NLP - identifying and replacing words (synonyms) in R - r

Question

I have problem with code in R.
I have a data-set(questions) with 4 columns and over 600k observation, of which one column is named 'V3'.
This column has questions like 'what is the day?'.
I have second data-set(voc) with 2 columns, of which one column name 'word' and other column name 'synonyms'. If In my first data-set (questions )exists word from second data-set(voc) from column 'synonyms' then I want to replace it word from 'word' column.
questions = cbind(V3=c("What is the day today?","Tom has brown eyes"))
questions <- data.frame(questions)
V3
1 what is the day today?
2 Tom has brown eyes
voc = cbind(word=c("weather", "a","blue"),synonyms=c("day", "the", "brown"))
voc <- data.frame(voc)
word synonyms
1 weather day
2 a the
3 blue brown
Desired output
V3 V5
1 what is the day today? what is a weather today?
2 Tom has brown eyes Tom has blue eyes
I wrote simple code but it doesn't work.
for (k in 1:nrow(question))
{
for (i in 1:nrow(voc))
{
question$V5<- gsub(do.call(rbind,strsplit(question$V3[k]," "))[which (do.call(rbind,strsplit(question$V3[k]," "))== voc[i,2])], voc[i,1], question$V3)
}
}
Maybe someone will try to help me? :)
I wrote second code, but it doesn't work too..
for( i in 1:nrow(questions))
{
for( j in 1:nrow(voc))
{
if (grepl(voc[j,k],do.call(rbind,strsplit(questions[i,]," "))) == TRUE)
{
new=matrix(gsub(do.call(rbind,strsplit(questions[i,]," "))[which(do.call(rbind,strsplit(questions[i,]," "))== voc[j,2])], voc[j,1], questions[i,]))
questions[i,]=new
}
}
questions = cbind(questions,c(new))
}

Answer 1

First, it is important that you use the stringsAsFactors = FALSE option, either at the program level, or during your data import. This is because R defaults to making strings into factors unless you otherwise specify. Factors are useful in modeling, but you want to do analysis of the text itself, and so you should be sure that your text is not coerced to factors.
The way I approached this was to write a function that would "explode" each string into a vector, and then uses match to replace the words. The vector gets reassembled into a string again.
I'm not sure how performant this will be given your 600K records. You might look into some of the R packages that handle strings, like stringr or stringi, since they will probably have functions that do some of this. match tends to be okay on speed, but %in% can be a real beast depending on the length of the string and other factors.
# Start with options to make sure strings are represented correctly
# The rest is your original code (mildly tidied to my own standard)
options(stringsAsFactors = FALSE)
questions <- cbind(V3 = c("What is the day today?","Tom has brown eyes"))
questions <- data.frame(questions)
voc <- cbind(word = c("weather","a","blue"),
synonyms = c("day","the","brown"))
voc <- data.frame(voc)
# This function takes:
# - an input string
# - a vector of words to replace
# - a vector of the words to use as replacements
# It returns a list of the original input and the changed version
uFunc_FindAndReplace <- function(input_string,words_to_repl,repl_words) {
# Start by breaking the input string into a vector
# Note that we use [[1]] to get first list element of strsplit output
# Obviously this relies on breaking sentences by spacing
orig_words <- strsplit(x = input_string,split = " ")[[1]]
# If we find at least one of the words to replace in the original words, proceed
if(sum(orig_words %in% words_to_repl) > 0) {
# The right side selects the elements of orig_words that match words to be replaced
# The left side uses match to find the numeric index of those replacements within the words_to_repl vector
# This numeric vector is used to select the values from repl_words
# These then replace the values in orig_words
orig_words[orig_words %in% words_to_repl] <- repl_words[match(x = orig_words,table = words_to_repl,nomatch = 0)]
# We rebuild the sentence again, and return a list with original and new version
new_sent <- paste(orig_words,collapse = " ")
return(list(original = input_string,new = new_sent))
} else {
# Otherwise we return the original version since no changes are needed
return(list(original = input_string,new = input_string))
}
}
# Using do.call and rbind.data.frame, we can collapse the output of a lapply()
do.call(what = rbind.data.frame,
args = lapply(X = questions$V3,
FUN = uFunc_FindAndReplace,
words_to_repl = voc$synonyms,
repl_words = voc$word))
>
original new
1 What is the day today? What is a weather today?
2 Tom has brown eyes Tom has blue eyes