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)
name official_name
1 alblasserdam Alblasserdam
2 amsterdam Amsterdam
3 amsterdam-z.o. Amsterdam
4 amsterdam-zuidoost Amsterdam
5 amsterdam z-o Amsterdam
6 amsterdam z.o. AmsterdamWhen 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)
First data set: 584 records
Second data set: 584 records
Total number of pairs: 170 236 pairs
.x .y
1: 1 2
2: 1 3
3: 1 4
4: 1 5
5: 1 6
---
170232: 581 583
170233: 581 584
170234: 582 583
170235: 582 584
170236: 583 584We will compare the records on name and use a string similarity function.
> compare_pairs(pairs, on = "name", comparators = list(jaro_winkler()),
+ inplace = TRUE)
> print(pairs)
First data set: 584 records
Second data set: 584 records
Total number of pairs: 170 236 pairs
.x .y name
1: 1 2 0.6679894
2: 1 3 0.5753968
3: 1 4 0.5383598
4: 1 5 0.5882173
5: 1 6 0.5753968
---
170232: 581 583 0.5219298
170233: 581 584 0.5228070
170234: 582 583 0.5351852
170235: 582 584 0.7092593
170236: 583 584 0.9296296Now 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 -> amsterdamThat 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:
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)])
Total number of pairs: 16 pairs
threshold ftrue
1: 0.25 0.000000000
2: 0.30 0.001727116
3: 0.35 0.014810045
4: 0.50 0.034417054
5: 0.45 0.036163208
6: 0.00 0.055555556
7: 0.40 0.055960708
8: 0.55 0.057551227
9: 0.60 0.176160406
10: 0.65 0.365251342
11: 0.70 0.573084217
12: 0.75 0.772509347
13: 0.80 0.895522388
14: 0.85 0.941742311
15: 0.90 0.995657100
16: 0.95 1.000000000For 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)
name official_name group
1: alblasserdam Alblasserdam 541
2: amsterdam Amsterdam 427
3: amsterdam-z.o. Amsterdam 427
4: amsterdam-zuidoost Amsterdam 166
5: amsterdam z-o Amsterdam 427
---
580: amsterdam (duivendrecht) Amsterdam-Duivendrecht 580
581: amsterdam zzuidoost Amsterdam 166
582: hoovliet (rotterdam) Hoogvliet Rotterdam 568
583: roettrdam Rotterdam 496
584: srotterdam Rotterdam 496With 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:
This is quite large. We started with 584 town names and reduced that to 162 while there are actually 19 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$n
> qual[errors > 0]
Empty data.table (0 rows and 4 cols): group,errors,n,ferrorsSo we have a large number of groups and no errors: no town names have been classified in the same group while actually being different towns. We can check what happens when we decrease the threshold. We will probably introduce some errors while we decrease the number of groups:
> thresholds <- seq(0.5, 1, by = 0.02)
> sizes <- numeric(length(thresholds))
> nerrors <- numeric(length(thresholds))
> for (i in seq_along(thresholds)) {
+ threshold <- thresholds[i]
+ select_threshold(pairs, "select", "name", threshold = threshold,
+ inplace = TRUE)
+ res <- deduplicate_equivalence(pairs, "group", "select")
+ sizes[i] <- length(unique(res$group))
+ qual <- res[, .(errors = length(unique(official_name)) -
+ 1, n = .N), by = group]
+ nerrors[i] <- sum(qual$errors)
+ }The results are plotted in the figure below.
We can see that as the threshold decreases the number of errors increases and the number of groups decreases. We cannot get much less than the 161 groups we found without introducing some errors. How many errors and/or groups are acceptable depends on the application and the amount of time one s willing to spend in manually merging the groups. In this case manually inspecting the groups and merging them will probably take only a few hours and
With a threshold of 0.9 we should get approximately 100 groups and 5 errors which seems a reasonable trade-off. So, let’s rerun some of the previous code with a threshold of 0.90.
> select_threshold(pairs, "select", "name", threshold = 0.9,
+ inplace = TRUE)
> res <- deduplicate_equivalence(pairs, "group", "select")
> qual <- res[, .(errors = length(unique(official_name)) -
+ 1, n = .N), by = group]
> qual$ferrors <- qual$errors/qual$n
> qual[errors > 0]
group errors n ferrors
1: 543 1 256 0.00390625
2: 571 1 7 0.14285714
3: 441 1 78 0.01282051
4: 578 1 10 0.10000000
5: 296 1 6 0.16666667One 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)
name official_name group
1: alblasserdam Alblasserdam 453
2: amsterdam Amsterdam 543
3: amsterdam-z.o. Amsterdam 543
4: amsterdam-zuidoost Amsterdam 543
5: amsterdam z-o Amsterdam 543
---
580: amsterdam (duivendrecht) Amsterdam-Duivendrecht 580
581: amsterdam zzuidoost Amsterdam 543
582: hoovliet (rotterdam) Hoogvliet Rotterdam 156
583: roettrdam Rotterdam 441
584: srotterdam Rotterdam 441
assigned_name
1: Alblasserdam
2: Amsterdam
3: Amsterdam
4: Amsterdam
5: Amsterdam
---
580: Amsterdam-Duivendrecht
581: Amsterdam
582: Hoogvliet Rotterdam
583: Rotterdam
584: RotterdamWe can now also look at the errors:
> print(res[assigned_name != official_name])
name official_name group assigned_name
1: rotterdam hoogvliet Hoogvliet Rotterdam 441 Rotterdam
2: rotterdam-hoogvliet Hoogvliet Rotterdam 441 Rotterdam
3: nw. amsterdam Nieuw Amsterdam 543 Amsterdam
4: rotterdam - hoogvliet Hoogvliet Rotterdam 441 Rotterdam
5: amsterdam (driemond) Amsterdam 578 Diemen
6: nw-amsterdam Nieuw Amsterdam 543 Amsterdam
7: nw amsterdam Nieuw Amsterdam 543 Amsterdam
8: amsterdam- driemond Amsterdam 578 Diemen
9: amsterdam-driemond Amsterdam 578 Diemen
10: rotterdam -hoogvliet Hoogvliet Rotterdam 441 Rotterdam
11: rotterdam/hoogvliet Hoogvliet Rotterdam 441 Rotterdam
12: rotterdam hoogvliet Hoogvliet Rotterdam 441 Rotterdam
13: nw.amsterdam Nieuw Amsterdam 543 Amsterdam
14: amsterdam driemond Amsterdam 578 Diemen
15: veerdam Veendam 571 Leerdam
16: rotterdam(hoogvliet) Hoogvliet Rotterdam 441 Rotterdam
17: rotterdam (hoogvliet) Hoogvliet Rotterdam 441 Rotterdam
18: rotterdam-hoogvlie Hoogvliet Rotterdam 441 Rotterdam
19: rotterdam (prinsen Rotterdam 296 Pernis Rotterdam
20: rotterdam- hoogvliet Hoogvliet Rotterdam 441 Rotterdam
21: rotterdam hoogvlie Hoogvliet Rotterdam 441 Rotterdam
22: amsterdam-driemand Amsterdam 578 Diemen
name official_name group assigned_nameWe 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.