R approach for iterative querying - r

This is a question of a general approach in R, I'm trying to find a way into R language but the data types and loop approaches (apply, sapply, etc) are a bit unclear to me.
What is my target:
Query data from API with parameters from a config list with multiple parameters. Return the data as aggregated data.frame.
First I want to define a list of multiple vectors (colums)
site segment id
google.com Googleuser 123
bing.com Binguser 456
How to manage such a list of value groups (row by row)? data.frames are column focused, you cant write a data.frame row by row in an R script. So the only way I found to define this initial config table is a csv, which is really an approach I try to avoid, but I can't find a way to make it more elegant.
Now I want to query my data, lets say with this function:
query.data <- function(site, segment, id){
config <- define_request(site, segment, id)
result <- query_api(config)
return result
}
This will give me a data.frame as a result, this means every time I query data the same columns are used. So my result should be one big data.frame, not a list of similar data.frames.
Now sapply allows to use one parameter-list and multiple static parameters. The mapply works, but it will give me my data in some crazy output I cant handle or even understand exactly what it is.
In principle the list of data.frames is ok, the data is correct, but it feels cumbersome to me.
What core concepts of R I did not understand yet? What would be the approach?

If you have a lapply/sapply solution that is returning a list of dataframes with identical columns, you can easily get a single large dataframe with do.call(). do.call() inputs each item of a list as arguments into another function, allowing you to do things such as
big.df <- do.call(rbind, list.of.dfs)
Which would append the component dataframes into a single large dataframe.
In general do.call(rbind,something) is a good trick to keep in your back pocket when working with R, since often the most efficient way to do something will be some kind of apply function that leaves you with a list of elements when you really want a single matrix/vector/dataframe/etc.

Related

How to pass a name to a function like dplyr::distinct()

I have a list of five data frames full of user responses to a survey.
In each of these data frames, the second column is the user id number. Some of the users took the survey multiple times, and I am trying to weed out the duplicate responses and just keep the first record.
The naming conventions are fairly standard, so the column in the first data frame is called akin to survey1_id and the second is survey2_id, etc. with the exception being that the column in the third data frame is called survey3a_id.
So basically what I tried to do was this:
for (i in seq(1,5)) {
newdata <- distinct(survey_list[[i]], grep(names("^survey.*_id$", survey_list[[i]]), value = TRUE))
}
But this doesn't work.
I originally thought it was just because the grep output had quotes around it, but I tried to strip them with noquote() and that didn't work. I then realized that distinct() doesn't actually evaluate the second argument, it just takes it literally, so I tried to force it to evaluate using eval(), but that didn't work. (Not sure I really expected it to.)
So now I'm kind of stuck. I don't know if the best solution is just to write five individual lines of code or, for a more generalizable solution, to sort and compare item-by-item in a loop? Was just hoping for a cleaner solution. I'm kind of new to this stuff.

How to make loops in R that operate on and return multiple objects

This is my first post, and I think I have looked thoroughly for my answer with no luck, but I might not be typing in the right search terms, since I am relatively new to R. I apologize if this has been answered before and if it has a link would be greatly appreciated.
In essence, I am trying to make a loop that will operate on a set of data frames that I have read into R from .txt files using read.table. I am working with simulated vegetation data organized into many species by site matrices, so it would be best for me if I could create loops that will just operate on the objects I have read in using some functions I have made and then put out new objects into my workspace with a specific naming pattern (e.g. put "_av" on the end of the name of the object operated on when creating a new object).
for convenience sake, lets say I have only four matrices I want to work with, all which contain the phrase "mod" for model. I have read that I can put these data frames into a list of data frames by the following code:
list.mods=lapply(ls(pattern="mod"),get)
This does create a list which I have been having trouble on getting my functions to actually operate on. From what I read this is the best way to make a list of objects you want to operate on.
So lets say that list.mods is now my list of operable matrices - mod1, mod2, mod3, and mod4. Also, lets say I have a function that simply calculates Bray-Curtis dissimilarity as follows:
bc=function(x){
vegdist(x,method="bray")
}
I can use this by typing in:
mod1.bc=bc(mod1)
That works. But it seems like I should be able to apply my list of models to the function bc and have it output the models with a pattern mod1.bc, mod2.bc, mod3.bc, and mod4.bc. I cannot get my list of files to work in the function much less save each operation as a new object with a patterned name.
What am I doing wrong? In the end I might have as many as a hundred models or more and would really appreciate being able to create a list of items that I can run through loops.
Thanks in advance.
You can use lapply again:
new.list.mods <- lapply(list.mods, bc)
This will return a new list in which each element is the result of applying bc to the corresponding element of list.mods.
The 'apply' family of functions in R basically allows you to save typing. If that's easier for you to understand, you can use a 'for loop' instead. Of course you will need to know how to access elements in a list for that. There is a question about that.
How about collecting the names of the models/objects you want into a list:
mod_list <- sapply(ls(pattern = "mod"), as.name)
and then looping over them with your function:
output_list <- lapply(eval(mod_list), bc)
With this approach you avoid creating the potentially large and redundant list.mods object in your example. Also, I think this will result in conveniently named lists.

