I have a list of phrases, in which I want to replace certain words with a similar word, in case it is misspelled.
library(stringr)
a4 <- "I would like a cheseburger and friees please"
badwords.corpus <- c("cheseburger", "friees")
goodwords.corpus <- c("cheeseburger", "fries")
vect.corpus <- goodwords.corpus
names(vect.corpus) <- badwords.corpus
str_replace_all(a4, vect.corpus)
# [1] "I would like a cheeseburger and fries please"
everything works perfectly, until it finds a similar string, and replaces it with another word
if I have a pattern like the following:
"plea", the correct one is "please", but when I execute it removes it and replaces it with "pleased".
What I am looking for is that if a string is already correct, it is no longer modified, in case it finds a similar pattern.
Perhaps you need to perform progressive replace. e.g. you should have multiple set of badwords and goodwords. First replace with badwords having more letters so that matching pattern is not found and then got go for smaller ones.
From the list provided by you, I would create 2 sets as:
goodwords1<-c( "three", "teasing")
badwords1<- c("thre", "teeasing")
goodwords2<-c("tree", "testing")
badwords2<- c("tre", "tesing")
First replace with 1st set and then with 2nd set. You can create many such sets.
str_replace_all takes regex as the pattern, so you can paste0 word boundaries \\b around each badwords so that a replacement will only be made if the whole word is matched:
library(stringr)
string <- c("tre", "tree", "teeasing", "tesing")
goodwords <- c("tree", "three", "teasing", "testing")
badwords <- c("tre", "thre", "teeasing", "tesing")
# Paste word boundaries around badwords
badwords <- paste0("\\b", badwords, "\\b")
vect.corpus <- goodwords
names(vect.corpus) <- badwords
str_replace_all(string, vect.corpus)
[1] "tree" "tree" "teasing" "testing"
The advantage of this is that you don't have to keep track of which strings are the longer strings.
This is what badwords looks like after pasting:
> badwords
[1] "\\btre\\b" "\\bthre\\b" "\\bteeasing\\b" "\\btesing\\b"
Related
I have this regex to separate letters from numbers (and symbols) of a word: (?<=[a-zA-Z])(?=([[0-9]|[:punct:]])). My test string is: "CALLE15 CRA22".
I want to apply this regex only to the first word of that sentence (the word is defined with spaces). Namely, I want apply that only to "CALLE15".
One solution is split the string (sentence) into words and then apply the regex to the first word, but I want to do all in one regex. Other solution is to use r stringr::str_replace() (or sub()) that replace only the first match, but I need stringr::str_replace_all (or gsub()) for other reasons.
What I need is to insert a space between the two that I do with the replacement function. The outcome I want is "CALLE 15 CRA22" and with the posibility of "CALLE15 CRA 22". I try a lot of positions for the space and nothing, neither the ^ at the beginning.
https://rubular.com/r/7dxsHdOA3avTdX
Thanks for your help!!!!
I am unsure about your problem statement (see my comment above), but the following reproduces your expected output and uses str_replace_all
ss <- "CALLE15 CRA22"
library(stringr)
str_replace_all(ss, "^([A-Za-z]+)(\\d+)(\\s.+)$", "\\1 \\2\\3")
#[1] "CALLE 15 CRA22"
Update
To reproduce the output of the sample string from the comment above
ss <- "CLL.6 N 5-74NORTE"
pat <- c(
"(?<=[A-Za-z])(?![A-Za-z])",
"(?<![A-Za-z])(?=[A-Za-z])",
"(?<=[0-9])(?![0-9])",
"(?<![0-9])(?=[0-9])")
library(stringr)
str_split(ss, sprintf("(%s)", paste(pat, collapse = "|"))) %>%
unlist() %>%
.[nchar(trimws(.)) > 0] %>%
paste(collapse = " ")
#[1] "CLL . 6 N 5 - 74 NORTE"
i would like to get the count times that in a given string a word start with the letter given.
For example, in that phrase: "that pattern is great but pigs likes milk"
if i want to find the number of words starting with "g" there is only 1 "great", but right now i get 2 "great" and "pigs".
this is the code i use:
x <- "that pattern is great but pogintless"
sapply(regmatches(x, gregexpr("g", x)), length)
We need either a space or word boundary to avoid th letter from matching to characters other than the start of the word. In addition, it may be better to use ignore.case = TRUE as some words may begin with uppercase
lengths(regmatches(x, gregexpr("\\bg", x, ignore.case = TRUE)))
The above can be wrapped as a function
fLength <- function(str1, pat){
lengths(regmatches(str1, gregexpr(paste0("\\b", pat), str1, ignore.case = TRUE)))
}
fLength(x, "g")
#[1] 1
You can also do it with stringr library
library(stringr)
str_count(str_split(x," "),"\\bg")
I have a list of phrases, in which I want to replace certain words with a similar word, in case it is misspelled.
