I have a problem where I´m struggling to find a solution or an approach to solve it.
I have some model sentences, e.g.
model_sentences = data.frame("model_id" = c("model_id_1", "model_id_2"), "model_text" = c("Company x had 3000 employees in 2016.",
"Google makes 300 dollar in revenue in 2018."))
and some texts
data = data.frame("id" = c("id1", "id2"), "text" = c("Company y is expected to employ 2000 employees in 2020. This is an increase of 10%. Some stupid sentences.",
"Amazon´s revenue is 400 dollar in 2020. That is twice as much as last year."))
and I would like to extract sentences from those texts which are similar to the model sentences.
Something like this would be my desired solution
result = data.frame("id" = c("id1", "id2"), "model_id" = c("model_id_1", "model_id_2"), "sentence_from_data" = c("Company y is expected to employ 2000 employees in 2020.", "Amazon´s revenue is 400 dollar in 2020."), "score" = c(0.5, 0.4))
Maybe it is possible to find kind of a 'similarity_score'.
I use this function to split texts by sentence:
split_by_sentence <- function (text) {
result <-unlist(strsplit(text, "(?<=[[:alnum:]]{4}[?!.])\\s+", perl=TRUE))
result <- stri_trim_both(result)
result <- result [nchar (result) > 0]
if (length (result) == 0)
result <- ""
return (result)
}
But I have no idea how to compare each sentence to a model sentence.
I'm glad for any suggestions.
Check out this package stringdist
Example:
library(stringdist)
mysent = "This is a sentence"
apply(model_sentences, 1, function(row) {
stringdist(row['model_text'], mysent, method="jaccard")
})
It will return jaccard distance from mysent to model_text variable. The smaller the value is, the sentences are more similar in terms of given distance measure.
Related
I have a table that is shaped like this called df (the actual table is 16,263 rows):
title date brand
big farm house 2022-01-01 A
ranch modern 2022-01-01 A
town house 2022-01-01 C
Then I have a table like this called match_list (the actual list is 94,000 rows):
words_for_match
farm
town
clown
beach
city
pink
And I'm trying to filter the first table to just be rows where the title contains a word in the words_for_match list. So I do this:
match_list <- match_list$words_for_match
match_list <- paste(match_list, collapse = "|")
match_list <- sprintf("\\b(%s)\\b", match_list)
df %>%
filter(grepl(match_list, title))
But then I get the following error:
Problem while computing `..1 = grepl(match_list, subject)`.
Caused by error in `grepl()`:
! invalid regular expression, reason 'Out of memory'
If I filter the table with 94,000 rows to just 1,000 then it runs, so it appears to just be a memory issue. So I'm wondering if there's a less memory-intensive way to do this or if this is an example of needing to look beyond my computer for computation. Advice on either pathway (or other options) is welcome. Thanks!
You could keep titles sequentially, let's say you have 10 titles that match 'farm' you do not need to evaluate those titles with other words.
Here a simple implementation :
titles <- c("big farm house", "ranch modern", "town house")
words_for_match <- c("farm", "town", "clown", "beach", "city", "pink")
titles.to.keep <- c()
for(w in words_for_match)
{
w <- sprintf("\\b(%s)\\b", w)
is.match <- grepl(w, titles)
titles.to.keep <- c(titles.to.keep, titles[is.match])
titles <- titles[!is.match]
print(paste(length(titles), "remaining titles"))
}
titles.to.keep
If you have a prior on the frequency of words on match_list, it's better to start with the most frequent ones.
UPDATE
You can also make a mix with your previous strategy to make it faster :
gr.size <- 20
gr.words <- split(words_for_match, ceiling(seq_along(words_for_match) / gr.size))
gr.words <- sapply(gr.words, function(words)
{
words <- paste(words, collapse = "|")
sprintf("\\b(%s)\\b", words)
})
and then iterate on gr.words and not on words_for_match in the first code chunk.
I have a dataframe with a column that consists of politician names which are extracted of thousands of news articles. Each row is a specific article. I want to count which politicians are mentioned the most, but count each name only one time per article (row).
