vignettes/deduplication.md
In reclin2: Record Linkage Toolkit

title: Deduplication using reclin2 author: Jan van der Laan css: "style.css"

We are going to work with the dataset town_name included in the package. The dataset contains a collection of town names as observed in administrative dataset. The first column name contains the names as observed. The second column official_name the official town name. We are going to assume that the second column is not available (or only for a part of the observations). The goal is to recode the 584 town names into a smaller set of town names knowing that most of the observed town names are actually misspelled versions of a smaller set of town names. We could also have solved the problem differently by linking the observed town names to a dataset containing all official town names. Often cleaning up these kind of misspellings is a first step in an actual linkage process. By first cleaning up the town names, subsequent use of the variable is easier and can lead to better quality linkage.

library(reclin2)
data(town_names)
head(town_names)

When performing deduplication we will link a dataset to itself and will try to link different records belonging to the same object. When a dataset to itself, it is not necessary to both compare record i to j and j to i and we certainly do not want to compare a record to itself. The option deduplication of the pair_ functions makes sure that only the needed pairs are generated. This is a small dataset so we can easily generate all pairs:

pairs <- pair(town_names, deduplication = TRUE)
print(pairs)

We will compare the records on name and use a string similarity function.

compare_pairs(pairs, on = "name", 
  comparators = list(cmp_jarowinkler()), 
  inplace = TRUE)
print(pairs)

Now comes the difficult part: selecting a threshold. The problem is that it is not really possible to say beforehand what an appropriate threshold is. That depends on the exact problem and also depends on the number of different objects that are expected. To explain that, first a short explanation how the deduplicate_equivalence function that we are going to use later works. Let's assume we have two actual town names and using our string similarity function we select pairs that differ one letter from each other, so we end up with the following set of pairs as an example

rotterdam -> rottrdam
rotterdam -> rotterdm
rotterdm -> rottrdm
rtterdam -> rotterdam
amsterdam -> amstrdam
amstrdam -> amstdam
amsterdm -> amsterdam

That means that we are saying that rotterdam is the same object as rottrdam which is the same object as rottrdm. Therefore, rotterdam and rottrdm are the same object although we didn't select a pair rotterdam -> rottrdm. So all names rotterdam, rottrdam, rotterdm, rottrdm and rtterdam are going to be in one class. When the number of misspelled names increases and when the number of actual town names increases, the likelihood that two names that do not belong to the same object are linked by a chain of pairs increases. This is a bit like the game where you have to change one word into another in a given number of steps by changing one letter at a time (the words in between have to be valid words). When the vocabulary is bigger this becomes easier. Therefore, the optimal threshold depends on the number of actual town names and the number of misspellings.

We have the official names and can therefore measure how many errors we make. We make an error when we put two records from x in the same group while they actually belong to different object (official town names). First we add a variable indicating whether two pairs have the same official name:

compare_vars(pairs, "true", on_x = "official_name",
  inplace = TRUE)

In practice this information is not available, but it might be available for a subset of records, for example, after manual inspection of a subset of the pairs. We now round the similarity scores and count how many errors we make for each value of the similarity score threshold:

pairs$threshold <- trunc(pairs$name/0.05)*0.05
thresholds <- pairs[, .(ftrue = mean(true)), by = threshold]
print(thresholds[order(ftrue)])

For a threshold of 0.95 and 1.00 we make no errors. Below that we start making errors. So let's work with a threshold of 0.95 for now

select_threshold(pairs, "select", "name", threshold = 0.95,
  inplace = TRUE)
res <- deduplicate_equivalence(pairs, "group", "select")
print(res)

With deduplicate_equivalence we take all selected pairs (indicated by the column select) and put them in the same group. res now contains the original dataset with a group column added that indicates the unique objects (towns in this case). We can see how many towns we have in the resulting dataset:

length(unique(res$group))

This is quite large. We started with nrow(res){.R} town names and reduced that to length(unique(res$group)){.R} while there are actually length(unique(res$official_name)){.R} town names. We can measure the quality by counting how often we have more than one official town name in one group:

qual <- res[, .(errors = length(unique(official_name))-1, n = .N), by = group]
qual$ferrors <- qual$errors/qual$n
qual[errors > 0]

So we have a large number of groups and no errors: no town names have been classified in the same group while actually being different towns. We can check what happens when we decrease the threshold. We will probably introduce some errors while we decrease the number of groups:

# Create a sequence of thresholds and initialise the result vectors
thresholds <- seq(0.5, 1, by = 0.02)
sizes <- numeric(length(thresholds))
nerrors <- numeric(length(thresholds))

for (i in seq_along(thresholds)) {
  threshold <- thresholds[i]
  # Perform deduplication with the given threshold
  select_threshold(pairs, "select", "name", threshold = threshold, inplace = TRUE)
  res <- deduplicate_equivalence(pairs, "group", "select")
  # Count the number of unique groups
  sizes[i] <- length(unique(res$group))
  # Count the number of errors
  qual <- res[, .(errors = length(unique(official_name))-1, n = .N), by = group]
  nerrors[i] <- sum(qual$errors)
}

The results are plotted in the figure below.

```{.R fun=output_figure name="fig2"} opar = par(mfrow = c(2,2)) plot(thresholds, sizes) plot(thresholds, nerrors) plot(sizes, nerrors) par(opar)


We can see that as the threshold decreases the number of errors increases and
the number of groups decreases. We cannot get much less than the 161 groups we
found without introducing some errors. How many errors and/or groups are
acceptable depends on the application and the amount of time one s willing to
spend in manually merging the groups. In this case manually inspecting the
groups and merging them will probably take only a few hours and 

With a threshold of 0.9 we should get approximately 100 groups and 5 errors
which seems a reasonable trade-off. So, let's rerun some of the previous code
with a threshold of 0.90. 

```{.R}
select_threshold(pairs, "select", "name", threshold = 0.9,
  inplace = TRUE)
res <- deduplicate_equivalence(pairs, "group", "select")
qual <- res[, .(errors = length(unique(official_name))-1, n = .N), by = group]
qual$ferrors <- qual$errors/qual$n
qual[errors > 0]

One way of assigning names to the groups we derived, is to use the most frequent name used in the group. Assuming that most people will correctly spell the town names this should give us the official town name belonging to each group. In this example dataset each town name occurs only once so can't use that trick. However, we can use the most frequent official name. We first define a function that returns the most frequent value of a vector and use that to derive the name of the group.

most_frequent <- function(x) {
  t <- table(x)
  t <- sort(t)
  tail(names(t), 1)
}

res[, assigned_name := most_frequent(official_name), by = group]
print(res)

We can now also look at the errors:

print(res[assigned_name != official_name])

We see that we make a lot of errors with the town of Hoogvliet Rotterdam. The problem we have is a difficult one. For example, rotterdam charlois should be called Rotterdam while rotterdam hoogvliet should be called Hoogvliet Rotterdam. We can't really expect that a computer is able to distinguish between these two without additional information. One other way of solving this problem is actually consider this as a linkage problem: we want to link a set of written town names to an official set of town names.