c("workgroup01","workgroup03"))
)
Then access each field like a dataframe, for output to your js:
dataList$name
dataList$Communities
etc...
As per Frank's suggestion, if you want to access each entry via the email address, so you can access each entry like this:
data_list[["joe.schmoe#email.com"]]
...then build the list with the names of the email as the index, like so:
data_list = list(`joe.schmoe#email.com`=list(name="Joe",
location="Sao Paulo",
Communities=....),
`joe.bloe#email.com`=list(n‌​ame="Joe", ...))
Then, you can avoid the non-R style of using for() loops, and start the fun of the lapply() family of functions to work on all the entries in a vectorised manner. (See ?lapply for details)
Hope it helps.

Select a column from a dynamic variable

How can I select the second column of a dynamically named variable?
I create variables of the form "population.USA", "population.Mexico", "population.Canada". Each variable has a column for the year, and another column for the population value. I would like to select the second column from each of these variables during a loop.
I use this syntax:
sprintf("population.%s", country)[, 2]
R returns the error: Error in sprintf("population.%s", country)[, 2] : incorrect number of dimensions
Based on your sequence of questions over the last few minutes, I have two general recommendations for you as you get familiar with R:
Don't use sprintf.
Don't use assign.
Now, obviously, those functions are both useful at times. But you've learned about them too early, before you've mastered some basic stuff about R's data structures. Try to write code without those crutches (for the time being!), as they're just causing you problems.
Rather than creating separate individual variables for each nation's population, place them in a list.
population <- vector("list",3)
names(population) <- c('USA','Mexico','Russia')
Then you can access each using the string representation of the name of each country:
population[['USA']] <- 10000
Or,
region <- 'USA'
population[[region]]
In this example, I've assigned a single value to a list element, lists will hold any other data type, including matrices or data frames. It will be a lot less typing than using sprintf and assign, and a lot safer and more efficient as well.
See ?get. Here is an example:
> country <- "FOO"
> assign(sprintf("population.%s", country), data.frame(runif(5), runif(5)))
>
> get(sprintf("population.%s", country))[,2]
[1] 0.2241105 0.5640709 0.5945869 0.1830719 0.1895938
It is critically important to look at the object returned by a function if you get an error. It is immediately clear why your example fails if you just look at what it returns:
> sprintf("population.%s", country)
[1] "population.FOO"
At that point it would be immediately clear, if you didn't already know or have thought to read ?sprintf, that sprintf() returns a string not the object of that name. Armed with that knowledge you would have narrowed down the problem to how to recall an object from the computed name?

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