String update for each group in data frame - r

Question

I have a large data frame df like this:
firstname = c("John L", "Robert C", "John", "J L", "Tom F", "T F", "Tom")
lastname = c("Doe", "Doe", "Doe", "Doe", "Frost", "Frost", "Frost")
id = c(178, 649, 384, 479, 539, 261, 347)
df = data.frame(firstname, lastname, id)
Which looks as below in df view:
firstname lastname id
John L Doe 178
Robert C Doe 649
John Doe 384
J L Doe 479
Tom F Frost 539
T F Frost 261
Tom Frost 347
As you see, the firstname in data frame is inconsistent. Sometime it is just an initial for example. I would like to have consistent firstname. I would like to have an output data frame like this:
firstname lastname id
John L Doe 178
Robert C Doe 649
John L Doe 384
John L Doe 479
Tom F Frost 539
Tom F Frost 261
Tom F Frost 347
I have tried few approaches like grouping by lastname and then getting longest string for each group and then updating firstname in the if elseif statement by matching with other firstname in the group using below
> sapply(strsplit("John L Doe"," "), function(a) paste(a[1],a[3]))
[1] "John Doe"
> sapply(strsplit("John L Doe"," "), function(a) paste(substr(a[1],1,1),a[2],a[3]))
[1] "J L Doe"
It did not work as I realized taking a longest string in the group is not a good approach.
Mapping from initials of the firstname to the full form of firstname is always going to be correct. For example, there will be "John L Doe". But, he will have 3 variants in his firstname. For example, "John L", "John", and "J L". It is because these are list of authors on a very narrow subjects. There is a just inconsistencies in the formatting of the name which I would like to fix. Having one consistent name will help me to do more analysis on a wider scale.
How can I do this in R?

Answer 1

The following solution produces your expected result, but bear in mind if Jack L Doe and John L Doe exist, J L Doe will map to the first longest name.
firstname = c("John L", "Robert C", "John", "J L", "Tom F", "T F", "Tom", "Jack L", "Robert Can","R C", "R C")
lastname = c("Doe", "Doe", "Doe", "Doe", "Frost", "Frost", "Frost", "Doe","Frost","Doe", "Frost")
id = c(178, 649, 384, 479, 539, 261, 347,100,200,300,400)
df = data.frame(firstname, lastname, id,stringsAsFactors = FALSE)
df$Initials <- sapply(strsplit(as.vector(firstname), " "), function(x) paste(substr(x, 1,1), collapse=""))
df$LongName<-apply(df,1,function(x) {
if(sub("\\s","",x[["firstname"]]) == x[["Initials"]]){
choices<-df$firstname[ grepl(x[["Initials"]], df$Initials) & df$lastname == x[["lastname"]]]
}
else{
choices<-df$firstname[ grepl(x[["Initials"]], df$Initials) & grepl(x[["firstname"]], df$firstname) & df$lastname == x[["lastname"]]]
}
choices[which.max(nchar(choices))]
}
)
Result
> df
firstname lastname id Initials LongName
1 John L Doe 178 JL John L
2 Robert C Doe 649 RC Robert C
3 John Doe 384 J John L
4 J L Doe 479 JL John L
5 Tom F Frost 539 TF Tom F
6 T F Frost 261 TF Tom F
7 Tom Frost 347 T Tom F
8 Jack L Doe 100 JL Jack L
9 Robert Can Frost 200 RC Robert Can
10 R C Doe 300 RC Robert C
11 R C Frost 400 RC Robert Can

Answer 2

Your usecase is not entirely clear.
As mentioned, there are issues if you have people with same last name, same initial but different first name. If you are convinced that this will never be the case in your data, then the solution may be quite simple.
However, if what you're trying to do is to find out if the names refer to the same people, you'll need a lot more, and that means diving into the subject of Entity Reconciliation.
There are some neat R packages for this (I've worked on a project involving entity reconciliation) including RecordLinkage, but the bottom line is: if you want reliable record linkage, you'll need at least a little more than first name & last name

Answer 3

What you are trying to achieve is usually done with a dictionary matching every spelling variant to a preferred name. There are smart solutions out there based on text similarity and text mining. Except if you already have the dictionary linking c("JL", J L", J.L.", etc....) to John L. I would not do it in R.
Have a look at DataWrangler, Trifacta, Dataiku or Openrefine they all have a free version that will do what you are looking for. I know that Openrefine (was GoogleRefine before) can be scripted.