I already have tried to find a solutions on the internet for my problem, and I have the feeling I know all the small pieces but I am unable to put them together. I'm quite knew at programing so pleace be patient :D...
I have a (in reality much larger) text string which look like this:
string <- "Test test [438] test. Test 299, test [82]."
Now I want to replace the numbers in square brackets using a lookup table and get a new string back. There are other numbers in the text but I only want to change those in brackets and need to have them back in brackets.
lookup <- read.table(text = "
Number orderedNbr
1 270 1
2 299 2
3 82 3
4 314 4
5 438 5", header = TRUE)
I have made a pattern to find the square brackets using regular expressions
pattern <- "\\[(\\d+)\\]"
Now I looked all around and tried sub/gsub, lapply, merge, str_replace, but I find myself unable to make it work... I don't know how to tell R! to look what's inside the brackets, to look for that same argument in the lookup table and give out what's standing in the next column.
I hope you can help me, and that it's not a really stupid question. Thx
We can use a regex look around to match only numbers that are inside a square bracket
library(gsubfn)
gsubfn("(?<=\\[)(\\d+)(?=\\])", setNames(as.list(lookup$orderedNbr),
lookup$Number), string, perl = TRUE)
#[1] "Test test [5] test. Test [3]."
Or without regex lookaround by pasteing the square bracket on each column of 'lookup'
gsubfn("(\\[\\d+\\])", setNames(as.list(paste0("[", lookup$orderedNbr,
"]")), paste0("[", lookup$Number, "]")), string)
Read your table of keys and values (a 2 column table) into a data frame. If your source information be a flat text file, then you can easily use read.csv to obtain a data frame. In the example below, I hard code a data frame with just two entries. Then, I iterate over it and make replacements in the input string.
df <- data.frame(keys=c(438, 82), values=c(5, 3))
string <- "Test test [438] test. Test [82]."
for (i in 1:nrow(df)) {
string <- gsub(paste0("(?<=\\[)", df$keys[i], "(?=\\])"), df$values[i], string, perl=TRUE)
}
string
[1] "Test test 5 test. Test 3."
Demo
Note: As #Frank wisely pointed out, my solution would fail if your number markers (e.g. [438]) happen to have replacements which are numbers also appearing as other markers. That is, if replacing a key with a value results in yet another key, there could be problems. If this be a possibility, I would suggest using markers for which this cannot happen. For example, you could remove the brackets after each replacement.
You can use regmatches<- with a pattern containing lookahead/lookbehind:
patt = "(?<=\\[)\\d+(?=\\])"
m = gregexpr(patt, string, perl=TRUE)
v = as.integer(unlist(regmatches(string, m)))
`regmatches<-`(string, m, value = list(lookup$orderedNbr[match(v, lookup$Number)]))
# [1] "Test test [5] test. Test 299, test [3]."
Or to modify the string directly, change the last line to the more readable...
regmatches(string, m) <- list(lookup$orderedNbr[match(v, lookup$Number)])
Related
I have a sentence that may contain keywords. I search for them, if one is true, I want the word before and after the keyword.
cont <- c("could not","would not","does not","will not","do not","were not","was not","did not")
text <- "this failed to increase incomes and production did not improve"
str_extract(text,"([^\\s]+\\s+){1}names(which(sapply(cont,grepl,text)))(\\s+[^\\s]+){1}")
This fails when I dynamically search using the names function but if I input:
str_extract(text,"([^\\s]+\\s+){1}did not(\\s+[^\\s]+){1}")
it correctly returns: production did not improve.
How can I get this to function without directly inputing the keywords?
Final note: I do not completely understand the syntax used to get surrounding objects. Basic r books have not covered this. Can someone explain please?
You could use your cont vector to create a vector of regex strings:
targets <- paste0("([^\\s]+\\s+){1}", cont, "(\\s+[^\\s]+){1}")
Which you can feed into str_extract_all and then unlist:
unlist(stringr::str_extract_all(text, targets))
#> [1] "production did not improve"
If this is something you need to do quite frequently, you could wrap it in a function:
get_surrounding <- function(string, keywords) {
targets <- paste0("([^\\s]+\\s+){1}", keywords, "(\\s+[^\\s]+){1}")
unlist(stringr::str_extract_all(string, targets))
}
With which you can easily run the query on new strings:
new_text <- "The production did not increase because the manager would not allow it."
get_surrounding(new_text, cont)
#> [1] "manager would not allow" "production did not increase"
Perhaps we can try this
> regmatches(text, gregexpr(sprintf("\\w+\\s(%s)\\s\\w+", paste0(cont, collapse = "|")), text))[[1]]
[1] "production did not improve"
Each match of the following regular expression will save the preceding and following words in capture groups 1 and 2, respectively.
