The SoundexBR package provides an algorithm for decoding names into phonetic codes as pronounced in Portuguese. The goal is for homophone strings to be encoded with same alphanumeric representation, so that they can match despite minor differences in spelling.
The Soundex algorithm encodes mainly consonants by default. However, a
vowel will be encoded or counted if it’s the first letter. The resultant
code consists of a string four digits long, composed by one letter
followed by three numerical digits: [LETTER] [0-9] [0-9] [0-9].
The letter is the first letter of the name while the digits encode the
remaining consonants.
As one can imagine now, the SoundexBR resultant string can be very useful at identifying “close” matches that would typically fail due to variant spelling of names or transposition errors. For instance, the difference in the names Clair and Claire is enough to cause deterministic linkage to fail when comparing them, but the SoundexBR will return the same string “C460” for both names. A walkthrough in the vignette provides more information.
1 - From the CRAN repository:
install.packages('SoundexBR', dep=TRUE)
library(SoundexBR)2 - To get the current development version from Github:
## install devtools package if it's not already
if (!requireNamespace("devtools", quietly = TRUE)) {
install.packages("devtools")
}
install_github("danielmarcelino/SoundexBR")
library(SoundexBR)names <-
c(
'Ana Karolina Kuhnen',
'Ana Carolina Kuhnen',
'Ana Karolina',
'João Souza',
'João Sousa',
'Dilma Vana Rousseff',
'Dilma Rousef'
)
soundexBR(names)
[1] "A526" "A526" "A526" "J220" "J220" "D451" "D456"names2 <- c("HILBERT", "Heilbronn", "Gauss", "Kant") soundexBR(names2, BR=FALSE)
[1] "H416" "H416" "G200" "K530"soundexBR(names2)
[1] "I416" "E416" "G200" "C530"data1 <- data.frame(list(
first_name = c('Ricardo', 'Maria', 'Tereza', 'Pedro', 'José', 'Germano'),
last_name = c('Cunha', 'Andrade', 'Silva', 'Soares', 'Silva', 'Lima'),
age = c(67, 89, 78, 65, 68, 67),
birth = c(1945, 1923, 1934, 1947, 1944, 1945),
date = c(20120907, 20120703, 20120301, 20120805, 20121004, 20121209)
))data2 <-
data.frame(list(
first_name = c('Maria', 'Lúcia', 'Paulo', 'Marcos', 'Ricardo', 'Germânio'),
last_name = c('Andrada', 'Silva', 'Soares', 'Pereira', 'Cunha', 'Lima'),
age = c(67, 88, 78, 60, 67, 80),
birth = c(1945, 1924, 1934, 1952, 1945, 1932),
date = c(20121208, 20121103, 20120302, 20120105, 20120907, 20121209)
))pairs <- compare.linkage(
data1,
data2,
blockfld = list(c(1, 2, 4), c(1, 2)),
phonetic <- c(1, 2),
phonfun = soundexBR,
strcmp = FALSE,
strcmpfun <- jarowinkler,
exclude = FALSE,
identity1 = NA,
identity2 = NA,
n_match <- NA,
n_non_match = NA
)print(pairs)
$data1
first_name last_name age birth date
1 Ricardo Cunha 67 1945 20120907
2 Maria Andrade 89 1923 20120703
3 Tereza Silva 78 1934 20120301
4 Pedro Soares 65 1947 20120805
5 José Silva 68 1944 20121004
6 Germano Lima 67 1945 20121209
$data2
first_name last_name age birth date
1 Maria Andrada 67 1945 20121208
2 Lúcia Silva 88 1924 20121103
3 Paulo Soares 78 1934 20120302
4 Marcos Pereira 60 1952 20120105
5 Ricardo Cunha 67 1945 20120907
6 Germânio Lima 80 1932 20121209
$pairs
id1 id2 first_name last_name age birth date is_match
1 1 5 1 1 1 1 1 NA
2 6 6 0 1 0 0 1 NA
3 2 1 1 0 0 0 0 NA
$frequencies
first_name last_name age birth date
0.1000000 0.1428571 0.1250000 0.1250000 0.1000000
$type
[1] "linkage"
attr(,"class")
[1] "RecLinkData"editMatch(pairs)weights <- epiWeights(pairs, e = 0.01, f = pairs$frequencies)
hist(weights$Wdata, plot = FALSE) # Plot TRUE
$breaks
[1] 0.2 0.4 0.6 0.8 1.0
$counts
[1] 2 0 0 1
$density
[1] 3.333333 0.000000 0.000000 1.666667
$mids
[1] 0.3 0.5 0.7 0.9
$xname
[1] "weights$Wdata"
$equidist
[1] TRUE
attr(,"class")
[1] "histogram"
getPairs(pairs, max.weight = Inf, min.weight = -Inf)
id first_name last_name age birth date Weight
1 1 Ricardo Cunha 67 1945 20120907
2 5 Ricardo Cunha 67 1945 20120907 <NA>
3
4 6 Germano Lima 67 1945 20121209
5 6 Germânio Lima 80 1932 20121209 <NA>
6
7 2 Maria Andrade 89 1923 20120703
8 1 Maria Andrada 67 1945 20121208 <NA>Capitalize all letters in the word and drop all punctuation marks. Pad the word with rightmost blanks as needed during each procedure step. Retain the first letter of the word. However, if the first letter of the word is H, retain the second letter. If the first letter of the word is Y, change to I. If the combination of the first and the second letters is: WA, change to VA. If the combination of the first and the second letters is: KA, change to CA. If the combination of the first and the second letters is: KO, change to CO. If the combination of the first and the second letters is: KU, change to CU. If the combination of the first and the second letters is: CI, change to SI. If the combination of the first and the second letters is: CE, change to SE. If the combination of the first and the second letters is: GE, change to JE. If the combination of the first and the second letters is: GI, change to JI.
Change all occurrence of the following letters to ‘0’ (zero):
A, E, I, O, U, H, W, Y.
Change letters from the following sets into the digit given:
1 = B, F, P, V
2 = C, G, J, K, Q, S, X, Z
3 = D, T
4 = L
5 = M, N
6 = R
Remove all pairs of digits which occur beside each other from the string
that resulted after step (4). Remove all zeros from the string that
results from step 5.0 (computed in step 3). Pad the resultant string
from step (6) with trailing zeros and return only the first four
positions, which will be of the form [ALPHA] [0-9] [0-9] [0-9].