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Could anyone please help to achieve the following with
gsub in R?
input string: a=5.00,b=120,c=0.0003,d=0.02,e=5.20, f=1200.0,g=850.02 desired output: a=5,b=120,c=0.0003,d=0.02,e=5.2, f=1200, g=850.02
Practically, removing the redundant 0s after the decimal point if they are all just 0s, don't remove if real fractions exist.
I couldn't get this to work using
gsub alone, but we can try splitting your input vector on comma, and then using an
apply function with
x <- "a=5.00,b=120,c=0.0003,d=0.02,e=5.20, f=1200.0,g=850.02" input <- sapply(unlist(strsplit(x, ",")), function(x) gsub("(?<=\\d)\\.$", "", gsub("(\\.[1-9]*)0+$", "\\1", x), perl=TRUE)) input <- paste(input, collapse=",") input  "a=5,b=120,c=0.0003,d=0.02,e=5.2, f=1200,g=850.02"
I actually make two calls to
gsub. The first call strips off all trailing zeroes appearing after a decimal point, should the number have one. And the second call removes stray decimal points, in the case of a number like
5.00, which the first call would leave as
5. and not
5, the latter which we want.
How can I remove non-numeric characters from strings using gsub , I use the gsub function in R to remove unwanted characters in numbers. So I should remove from the strings every character that is not a number, . , and - . I use the gsub function in R to remove unwanted characters in numbers. So I should remove from the strings every character that is not a number, ., and -.My problem is that the regular expression is not removing some non-numeric characters like d, +, and <.
To remove trailing 0s after the decimal, try this:
x = c('5.00', '0.500', '120', '0.0003', '0.02', '5.20', '1200', '850.02') gsub("\\.$" "", gsub("(\\.(|[1-9]+))0+$", "\\1", x)) #  "5" "0.5" "120" "0.0003" "0.02" "5.2" "1200" "850.02"
HT @TimBiegeleisen: I misread input as a vector of strings. For a single-string input, convert to vector of strings, which you can call
gsub on, then collapse output back to a single string:
paste( gsub("\\.$", "", gsub("(\\.(|[1-9]+))0+$", "\\1", unlist(strsplit(x, ", ")))), collapse=", ")
 "a=5, b=0.5, c=120, d=0.0003, e=0.02, f=5.2, g=1200, h=850.02"
Cleaning Up Ruby Strings 13 Times Faster, To work with it, we want to remove the spaces. #gsub is not only slower, but it also requires an extra effort for the reader to The specialized methods are like precision tools—suitable for a specific task in a specific context. According to Dr. Rabach, at-home dermaplaning is “a good way to remove fine vellus hairs (aka peach fuzz) and also remove the thin top layer of dead skin.” Like almost every option on the list
#gsub hashtag on Twitter, "CE COMEBACK !!! #GSUB" - read what others are saying and join the conversation. Removing StopWords from a Character using R gsub to remove unwanted precision http://stackoverflow.com/questions/50332574 … #r #string #gsub gsub() function replaces all matches of a string, if the parameter is a string vector, returns a string vector of the same length and with the same attributes (after possible coercion to character). Elements of string vectors which are not substituted will be returned unchanged (including any declared encoding).
Removing Space From Strings in R Programming, This lesson will discuss the use of removing spaces from a string. For example, here is a string with an extra space at the beginning and the end: The gsub() function is just like sub(), except it replaces all occurrences. In this case, all spaces in the string must be removed to get a more accurate 'human' count. Below is I am looking for a way to tell awk to do high-precision arithmetic in a substitution operation. This involves, reading a field from a file and substituting it with a 1% increment on that value. However, I am losing precision there. Here is a simplified reproduction of the problem:
[PDF] Handling and Processing Strings in R, You want to remove a given character in the names of your variables the regex functions in the base package as well as in the stringr package. To be more precise, print() is a generic function, which means that you should We succesfully replaced the punctuation symbols with empty strings, but now we have extra. How to remove the dollar signs from column in R. One way to do it is with the gsub() function, in conjunction with as.numeric(). gsub() is used to substitute specific text from a string with other text, and as.numeric() can coerce a variable to numeric.
trimws: Remove Leading/Trailing Whitespace, x. a character vector. which. a character string specifying whether to remove both leading and trailing whitespace (default), or only leading ( "left" ) or trailing If the event has field "somefield" == "hello" this filter, on success, would remove the tag foo_hello if it is present. The second example would remove a sad, unwanted tag as well. The second example would remove a sad, unwanted tag as well.
Understanding and Writing your first Text Mining Script with R., We shall remove URLs, emojis,non-English words, punctuations, numbers, These words have to be eliminated so as to render the results more accurate.It is also content_transformer(gsub), pattern="\\W",replace=" ") # remove URLs removeWords, myStopwords) # remove extra whitespace mydata Dear R Users, I am working with gsub for the first time. I am trying to remove some characters from a string. I have hit the problem where the period is the shorthand for 'everything' in the R language when what I want to remove is the actual periods.
- My feeling is that this is something which should be handled by, e.g. casting, your numerical values and then rendering them the way you want.
gsubis probably not the best tool to use here.
- Something like
gsub("(\\.[1-9]+)0+$", "\\1", input_string)?
5with no decimal point.
- Yep, caught and edited as I posted
- @ Juan: this outputs: "a=5.00,b=120,c=0.0003,d=0.02,e=5.20, f=1200.0,g=850.02"
- Nice use of
applyand look behind regex
- You left out the
5.00in the OP's original data. That should become
5but does not with this code.
- This still has the problem that you are operating on a vector, and not the original string input.
- Ah, ok. I misunderstood. I didn't realize it was a single string!
- But I'll upvote you, because your replacement logic looks sound if we only wanted to tweak a single number. I did an
strsplitto get the vector your answer assumes, then joined it back using
- Thanks @TimBiegeleisen Yep, if I'd read more carefully, I would've thrown a