How can I search a string, a word that matches and replace it?
The expected result is the following example:
a1<- c(" the classroom is ful ")
a2<- c(" full")
In this case I would be replacing ful for full in a1
Take a look at the hunspell package. As the comments have already suggested, your problem is much more difficult than it seems, unless you already have a dictionary of misspelled words and their correct spelling.
library(hunspell)
a1 <- c(" the classroom is ful ")
bads <- hunspell(a1)
bads
# [[1]]
# [1] "ful"
hunspell_suggest(bads[[1]])
# [[1]]
# [1] "fool" "flu" "fl" "fuel" "furl" "foul" "full" "fun" "fur" "fut" "fol" "fug" "fum"
So even in your example, would you want to replace ful with full, or many of the other options here?
The package does let you use your own dictionary. Let's say you're doing that, or at least you're happy with the first returned suggestion.
library(stringr)
str_replace_all(a1, bads[[1]], hunspell_suggest(bads[[1]])[[1]][1])
# [1] " the classroom is fool "
But, as the other comments and answers have pointed out, you do need to be careful with the word showing up within other words.
a3 <- c(" the thankful classroom is ful ")
str_replace_all(a3,
paste("\\b",
hunspell(a3)[[1]],
"\\b",
collapse = "", sep = ""),
hunspell_suggest(hunspell(a3)[[1]])[[1]][1])
# [1] " the thankful classroom is fool "
Update
Based on your comment, you already have a dictionary, structured as a vector of badwords and another vector of their replacements.
library(stringr)
a4 <- "I would like a cheseburger and friees please"
badwords.corpus <- c("cheseburger", "friees")
goodwords.corpus <- c("cheeseburger", "fries")
vect.corpus <- goodwords.corpus
names(vect.corpus) <- badwords.corpus
str_replace_all(a4, vect.corpus)
# [1] "I would like a cheeseburger and fries please"
Update 2
Addressing your comment, with your new example the issue is back to having words showing up in other words. The solutions is to use \\b. This represents a word boundary. Using pattern "thin" it will match to "thin", "think", "thinking", etc. But if you bracket with \\b it anchors the pattern to a word boundary. \\bthin\\b will only match "thin".
Your example:
a <- c(" thin, thic, thi")
badwords.corpus <- c("thin", "thic", "thi" )
goodwords.corpus <- c("think", "thick", "this")
The solution is to modify badwords.corpus
badwords.corpus <- paste("\\b", badwords.corpus, "\\b", sep = "")
badwords.corpus
# [1] "\\bthin\\b" "\\bthic\\b" "\\bthi\\b"
Then create the vect.corpus as I describe in the previous update, and use in str_replace_all.
vect.corpus <- goodwords.corpus
names(vect.corpus) <- badwords.corpus
str_replace_all(a, vect.corpus)
# [1] " think, thick, this"
I think the function you are looking for is gsub():
gsub (pattern = "ful", replacement = a2, x = a1)
Create a list of the corrections then replace them using gsubfn which is a generalization of gsub that can also take list, function and proto object replacement objects. The regular expression matches a word boundary, one or more word characters and another word boundary. Each time it finds a match it looks up the match in the list names and if found replaces it with the corresponding list value.
library(gsubfn)
L <- list(ful = "full") # can add more words to this list if desired
gsubfn("\\b\\w+\\b", L, a1, perl = TRUE)
## [1] " the classroom is full "
For a kind of ordered replacement, you can try this
a1 <- c("the classroome is ful")
# ordered replacement
badwords.corpus <- c("ful", "classroome")
goodwords.corpus <- c("full", "classroom")
qdap::mgsub(badwords.corpus, goodwords.corpus, a1) # or
stringi::stri_replace_all_fixed(a1, badwords.corpus, goodwords.corpus, vectorize_all = FALSE)
For unordered replacement you can use an approximate string matching (see stringdist::amatch). Here is an example
a1 <- c("the classroome is ful")
a1
[1] "the classroome is ful"
library(stringdist)
goodwords.corpus <- c("full", "classroom")
badwords.corpus <- unlist(strsplit(a1, " ")) # extract words
for (badword in badwords.corpus){
patt <- paste0('\\b', badword, '\\b')
repl <- goodwords.corpus[amatch(badword, goodwords.corpus, maxDist = 1)] # you can change the distance see ?amatch
final.word <- ifelse(is.na(repl), badword, repl)
a1 <- gsub(patt, final.word, a1)
}
a1
[1] "the classroom is full"
I have an atomic vector like:
col_names_to_be_changed <- c("PRODUCTIONDATE", "SPEEDRPM", "PERCENTLOADATCURRENTSPEED", sprintf("SENSOR%02d", 1:18))
I'd like to have _ between words, have them all lower case, except first letters of words (following R Style for dataframes from advanced R). I'd like to have something like this:
new_col_names <- c("Production_Date", "Percent_Load_At_Current_Speed", sprintf("Sensor_%02d", 1:18))
Assume that my words are limited to this list:
list_of_words <- c('production', 'speed', 'percent', 'load', 'at', 'current', 'sensor')
I am thinking of an algorithm that uses gsub, puts _ wherever it finds a word from the above list and then Capitalizes the first letter of each word. Although I can do this manually, I'd like to learn how this can be done more beautifully using gsub. Thanks.
