I have a large text that contains expressions such as: "aaaahahahahaha that was a good joke". after processing, I want the "aaaaahahahaha" to disappear, or at least, change it to simply "ha".
At the moment, I am using this:
gsub('(.+?)\\1', '', str)
This works when the string with the pattern is at the beginning of the sentence, but not where is located anywhere else. So:
str <- "aaaahahahahaha that was a good joke"
gsub('(.+?)\\1', '', str)
#[1] "ha that was a good joke"`
But
str <- "that was aaaahahahahaha a good joke"
gsub('(.+?)\\1', '', str)
#[1] "that was aaaahahahahaha a good joke"
This question might relate to this: find repeated pattern in python, but I can't find the equivalence in R.
I am assuming is very simple and perhaps I am missing something trivial, but since regular expressions are not my strength and I have already tried a bunch of things that have not worked, I was wondering if someone could help me. The question is: How to find, and substitute, repeated patterns in a string of characters in R?
Thanks in advance for your time.
\b(\S+?)\1\S*\b
Use this.See demo.
https://regex101.com/r/sJ9gM7/46
For r use \\b(\\S+?)\\1\\S*\\b with perl=TRUE option.
Related
I need split sentence to words removing redundant characters.
I prepared regexp for that:
val wordCharacters = """[^A-z'\d]""".r
right now I have rule which can be used to handle task in next way:
wordCharacters.split(words)
.filterNot(_.isEmpty)
where words any sentence I need to parse.
But issue is that in case I try to handle "car: carpet, as,,, java: javascript!!&#$%^&" I get one more word ^. Trying to change my regex and without ^ I'm getting much more issues for different cases...
Is any ideas how to solve it?
P.S.
If somebody want to play with it try link or code below please:
val wordCharacters = """[^A-z'\d]""".r
val stringToInt =
wordCharacters.split("car: carpet, as,,, java: javascript!!&#$%^&")
.filterNot(_.isEmpty)
.toList
println(stringToInt)
Expected result is:
List(car, carpet, as, java, javascript)
The part A-z is not exactly what you want. Probably you assume that lower a comes immediately after upper Z, but there are some other characters in between, and one of them is ^.
So, correcting the regex as
"""[^A-Za-z'\d]""".r
would fix the issue.
Have a look at the order of characters:
https://en.wikipedia.org/wiki/List_of_Unicode_characters
I'd be tempted to start with \W and expand from there.
"\\W+".r.split("car: carpet, as,,, java: javascript!!&#$%^&")
//res0: Array[String] = Array(car, carpet, as, java, javascript)
I have text that looks like this:-
"I am an engineer. I am skilled in ASP.NET. I also know Node.js.But I don't have much experience. "
Here, "ASP.NET" and "Node.js" are to be treated as words. Also, there is no space before "But I...", but it should be treated as a separate sentence.
The expected output is:
["I am an engineer"," I am skilled in ASP.NET","I also know Node.js","But I don't have much experience"]
Is there a way of doing this?
For your current input you may use the following approach with re.split() function and specific regex pattern:
import re
s = "I am an engineer. I am skilled in ASP.NET. I also know Node.js.But I don't have much experience. "
result = re.split(r'\.(?=\s?[A-Z][^.]*? )', s)
print(result)
The output:
['I am an engineer', ' I am skilled in ASP.NET', ' I also know Node.js', "But I don't have much experience. "]
(?=\s?[A-Z][^.]*? ) - lookahead positive assertion, ensures that sentence delimiter . is followed by word from next sentence
I have an array of strings (n=1000) in this format:
strings<-c("GSM1264936_2202_4866_28368_150cGy-GCSF6-m3_Mouse430A+2.CEL.gz",
"GSM1264937_2202_4866_28369_150cGy-GCSF6-m4_Mouse430A+2.CEL.gz",
"GSM1264938_2202_4866_28370_150cGy-GCSF6-m5_Mouse430A+2.CEL.gz")
I'm wondering what may be a easy way to get this:
strings2<-c(2201_4866_28368_150cGy-GCSF6-m3_Mouse430A+2.CEL,
2202_4866_28369_150cGy-GCSF6-m4_Mouse430A+2.CEL,
2203_4866_28370_150cGy-GCSF6-m5_Mouse430A+2.CEL)
which means to trim off "GSM1234567" from the front and ".gz" from the end.
