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Soundex (Phonetic) Algorithm for Portuguese Spelling Strings
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README.md

SoundexBR

lifecycle Build Status CRAN Version

Phonetic-Coding For Portuguese

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.

Installation

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)

Usage

A silly example

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"

The SoundexBR vs the original Soundex values

names2 <- c("HILBERT", "Heilbronn", "Gauss", "Kant")
Original Soundex outcome
 soundexBR(names2, BR=FALSE) 
[1] "H416" "H416" "G200" "K530"
The SoundexBR outcome
soundexBR(names2)
[1] "I416" "E416" "G200" "C530"

Example with RecordLinkage:

Some data

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)
  ))

Must call RecordLinkage package

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"

Editing correspondences

editMatch(pairs)

Accessing information within object:

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>

The Algorithm in a Nutshell

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].

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