The entities recognition algorithm returned these results. Now I have to convert the names in a standard form to be able to summarise and compare them.
Because there are maximal 20 people I am interested in, I knew the names and thought despite the effort, manually coding the patterns for each name might be the fastest way (I am happy for other ideas).
#example-data
persons <- c("Merkel,Angela Merkel,Trump,Ursula,Merkels", "Ursula von,Trumps,Donald Trump,Leyen")
df <- data.frame(persons)
#change pattern
df <- df %>%
mutate(
persons= paste(" ",str_replace_all(df$persons,",", " , "), sep = "")
)
#example of exctracting the names.. and so on, you get the idea
str_replace_all(df$persons, c(" Trump(s)?" = "Donald Trump", ", Trump(s)?" = ", Donald Trump", "Donald Trumps" = "Donald Trump",
" Merkel(s)?") = "Angela Merkel")
My desired output is to have for each row just the full names. In the end I would remove the duplicated names per row and then I could count the dataset like desired.
The data would should look like this in the end:
persons <- c("Angela Merkel,Angela Merkel,Donald Trump,Ursula von der Leyen,Angela Merkel", "Ursula von der Leyen,Donald Trump,Donald Trump,Ursula von der Leyen")
I have especially a hard time with patterns for names which consists of more than two parts like Ursula von der Leyen. What would the best way to do convert the names and how would the pattern for replacement look like?
Edit
I wrote now a function, witch takes care that there is only one instance of a name in my dataframe for each row. Not really elegant and nice code but its working.
clean_name <- function(x) {
b <- unlist(strsplit(x, '[,]')) %>%
str_squish(.)
c <- b[!duplicated(b)]
#Lists mit forename and surname
ganzer_name <- vector()
nachname<-vector()
for (person in c){
if(any(str_count(person," ") == 0)){
nachname <- append(nachname,person)
} else{
ganzer_name <- append(ganzer_name,person)
}
}
#chckes if therese constructions like s wie Angela Merkels
ganzer_name <- ganzer_name%>%
str_sort() %>% str_replace(.,"\\+","")
i <-0
aussortieren <- c("","")
while(i < (length(ganzer_name)-1) ){
i <- i+1
if(str_detect(ganzer_name[i+1],
paste0(ganzer_name[i],"*")
)){
aussortieren <- append(aussortieren, ganzer_name[i+1] )
}else{ }
}
ganzer_name <- ganzer_name[!ganzer_name %in% aussortieren]
#check if surname is already in a full name
for( person in nachname ){
#check construction with s like Merkels
if(any(str_detect(ganzer_name, paste0(
"\\Q",
str_sub(person,end = nchar(person)-1),
"\\E" )
)
)
) {
} else {
ganzer_name <- append(ganzer_name,person)
}
}
return(paste(ganzer_name, collapse=","))
}
Instead of turning various combination of names into standard format, removing duplicates and then counting here is a different approach.
We can use grepl for pattern matching and count how many times a politician occurs in different news articles.
name <- c('Trump', 'Merkel')
sapply(name, function(x) sum(grepl(x, df$persons)))
# Trump Merkel
# 2 1
Use ignore.case = TRUE in grepl if you want to make the comparison case insensitive.
Given the below two vectors is there a way to produce the desired data frame? This represents a real world situation which I have to data frames the first contains a col with database values (keys) and the second contains a col of 1000+ rows each a file name (potentials) which I need to match. The problem is there can be multiple files (potentials) matched to any given key. I have worked with grep, merge, inner join etc. but was unable to incorporate them into one solution. Any advise is appreciated!