Just a gsub solution that matches strings that starts ^ with digits and alphabetical symbols, zero or more times *, until a _ is encountered and (more precisely "or") pieces or strings that have .gz at the end $.
gsub("^([[:alnum:]]*_)|(\\.gz)$", "", strings)
[1] "2202_4866_28368_150cGy-GCSF6-m3_Mouse430A+2.CEL"
[2] "2202_4866_28369_150cGy-GCSF6-m4_Mouse430A+2.CEL"
[3] "2202_4866_28370_150cGy-GCSF6-m5_Mouse430A+2.CEL"
Edit
I forget to escape the second point.
strings <- c("GSM1264936_2202_4866_28368_150cGy-GCSF6-m3_Mouse430A+2.CEL.gz", "GSM1264937_2202_4866_28369_150cGy-GCSF6-m4_Mouse430A+2.CEL.gz", "GSM1264938_2202_4866_28370_150cGy-GCSF6-m5_Mouse430A+2.CEL.gz")
strings2 <- lapply(strings, function (x) substr(x, 12, 58))
You can do this using sub:
sub('[^_]+_(.*)\\.gz', '\\1', strings)
# [1] "2202_4866_28368_150cGy-GCSF6-m3_Mouse430A+2.CEL"
# [2] "2202_4866_28369_150cGy-GCSF6-m4_Mouse430A+2.CEL"
# [3] "2202_4866_28370_150cGy-GCSF6-m5_Mouse430A+2.CEL"
Try:
gsub('^[^_]+_|\\.[^.]*$','',strings)
I strongly suggest doing this in two steps. The other solutions work but are completely unreadable: they don’t express the intent of your code. Here it is, clearly expressed:
trimmed_prefix = sub('^GSM\\d+_', '', strings)
strings2 = sub('\\.gz$', '', trimmed_prefix)
But admittedly this can be expressed in one step, and wouldn’t look too badly, as follows:
strings2 = sub('^GSM\\d+_(.*)\\.gz$', '\\1', strings)
In general, think carefully about the patterns you actually want to match: your question says to match the prefix “GSM1234567” but your example contradicts that. I’d generally choose a pattern that’s as specific as possible to avoid accidentally matching faulty input.
Suppose I have this test string:
test.string <- c("This is just a <test> string. I'm trying to see, if a FN will remove certain things like </HTML tags>, periods; but not the one in ASP.net, for example.")
I want to:
Remove anything contained within an html tag
Remove certain punctuation (,:;)
Period that end a sentence.
So the above should be:
c("This is just a string I'm trying to see if a FN will remove certain things like periods but not the one in ASP.net for example")
For #1, I've tried the following:
gsub("<.*?>", "", x, perl = FALSE)
And that seems to work OK.
For #2, I think it's simply:
gsub("[:#$%&*:,;^():]", "", x, perl = FALSE)
Which works.
For #3, I tried:
gsub("+[:alpha:]?[.]+[:space:]", "", test.string, perl = FALSE)
But that didn't work...
Any ideas on where I went wrong? I totally suck at RegExp, so any help would be much appreciated!!
Based on your provided input and rules for what you want removed, the following should work.
gsub('\\s*<.*?>|[:;,]|(?<=[a-zA-Z])\\.(?=\\s|$)', '', test.string, perl=T)
See Working Demo
Try this:
test.string <- "There is a natural aristocracy among men. The grounds of this are virtue and talents. "
gsub("\\.\\s*", "", gsub("([a-zA-Z0-9]). ([A-Z])", "\\1 \\2", test.string))
# "There is a natural aristocracy among men The grounds of this are virtue and talents
I want to use a regular expression that would do the following thing ( i extracted the part where i'm in trouble in order to simplify ):
any character for 1 to 5 first characters, then an "underscore", then some digits, then an "underscore", then some digits or dot.
With a restriction on "underscore" it should give something like that:
^([^_]{1,5})_([\\d]{2,3})_([\\d\\.]*)$
But i want to allow the "_" in the 1-5 first characters in case it still match the end of the regular expression, for example if i had somethink like:
to_to_123_12.56
I think this is linked to an eager problem in the regex engine, nevertheless, i tried to do some lazy stuff like explained here but without sucess.
Any idea ?
I used the following regex and it appeared to work fine for your task. I've simply replaced your initial [^_] with ..
^.{1,5}_\d{2,3}_[\d\.]*$
It's probably best to replace your final * with + too, unless you allow nothing after the final '_'. And note your final part allows multiple '.' (I don't know if that's what you want or not).
For the record, here's a quick Python script I used to verify the regex:
import re
strs = [ "a_12_1",
"abc_12_134",
"abcd_123_1.",
"abcde_12_1",
"a_123_123.456.7890.",
"a_12_1",
"ab_de_12_1",
]
myre = r"^.{1,5}_\d{2,3}_[\d\.]+$"
for str in strs:
m = re.match(myre, str)
if m:
print "Yes:",
if m.group(0) == str:
print "ALL",
else:
print "No:",
print str
Output is:
Yes: ALL a_12_1
Yes: ALL abc_12_134
Yes: ALL abcd_134_1.
Yes: ALL abcde_12_1
Yes: ALL a_123_123.456.7890.
Yes: ALL a_12_1
Yes: ALL ab_de_12_1
^(.{1,5})_(\d{2,3})_([\d.]*)$
works for your example. The result doesn't change whether you use a lazy quantifier or not.
While answering the comment ( writing the lazy expression ), i saw that i did a mistake... if i simply use the folowing classical regex, it works:
^(.{1,5})_([\\d]{2,3})_([\\d\\.]*)$
Thank you.