You can take the list of words and paste them with a look-behind ((?<=)). I added the (?=.{2,}) because this will also match the "AT" in "DATE" since "AT" is in the list of words, so whatever is in the list of words will need to be followed by 2 or more characters to be split with an underscore.
The second gsub just does the capitalization
list_of_words <- c('production', 'speed', 'percent', 'load', 'at', 'current', 'sensor')
col_names_to_be_changed <- c("PRODUCTIONDATE", "SPEEDRPM", "PERCENTLOADATCURRENTSPEED", sprintf("SENSOR%02d", 1:18))
(pattern <- sprintf('(?i)(?<=%s)(?=.{2,})', paste(list_of_words, collapse = '|')))
# [1] "(?i)(?<=production|speed|percent|load|at|current|sensor)(?=.{2,})"
(split_words <- gsub(pattern, '_', tolower(col_names_to_be_changed), perl = TRUE))
# [1] "production_date" "speed_rpm" "percent_load_at_current_speed"
# [4] "sensor_01" "sensor_02" "sensor_03"
gsub('(?<=^|_)([a-z])', '\\U\\1', split_words, perl = TRUE)
# [1] "Production_Date" "Speed_Rpm" "Percent_Load_At_Current_Speed"
# [4] "Sensor_01" "Sensor_02" "Sensor_03"
I have names of some 7 countries which is stored somewhere like:
Random <- c('norway', 'india', 'china', 'korea', 'france','japan','iran')
Now, I have to find out using r if a given sentence has these words.
Sometimes the name of a country is hiding in the consecutive letters within a sentence.
for ex:
You all must pay it bac**k, or ea**ch of you will be in trouble.
If this sentence is passed it should return "korea"
I have tried:
grep('You|all|must|pay|it|back|or|each|of|you|will|be|in|trouble',Random, value = TRUE,ignore.case=TRUE,
fixed = FALSE)
it should return korea
but it's not working. Perhaps I should not use Partial Matching, but i dont have much knowledge regarding it.
Any help is appreciated.
You can use the handy stringr library for this. First, remove all the punctuation and spaces from your sentence that we want to match.
> library(stringr)
> txt <- "You all must pay it back, or each of you will be in trouble."
> g <- gsub("[^a-z]", "", tolower(txt))
# [1] "Youallmustpayitbackoreachofyouwillbeintrouble"
Then we can use str_detect to find the matches.
> Random[str_detect(g, Random)]
# [1] "korea"
Basically you're just looking for a sub-string within a sentence, so collapsing the sentence first seems like a good way to go. Alternatively, you could use str_locate with str_sub to find the relevant sub-strings.
> no <- na.omit(str_locate(g, Random))
> str_sub(g, no[,1], no[,2])
# [1] "korea"
Edit Here's one more I came up with
> Random[Vectorize(grepl)(Random, g)]
# [1] "korea"
Using base functions only:
Random <- c('norway', 'india', 'china', 'korea', 'france','japan','iran')
Random2=paste(Random,collapse="|") #creating pattern for match
text="bac**k, or ea**ch of you will be in trouble."
text2=gsub("[[:punct:][:space:]]","",text,perl=T) #removing punctuations and space characters
regmatches(text2,gregexpr(Random2,text2))
[[1]]
[1] "korea"
You could use stringi which is faster for these operations
library(stringi)
Random[stri_detect_regex(gsub("[^A-Za-z]", "", txt), Random)]
#[1] "korea"
#data
Random <- c('norway', 'india', 'china', 'korea', 'france','japan','iran')
txt <- "You all must pay it back, or each of you will be in trouble."
Try:
Random <- c('norway', 'india', 'china', 'korea', 'france','japan','iran')
txt <- "You all must pay it back, or each of you will be in trouble."
tt <- gsub("[[:punct:]]|\\s+", "", txt)
unlist(sapply(Random, function(r) grep(r, tt)))
korea
1