\\b([a-z]+) +(?:could|would|does|will|do|were|was|did) +not +([a-z]+)\\b
You will of course have to form this expression programmatically, but that should be straightforward.
Hover the cursor over each element of the expression at this demo to obtain an explanation of its function.
For the string
"she could not believe that production did not improve"
there are two matches. For the first ("she could not believe") "she" and "believe" are saved to capture groups 1 and 2, respectively. For the second ("production did not improve") "production" and "improve" are saved to capture groups 1 and 2, respectively.
I just learnt R and was trying to clean data for analysis using R using string manipulation using the code given below for Amount_USD column of a table. I could not find why changes were not made. Please help.
Code:
csv_file2$Amount_USD <- ifelse(str_sub(csv_file$Amount_USD,1,10) == "\\\xc2\\\xa0",
str_sub(csv_file$Amount_USD,12,-1),csv_file2$Amount_USD)
Result:
\\xc2\\xa010,000,000
\\xc2\\xa016,200,000
\\xc2\\xa019,350,000
Expected Result:
10,000,000
16,200,000
19,350,000
You could use the following code, but maybe there is a more compact way:
vec <- c("\\xc2\\xa010,000,000", "\\xc2\\xa016,200,000", "\\xc2\\xa019,350,000")
gsub("(\\\\x[[:alpha:]]\\d\\\\x[[:alpha:]]0)([d,]*)", "\\2", vec)
[1] "10,000,000" "16,200,000" "19,350,000"
A compact way to extract the numbers is by using str_extract and negative lookahead:
library(stringr)
str_extract(vec, "(?!0)[\\d,]+$")
[1] "10,000,000" "16,200,000" "19,350,000"
How this works:
(?!0): this is negative lookahead to make sure that the next character is not 0
[\\d,]+$: a character class allowing only digits and commas to occur one or more times right up to the string end $
Alternatively:
str_sub(vec, start = 9)
There were a few minor issues with your code.
The main one being two unneeded backslashes in your matching statement. This also leads to a counting error in your first str_sub(), where you should be getting the first 8 characters not 10. Finally, you should be getting the substring from the next character after the text you want to match (i.e. position 9, not 12). The following should work:
csv_file2$Amount_USD <- ifelse(str_sub(csv_file$Amount_USD,1,8) == "\\xc2\\xa0", str_sub(csv_file$Amount_USD,9,-1),csv_file2$Amount_USD)
However, I would have done this with a more compact gsub than provided above. As long as the text at the start to remove is always going to be "\\xc2\\xa0", you can simply replace it with nothing. Note that for gsub you will need to escape all the backslashes, and hence you end up with:
csv_file2$Amount_USD <- gsub("\\\\xc2\\\\xa0", replacement = "", csv_file2$Amount_USD)
Personally, especially if you plan to do any sort of mathematics with this column, I would go the additional step and remove the commas, and then coerce the column to be numeric:
csv_file2$Amount_USD <- as.numeric(gsub("(\\\\xc2\\\\xa0)|,", replacement = "", csv_file2$Amount_USD))
Minimal Reprex
Suppose I have the string as1das2das3D. I want to extract everything from the letter a to the letter D. There are three different substrings that match this - I want the shortest / right-most match, i.e. as3D.
One solution I know to make this work is stringr::str_extract("as1das2das3D", "a[^a]+D")
Real Example
Unfortunately, I can't get this to work on my real data. In my real data I have string with (potentially) two URLs and I'm trying to extract the one that's immediately followed by rel=\"next\". So, in the below example string, I'd like to extract the URL https://abc.myshopify.com/ZifQ.
foo <- "<https://abc.myshopify.com/YifQ>; rel=\"previous\", <https://abc.myshopify.com/ZifQ>; rel=\"next\""
# what I've tried
stringr::str_extract(foo, '(?<=\\<)https://.*(?=\\>; rel\\="next)') # wrong output
stringr::str_extract(foo, '(?<=\\<)https://(?!https)+(?=\\>; rel\\="next)') # error
You could do:
stringr::str_extract(foo,"https:[^;]+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
or even
stringr::str_extract(foo,"https(?:(?!https).)+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
Would this be an option?