potentials <- c("tigerINTHENIGHT",
"tigerWALKINGALONE",
"bearOHMY",
"bearWITHME",
"rat",
"imatchnothing")
keys <- c("tiger",
"bear",
"rat")
desired <- data.frame(keys, c("tigerINTHENIGHT, tigerWALKINGALONE", "bearOHMY, bearWITHME", "rat"))
names(desired) <- c("key", "matches")
Psudo code for what I think of as the solution:
#new column which is comma separated potentials
# x being the substring length i.e. x = 4 means true if first 4 letters match
function createNewColumn(keys, potentials, x){
str result = na
foreach(key in keys){
if(substring(key, 0, x) == any(substring(potentals, 0 ,x))){ //search entire potential vector
result += potential that matched + ', '
}
}
return new column with result as the value on the current row
}
We can write a small functions to extract matches and then loop over the keys:
return_matches <- function(keys, potentials, fixed = TRUE) {
vapply(keys, function(k) {
paste(grep(k, potentials, value = TRUE, fixed = fixed), collapse = ", ")
}, FUN.VALUE = character(1))
}
vapply is just a typesafe version of sapply meaning it will never return anything but a character vector. When you set fixed = TRUE the function will run a lot faster but does not recognise regular expressions anymore. Then we can easily make the desired data.frame:
df <- data.frame(
key = keys,
matches = return_matches(keys, potentials),
stringsAsFactors = FALSE
)
df
#> key matches
#> tiger tiger tigerINTHENIGHT, tigerWALKINGALONE
#> bear bear bearOHMY, bearWITHME
#> rat rat rat
The reason for putting the loop in a function instead of running it directly is just to make the code look cleaner.
You can interate using grep
> Match <- sapply(keys, function(item) {
paste0(grep(item, potentials, value = TRUE), collapse = ", ")
} )
> data.frame(keys, Match, row.names = NULL)
keys Match
1 tiger tigerINTHENIGHT, tigerWALKINGALONE
2 bear bearOHMY, bearWITHME
3 rat rat
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))
}
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$word))
>
original new
1 What is the day today? What is a weather today?
2 Tom has brown eyes Tom has blue eyes
New to R ... struggling to produce results on 10,000 lines; Data model actually has about 1M lines. Is there a better option than a Loop? Read about vectorization and attempted tapply with no success.
Data set has a column of free form text and a category associated to the text. I need to parse the text into distinct words to then perform statistics on the frequency of words being able to predict the category with a certain degree of accuracy. I read in the data via read.table and create a data.frame called data.
Function attempts to parse Text, and count occurrences of each word:
data <- data.frame(category = c("cat1","cat2","cat3", "cat4"),
text = c("The quick brown fox",
"Jumps over the fence",
"The quick car hit a fence",
"Jumps brown"))
parsefunc <- function(data){
finalframe <- data.frame()
for (i in 1:nrow(data)){
description <- strsplit(as.character(data[i,2]), " ")[[1]]
category <- rep(data[i,1], length(description))
worddataframe <- data.frame(description, category)
finalframe <- rbind(finalframe, worddataframe)
}
m1<- ddply(finalframe, c("description","category"), nrow)
m2<- ddply(m1, 'description', transform, totalcount = sum(nrow), percenttotal = nrow/sum(nrow))
m3 <- m2[(m2$totalcount>10) & (m2$percenttotal>0.8), ]
m3
}
This will get your finalframe and do something close to your m1,2, and 3 part. You'll have to edit it to do exactly what you want. I used a longer data set of 40k rows to make sure it performs alright:
# long data set
data <- data.frame(Category = rep(paste0('cat',1:4),10000),
Text = rep(c('The quick brown fox','Jumps over the fence','The quick car hit a fence','Jumps brown cars'),10000),stringsAsFactors = F)
# split into words
wordbag <- strsplit(data$Text,split = ' ')
# find appropriate category for each word
categoryvar <- rep(data$Category,lapply(wordbag,length))
# stick them in a data frame and aggregate
newdf <- data.frame(category = categoryvar,word = tolower(unlist(wordbag)))
agg <- aggregate(list(wordcount = rep(1,nrow(newdf))),list(category = newdf$category,word =newdf$word),sum)
# find total count in entire data set and put in data set
wordagg <- aggregate(list(totalwordcount = rep(1,nrow(newdf))),list(word =newdf$word),sum)
agg <- merge(x = agg,y = wordagg,by = 'word')
# find percentages and do whatever else you need
agg$percentageofword <- agg$wordcount/agg$totalwordcount