如何在 R 中逐字比较两个字符串

Question

I have a dataset, let's call it "ORIGINALE", composed by several different rows for only two columns, the first called "DESCRIPTION" and the second "CODICE". The description column has the right information while the column codice, which is the key, is almost always empty, therefore I'm tryng to search for the corresponding codice in another dataset, let's call it "REFERENCE". I am using the column desciption, which is in natural language, and trying to match it with the description in the second dataset. I have to match word by word since there may be a different order of words, synonims or abbreviations. Then I calcolate the similarity score to keep only the best match and accept those above a certain score. Is there a way to improve it? I'm working with around 300000 rows and, even though I know is always going to take time, perhaps there could be a way to make it even just slightly faster.

ORIGINALE <- data.frame(DESCRIPTION = c("mr peter 123 rose street 3b LA"," 4c flower str jenny jane Chicago", "washington miss sarah 430f name strt"), CODICE = (NA, NA, NA))
REFERENE <- dataframe (DESCRIPTION = c("sarah brown name street 430f washington", "peter green 123 rose street 3b LA", "jenny jane flower street 4c Chicago"), CODICE = c("135tg67","aw56", "83776250"))

algoritmo <- function(ORIGINALE, REFERENCE) {
   split1 <- strsplit(x$DESCRIPTION, " ")
   split2 <- strsplit(y$DESCRIPTION, " ")
   risultato <- vector()
   distanza <- vector()
      for(i in 1:NROW(split1)) {
      best_dist <- -5
      closest_match <- -5
        for(j in 1:NROW(split2)) {
          dist <- stringsim(as.character(split1[i]), as.character(split2[j]))
            if (dist > best_dist) {
              closest_match <- y$DESCRIPTION[j]
              best_dist <- dist 
            } 
        } 
      distanza <- append(distanza, best_dist)    
      risultato <- append(risultato, closest_match)
      }
    confronto <<- tibble(x$DESCRIPTION, risultato, distanza)
  }

match <- subset.data.frame(confronto, confronto$distanza >= "0.6")
missing <- subset.data.frame(confronto, confronto$distanza <"0.6")

Answer 1

Good question. for loops are slow in R:

for(i in 1:NROW(split1)) {
for(j in 1:NROW(split2)) {

For fast R, you need to vectorize your algorithm. I'm not that handy with data.frame anymore, so I'll use its successor, data.table .

library(data.table)
ORIGINALE = data.table(DESCRIPTION = c("mr peter 123 rose street 3b LA"," 4c flower str jenny jane Chicago", "washington miss sarah 430f name strt"), CODICE = c(NA, NA, NA))
REFERENCE = data.table(DESCRIPTION = c("sarah brown name street 430f washington", "peter green 123 rose street 3b LA", "jenny jane flower street 4c Chicago"), CODICE = c("135tg67","aw56", "83776250"))

# split DESCRIPTION to make tables that have one word per row
ORIGINALE_WORDS = ORIGINALE[,.(word=unlist(strsplit(DESCRIPTION,' ',fixed=T))),.(DESCRIPTION,CODICE)]
REFERENCE_WORDS = REFERENCE[,.(word=unlist(strsplit(DESCRIPTION,' ',fixed=T))),.(DESCRIPTION,CODICE)]

# remove empty words introduced by extra spaces in your DESCRIPTIONS
ORIGINALE_WORDS = ORIGINALE_WORDS[word!='']
REFERENCE_WORDS = REFERENCE_WORDS[word!='']

# merge the tables by word
merged = merge(ORIGINALE_WORDS,REFERENCE_WORDS,by='word',all=F,allow.cartesian=T)

# count matching words for each combination of ORIGINALE DESCRIPTION and REFERENCE DESCRIPTION and CODICE
counts = merged[,.N,.(DESCRIPTION.x,DESCRIPTION.y,CODICE.y)]

# keep only the highest N CODICE.y for each DESCRIPTION.x
topcounts = merged[order(-N)][!duplicated(DESCRIPTION.x)]

# merge the counts back to ORIGINALE
result = merge(ORIGINALE,topcounts,by.x='DESCRIPTION',by.y='DESCRIPTION.x',all.x=T,all.y=F)

Here is result:

                            DESCRIPTION CODICE                           DESCRIPTION.y CODICE.y N
1:     4c flower str jenny jane Chicago     NA     jenny jane flower street 4c Chicago 83776250 5
2:       mr peter 123 rose street 3b LA     NA       peter green 123 rose street 3b LA     aw56 6
3: washington miss sarah 430f name strt     NA sarah brown name street 430f washington  135tg67 4

PS: There are more memory-efficient ways to do this, and this code could cause your machine to crash due to an out-of-memory error or go slowly due to needing virtual memory, but if not, it should be faster than the for loops.

