Identify multiple spellings in R

pairs <- expand.grid(x = seq_len(nrow(dta)), y = seq_len(nrow(dta))) %>% 
  # Only need to compare i to  all records j, with j > i 
  filter(y > x) %>%
  left_join(dta, by = c(x = 'i')) %>% rename(cityx = city, nx = n) %>%
  left_join(dta, by = c(y = 'i')) %>% rename(cityy = city, ny = n) %>%
  mutate(similarity = stringsim(cityx, cityy, method = "jw")) %>% 
  arrange(desc(similarity))

Source: stackoverflow.com

Identify multiple spellings in R

cities <- data.frame(city = c('bangalore','bengaluru','banglore',
  'bangalore', 'bangalore', 'bangalore', 'new york', 'newyork', 
  'nyork', 'new york', 'new york'))

Source: stackoverflow.com

Identify multiple spellings in R

> pairs
#    x y     cityx nx     cityy ny similarity
# 1  1 2 bangalore  4  banglore  1  0.9629630
# 2  4 5  new york  3   newyork  1  0.9583333
# 3  5 6   newyork  1     nyork  1  0.9047619
# 4  4 6  new york  3     nyork  1  0.8750000
# 5  2 3  banglore  1 bengaluru  1  0.7222222
# 6  1 3 bangalore  4 bengaluru  1  0.6944444
# 7  2 6  banglore  1     nyork  1  0.6583333
# 8  2 5  banglore  1   newyork  1  0.6011905
# 9  1 5 bangalore  4   newyork  1  0.5873016
# 10 2 4  banglore  1  new york  3  0.5833333
# 11 1 4 bangalore  4  new york  3  0.5694444
# 12 3 5 bengaluru  1   newyork  1  0.4761905
# 13 3 4 bengaluru  1  new york  3  0.4583333
# 14 1 6 bangalore  4     nyork  1  0.4370370
# 15 3 6 bengaluru  1     nyork  1  0.4370370

Source: stackoverflow.com

Identify multiple spellings in R

dta <- cities %>% group_by(city) %>% count() %>% 
  ungroup() %>% mutate(i = row_number())

Source: stackoverflow.com

Identify multiple spellings in R

library(dplyr)
library(stringdist)

Source: stackoverflow.com

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