I am using the following code for finding number of occurrences of a word memory in a file and I am getting the wrong result. Can you please help me to know what I am missing?
NOTE1: The question is looking for exact occurrence of word "memory"!
NOTE2: What I have realized they are exactly looking for "memory" and even something like "memory," is not accepted! That was the part which has brought up the confusion I guess. I tried it for word "action" and the correct answer is 7! You can try as well.
#names=scan("hamlet.txt", what=character())
names <- scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character())
Read 28230 items
> length(grep("memory",names))
[1] 9
Here's the file
The problem is really Shakespeare's use of punctuation. There are a lot of apostrophes (') in the text. When the R function scan encounters an apostrophe it assumes it is the start of a quoted string and reads all characters up until the next apostrophe into a single entry of your names array. One of these long entries happens to include two instances of the word "memory" and so reduces the total number of matches by one.
You can fix the problem by telling scan to regard all quotation marks as normal characters and not treat them specially:
names <- scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character(), quote=NULL )
Be careful when using the R implementation of grep. It does not behave in exactly the same way as the usual GNU/Linux program. In particular, the way you have used it here WILL find the number of matching words and not just the total number of matching lines as some people have suggested.
As pointed by #andrew, my previous answer would give wrong results if a word repeats on the same line. Based on other answers/comments, this one seems ok:
names = scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character(), quote=NULL )
idxs = grep("memory", names, ignore.case = TRUE)
length(idxs)
# [1] 10
Related
I'm using the following code to find if the word "assist" is used in a string variable.
string<- c("assist")
`assist <-
(1:nrow(df) %in% c(sapply(string, grep, df$textvariable, fixed = TRUE)))+0`
`sum(assist)`
If I also wanted to check if synonyms such as "help" and "support" are used in the string, how can I update the code? So if either of these synonyms are used, I want to code it as 1. If neither of these words are used, I want to code it as 0. It doesn't matter if all of the words appear in the string or how many times they are used.
I tried changing it to
string<- c("assist", "help", "support")
But it looks like it is searching for strings in which all of these words are used?
I'd appreciate your help!
Thank you
I'm working with the following code:
Y_Columns <- c("Y.1.1")
paste('{"ImportId":"', Y_Columns, '"}', sep = "")
The paste function produces the following output:
"{\"ImportId\":\"Y.1.1\"}"
How do I get the paste function to omit the \? Such that, the output is:
"{"ImportId":"Y.1.1"}"
Thank you for your help.
Note: I did do a search on SO to see if there were any Q's that asked "what is an escape character in R". But I didn't review all the 160 answers, only the first 20.
This is one way of demonstrating what I wrote in my comment:
out <- paste('{"ImportId":"', Y_Columns, '"}', sep = "")
out
#[1] "{\"ImportId\":\"Y.1.1\"}"
?print
print(out,quote=FALSE)
#[1] {"ImportId":"Y.1.1"}
Both R and regex patterns use escape characters to allow special characters to be displayed in print output or input. (And sometimes regex patterns need to have doubled escapes.) R has a few characters that need to be "escaped" in certain situation. You illustrated one such situation: including double-quote character inside a result that will be printed with surrounding double-quotes. If you were intending to include any single quotes inside a character value that was delimited by single quotes at the time of creation, they would have needed to be escaped as well.
out2 <- '\'quoted\''
nchar(out2)
#[1] 8 ... note that neither the surround single-quotes nor the backslashes get counted
> out2
[1] "'quoted'" ... and the default output quote-char is a double-quote.
Here's a good Q&A to review:How to replace '+' using gsub() function in R
It has two answers, both useful: one shows how to double escape a special character and the other shows how to use teh fixed argument to get around that requirement.
And another potentially useful Q&A on the topic of handling Windows paths:
File path issues in R using Windows ("Hex digits in character string" error)
And some further useful reading suggestions: Look at the series of help pages that start with capital letters. (Since I can never remember which one has which nugget of essential information, I tried ?Syntax first and it has a "See Also" list of essential reading: Arithmetic, Comparison, Control, Extract, Logic, NumericConstants, Paren, Quotes, Reserved. and I then realized what I wanted to refer you to was most likely ?Quotes where all the R-specific escape sequence letters should be listed.
I have been mucking around with regex strings and strsplit but can't figure out how to solve my problem.
I have a collection of html documents that will always contain the phrase "people own these". I want to extract the number immediately preceding this phrase. i.e. '732,234 people own these' - I'm hoping to capture the number 732,234 (including the comma, though I don't care if it's removed).
The number and phrase are always surrounded by a . I tried using Xpath but that seemed even harder than a regex expression. Any help or advice is greatly appreciated!
example string: >742,811 people own these<
-> 742,811
Could you please try following.
val <- "742,811 people own these"
gsub(' [a-zA-Z]+',"",val)
Output will be as follows.
[1] "742,811"
Explanation: using gsub(global substitution) function of R here. Putting condition here where it should replace all occurrences of space with small or capital alphabets with NULL for variable val.
Try using str_extract_all from the stringr library:
str_extract_all(data, "\\d{1,3}(?:,\\d{3})*(?:\\.\\d+)?(?= people own these)")
(strap in!)