Answer 2

What about:

library(stringdist)
library(dplyr)
library(tidyr) 

data_o <- ORIGINALE %>% mutate(desc_o = DESCRIPTION) %>% select(desc_o)
data_r <- REFERENE %>% mutate(desc_r = DESCRIPTION) %>% select(desc_r)
data <- crossing(data_o,data_r)
data %>% mutate(dist= stringsim(as.character(desc_o),as.character(desc_r))) %>%
         group_by(desc_o) %>% 
         filter(dist==max(dist))

  desc_o                                 desc_r                                   dist
  <chr>                                  <chr>                                   <dbl>
1 " 4c flower str jenny jane Chicago"    jenny jane flower street 4c Chicago     0.486
2 "mr peter 123 rose street 3b LA"       peter green 123 rose street 3b LA       0.758
3 "washington miss sarah 430f name strt" sarah brown name street 430f washington 0.385

Answer 3

The R tm (text mining) library can help here:

library(tm)
library(proxy) # for computing cosine similarity
library(data.table)

ORIGINALE = data.table(DESCRIPTION = c("mr peter 123 rose street 3b LA"," 4c flower str jenny jane Chicago", "washington miss sarah 430f name strt"), CODICE = c(NA, NA, NA))
REFERENCE = data.table(DESCRIPTION = c("sarah brown name street 430f washington", "peter green 123 rose street 3b LA", "jenny jane flower street 4c Chicago"), CODICE = c("135tg67","aw56", "83776250"))

# combine ORIGINALE and REFERENCE into one data.table
both = rbind(ORIGINALE,REFERENCE)

# create "doc_id" and "text" columns (required by tm)
both[,doc_id:=1:.N]
names(both)[1] = 'text'

# convert to tm corpus
corpus = SimpleCorpus(DataframeSource(both))

# convert to a tm document term matrix
dtm = DocumentTermMatrix(corpus)

# convert to a regular matrix
dtm = as.matrix(dtm)

# look at it (t() transpose for readability)
t(dtm)
            Docs
Terms        1 2 3 4 5 6
  123        1 0 0 0 1 0
  peter      1 0 0 0 1 0
  rose       1 0 0 0 1 0
  street     1 0 0 1 1 1
  chicago    0 1 0 0 0 1
  flower     0 1 0 0 0 1
  jane       0 1 0 0 0 1
  jenny      0 1 0 0 0 1
  str        0 1 0 0 0 0
  430f       0 0 1 1 0 0
  miss       0 0 1 0 0 0
  name       0 0 1 1 0 0
  sarah      0 0 1 1 0 0
  strt       0 0 1 0 0 0
  washington 0 0 1 1 0 0
  brown      0 0 0 1 0 0
  green      0 0 0 0 1 0

# compute similarity between each combination of documents 1:3 and documents 4:6
similarity = proxy::dist(dtm[1:3,], dtm[4:6,], method="cosine")

# result:
ORIGINALE               REFERENCE document
 document              4         5         6
        1      0.7958759 0.1055728 0.7763932   <-- difference (smaller = more similar)
        2      1.0000000 1.0000000 0.2000000
        3      0.3333333 1.0000000 1.0000000

# make a table of which REFERENCE document is most similar
most_similar = rbindlist(
  apply(
    similarity,1,function(x){
      data.table(i=which.min(x),distance=min(x))
    }
  )
)

# result:
   i  distance
1: 2 0.1055728
2: 3 0.2000000
3: 1 0.3333333
# rows 1, 2, 3 or rows of ORIGINALE; i: 2 3 1 are rows of REFERENCE

# add the results back to ORIGINALE
ORIGINALE1 = cbind(ORIGINALE,most_similar)
REFERENCE[,i:=1:.N]
ORIGINALE2 = merge(ORIGINALE1,REFERENCE,by='i',all.x=T,all.y=F)

# result:
   i                        DESCRIPTION.x CODICE.x  distance                           DESCRIPTION.y CODICE.y
1: 1 washington miss sarah 430f name strt       NA 0.3333333 sarah brown name street 430f washington  135tg67
2: 2       mr peter 123 rose street 3b LA       NA 0.1055728       peter green 123 rose street 3b LA     aw56
3: 3     4c flower str jenny jane Chicago       NA 0.2000000     jenny jane flower street 4c Chicago 83776250

# now the documents are in a different order than in ORIGINALE2.
# this is caused by merging by i (=REFERENCE document row).
# if order is important, then add these two lines around the merge line:
ORIGINALE1[,ORIGINALE_i:=1:.N]
ORIGINALE2 = merge(...
ORIGINALE2 = ORIGINALE2[order(ORIGINALE_i)]