Splitting string on ; or , comparing it with target string and take url from its previous index.
urls <- strsplit(foo, ";\\s+|,\\s+")[[1]]
urls[which(urls == "rel=\"next\"") - 1]
#[1] "<https://abc.myshopify.com/ZifQ>"
Here may be an option.
gsub(".+\\, <(.+)>; rel=\"next\"", "\\1", foo, perl = T)
#[1] "https://abc.myshopify.com/ZifQ"
I am interested to assign names to list elements. To do so I execute the following code:
file_names <- gsub("\\..*", "", doc_csv_names)
print(file_names)
"201409" "201412" "201504" "201507" "201510" "201511" "201604" "201707"
names(docs_data) <- file_names
In this case the name of the list element appears with ``.
docs_data$`201409`
However, in this case the name of the list element appears in the following way:
names(docs_data) <- paste("name", 1:8, sep = "")
docs_data$name1
How can I convert the gsub() result to receive the latter naming pattern without quotes?
gsub() and paste () seem to produce the same class () object. What is the difference?
Both gsub and paste return character objects. They are different because they are completely different functions, which you seem to know based on their usage (gsub replaces instances of your pattern with a desired output in a string of characters, while paste just... pastes).
As for why you get the quotations, that has nothing to do with gsub and everything to do with the fact that you are naming variables/columns with numbers. Indeed, try
names(docs_data) <- paste(1:8)
and you'll realize you have the same problem when invoking the naming pattern. It basically has to do with the fact that R doesn't want to be confused about whether a number is really a number or a variable because that would be chaos (how can 1 refer to a variable and also the number 1?), so what it does in such cases is change a number 1 into the character "1", which can be given names. For example, note that
> 1 <- 3
Error in 1 <- 3 : invalid (do_set) left-hand side to assignment
> "1" <- 3 #no problem!
So R is basically correcting that for you! This is not a problem when you name something using characters. Finally, an easy fix: just add a character in front of the numbers of your naming pattern, and you'll be able to invoke them without the quotations. For example:
file_names <- paste("file_",gsub("\\..*", "", doc_csv_names),sep="")
Should do the trick (or just change the "file_" into whatever you want as long as it's not empty, cause then you just have numbers left and the same problem)!
I have millions of Keywords in a column labeled Keyword.text. Each factor or Keyword can contains multiple words (or shall we say token). Here is an example with 4 keywords
Keyword.text
The quick brown fox the
.8 .crazy lazy dog
dog
jumps over+the 9
I'd like to count the number of tokens in each Keyword, so as to obtain:
Keyword.length
5
4
1
4
I installed the Tau package but I haven't gotten very far...
textcnt(Mydf$Keyword.text, split = "[[:space:][:punct:]]+", method = "string", n = 1L)
returns an error I don't understand. Maybe it's due to having factors; it worked fine when practicing with a string.
I know how to do it in excel, but it doesn't work for the last line. If A2 has the keywords then: =LEN(TRIM(A2))-LEN(SUBSTITUTE(A2," ",""))+1 would do
Edit : For a dataframe and the total number of keywords, just use strsplit. There's no need to use strcnt if you're not interested in the counts per keyword. That's where I got you wrong :
tt <- data.frame(
a=rnorm(3),
b=rnorm(3),
c=c("the quick fox lazy","rbrown+fr even","what what goes & around"),
stringsAsFactors=F
)
sapply(tt$c, function(n){
length(strsplit(n, split = "[[:space:][:punct:]]+")[[1]])
})
To read the data, take also a look at ?readLines and/or ?scan. This preserves the string format and allows you to process the file line by line (or row per row). If you use a file connection, you can even load the file in parts, which helps you when you hit memory limits.
A simple example using readLines :
con <- textConnection("
The lazy fog+fog fog
never ended for fog jumping over the
fog whatever . $ plus.
")
# You use con <- file("myfile.txt")
Text <- readLines(con)
sapply(Text,textcnt, split = "[[:space:][:punct:]]+", method = "string", n = 1L)
On a sidenote, using the option Dirk mentioned (stringsAsFactors=F) won't slow down performance compared to the usual read.table command. In contrary actually. You should use the sapply as mentioned above, but replace Text with as.character(Mydf$Keyword.text) (or use the stringsAsFactors=F option and drop the as.character().
Please show the error.
Also try:
require(tau)
textcnt(as character(Mydf$Keyword.txt), split, ....)
... to force character mode.
Or load your data with stringsAsFactors=FALSE -- the same question has come up here before.
What about a nice little function that let us also decide which kind of words we would like to count and which works on whole vectors as well?
require(stringr)
nwords <- function(string, pseudo=F){
ifelse( pseudo,
pattern <- "\\S+",
pattern <- "[[:alpha:]]+"
)
str_count(string, pattern)
}
nwords("one, two three 4,,,, 5 6")
# 3
nwords("one, two three 4,,,, 5 6", pseudo=T)
# 6