Hi, I'm running into issues involving Unicode encoding in R.
Basically, I'm importing data sets that contain Unicode (UTF-8) characters, and then running grep() searches to match values. For example, say I have:
bigData <- c("foo","αβγ","bar","αβγγ (abgg)", ...)
smallData <- c("αβγ","foo", ...)
What I'm trying to do is take the entries in smallData and match them to entries in bigData. (The actual sets are matrixes with columns of values, so what I'm trying to do is find the indexes of the matches, so I can tell what row to add the values to.) I've been using
matches <- grepl(smallData[i], bigData, fixed=T)
which usually results in a vector of matches. For i=2, it would return 1, since "foo" is element 1 of bigData. This is peachy and all is well. But RStudio seems to not be dealing with unicode characters properly. When I import the sets and view them, they use the character IDs.
dataset <- read_csv("[file].csv", col_names = FALSE, locale = locale())
Using View(dataset) shows "aß<U+03B3>" instead of "αβγ." The same goes for
dataset[1]
A tibble: 1x1 <chr>
[1] aß<U+03B3>
print(dataset[1])
A tibble: 1x1 <chr>
[1] aß<U+03B3>
However, and this is why I'm stuck rather than just adjusting the encoding:
paste(dataset[1])
[1] "αβγ"
Encoding(toString(dataset[1]))
[1] "UTF-8"
So it appears that R is recognizing in certain contexts that it should display Unicode characters, while in others it just sticks to--ASCII? I'm not entirely sure, but certainly a more limited set.
In any case, regardless of how it displays, what I want to do is be able to get
grep("αβγ", bigData)
[1] 2 4
However, none of the following work:
grep("αβ", bigData) #(Searching the two letters that do appear to convert)
grep("<U+03B3>",bigData,fixed=T) #(Searching the code ID itself)
grep("αβ", toString(bigData)) #(converts the whole thing to one string)
grep("\\β", bigData) #(only mentioning because it matches, bizarrely, to ß)
The only solution I've found is:
grep("\u03B3", bigData)
[1] 2 4
Which is not ideal for a couple reasons, most jarringly that it doesn't look like it's possible to just take every <U+####> and replace it with \u####, since not every Unicode character is converted to the <U+####> format, but none of them can be searched. (i.e., α and ß didn't turn into their unicode keys, but they're also not searchable by themselves. So I'd have to turn them into their keys, then alter their keys to a form that grep() can use, then search.)
That means I can't just regex the keys into a searchable format--and even if I could, I have a lot of entries including characters that'd need to be escaped (e.g., () or ), so having to remove the fixed=T term would be its own headache involving nested escapes.
Anyway...I realize that a significant part of the problem is that my set apparently involves every sort of character under the sun, and it seems I have thoroughly entrapped myself in a net of regular expressions.
Is there any way of forcing a search with (arbitrary) unicode characters? Or do I have to find a way of using regular expressions to escape every ( and α in my data set? (coordinate to that second question: is there a method to convert a unicode character to its key? I can't seem to find anything that does that specific function.)
I looked around both here and elsewhere, I found many similar questions but none which exactly answer mine. I need to clean up naming conventions, specifically replace/remove certain words and phrases from a specific column/variable, not the entire dataset. I am migrating from SPSS to R, I have an example of the code to do this in SPSS below, but I am not sure how to do it in R.
EG:
"Acadia Parish" --> "Acadia" (removes Parish and space before Parish)
"Fifth District" --> "Fifth" (removes District and space before District)
SPSS syntax:
COMPUTE county=REPLACE(county,' Parish','').
There are only a few instances of this issue in the column with 32,000 cases, and what needs replacing/removing varies and the cases can repeat (there are dozens of instances of a phrase containing 'Parish'), meaning it's much faster to code what needs to be removed/replaced, it's not as simple or clean as a regular expression to remove all spaces, all characters after a specific word or character, all special characters, etc. And it must include leading spaces.
I have looked at the replace() gsub() and other similar commands in R, but they all involve creating vectors, or it seems like they do. What I'd like is syntax that looks for characters I specify, which can include leading or trailing spaces, and replaces them with something I specify, which can include nothing at all, and if it does not find the specific characters, the case is unchanged.
Yes, I will end up repeating the same syntax many times, it's probably easier to create a vector but if possible I'd like to get the syntax I described, as there are other similar operations I need to do as well.
Thank you for looking.
> x <- c("Acadia Parish", "Fifth District")
> x2 <- gsub("^(\\w*).*$", "\\1", x)
> x2
[1] "Acadia" "Fifth"
Legend:
^ Start of pattern.
() Group (or token).
\w* One or more occurrences of word character more than 1 times.
.* one or more occurrences of any character except new line \n.
$ end of pattern.
\1 Returns group from regexp
Maybe I'm missing something but I don't see why you can't simply use conditionals in your regex expression, then trim out the annoying white space.
string <- c("Arcadia Parish", "Fifth District")
bad_words <- c("Parish", "District") # Write all the words you want removed here!
bad_regex <- paste(bad_words, collapse = "|")
trimws( sub(bad_regex, "", string) )
# [1] "Arcadia" "Fifth"
dataframename$varname <- gsub(" Parish","", dataframename$varname)