How to order a matrix by all columns

Ok, I'm stuck in a dumbness loop. I've read thru the helpful ideas at How to sort a dataframe by column(s)? , but need one more hint. I'd like a function that takes a matrix with an arbitrary number of columns, and sorts by all columns in sequence. E.g., for a matrix foo with N columns,
does the equivalent of foo[order(foo[,1],foo[,2],...foo[,N]),] . I am happy to use a with or by construction, and if necessary define the colnames of my matrix, but I can't figure out how to automate the collection of arguments to order (or to with) .
Or, I should say, I could build the entire bloody string with paste and then call it, but I'm sure there's a more straightforward way.
The most elegant (for certain values of "elegant") way would be to turn it into a data frame, and use do.call:
foo[do.call(order, as.data.frame(foo)), ]
This works because a data frame is just a list of variables with some associated attributes, and can be passed to functions expecting a list.

Best way of storing data in 100 objects for later retrieval?

When doing sequencing, I normally apply TraMineR's seqdef function on a dataset to generate a single sequence object:
sequence_object <- seqdef(data)
However, let's say I want to loop through a dataframe and generate 1 sequence object per every chunk of 10 columns. Then I would do something like this:
colpicks <- seq(10,1000,by=10)
mapply(function(start,stop) seqdef(df[,start:stop]), colpicks-9, colpicks)
Now, I want to store these objects in some suitable manner. Two questions:
What is the most suitable way of storing (or maybe just automatically naming) 100 objects, so that I can easily loop through each of them at a later point?
How can I modify my code above so that it stores the data per your answer to (1)?
"Most suitable" is completely subjective and dependent on your goal.
I'm assuming this question is related to your previous question, and thus I would suggest setting the simplify argument of mapply to FALSE
myMatrixList <- mapply(.... , simplify=FALSE)
However, even that is not necessary, as you can just combine the sapply from the previous question and skip the middle step

Saving many subsets as dataframes using "for"-loops

this question might be very simple, but I do not find a good way to solve it:
I have a dataset with many subgroups which need to be analysed all-together and on their own. Therefore, I want to use subsets for the groups and use them for the later analysis. As well, the defintion of the subsets as the analysis should be partly done with loops in order to save space and to ensure that the same analysis has been done with all subgroups.
Here is an example of my code using an example dataframe from the boot package:
data(aids)
qlist <- c("1","2","3","4")
for (i in length(qlist)) {
paste("aids.sub.",qlist[i],sep="") <- subset(aids, quarter==qlist[i])
}
The variable which contains the subgroups in my dataset is stored as a string, therefore I added the qlist part which would be not required otherwise.
Make a list of the subsets with lapply:
lapply(qlist, function(x) subset(aids, quarter==x))
Equivalently, avoiding the subset():
lapply(qlist, function(x) aids[aids$quarter==x,])
It is likely the case that using a list will make the subsequent code easier to write and understand. You can subset the list to get a single data frame (just as you can use one of the subsets, as created below). But you can also iterate over it (using for or lapply) without having to construct variable names.
To do the job as you are asking, use assign:
for (i in qlist) {
assign(paste("aids.sub.",i,sep=""), subset(aids, quarter==i))
}
Note the removal of the length() function, and that this is iterating directly over qlist.

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