Used to run R with numbers and matrix, when it comes to play with strings and characters I am lost. I want to analyze some data where the time is read into R as follow:
>my.time.char[1]
[1] "\"2011-10-05 15:55:00\""
I want to end up with a string containing only:
"2011-10-05 15:55:00"
Using the function sub() (that i barely understand...), I got the following result:
> sub("(\")","",my.time.char[1])
[1] "2011-10-05 15:55:00\""
This is closer to the format i am looking for, but I still need to get rid of the two last characters (\").
The second line from ?sub explains:
sub and gsub perform replacement of the first and all matches respectively.
which should tell you to use gsub instead.
Related
There is a vector of strings that looks like the following (text with two or more substrings in quotation marks):
vec <- 'ab"cd"efghi"j"kl"m"'
The text within the first pair of quotation marks (cd) contains a useful identifier (cd is the desired output). I have been studying how to use regular expressions but I haven't learned how to find the first and second occurrences of something like quotation marks.
Here's how I have been getting cd:
tmp <- strsplit(vec,split="")[[1]]
paste(tmp[(which(tmp=='\"')[1]+1):(which(tmp=='\"')[2]-1)],collapse="")
"cd"
My question is, is there another way to find "cd" using regular expressions? in order to learn more how to use them. I prefer base R solutions but will accept an answer using packages if that's the only way. Thanks for your help.
Match everything except " then capture everything upto next " and replace captured group by itself.
gsub( '[^"]*"([^"]*).*', '\\1', vec)
[1] "cd"
For detailed explanation of regex you can see this demo
I have a column within a data frame with a series of identifiers in, a letter and 8 numbers, i.e. B15006788.
Is there a way to remove all instances of B15.... to make them empty cells (there’s thousands of variations of numbers within each category) but keep B16.... etc?
I know if there was just one thing I wanted to remove, like the B15, I could do;
sub(“B15”, ””, df$col)
But I’m not sure on the how to remove a set number of characters/numbers (or even all subsequent characters after B15).
Thanks in advance :)
Welcome to SO! This is a case of regex. You can use base R as I show here or look into the stringR package for handy tools that are easier to understand. You can also look for regex rules to help define what you want to look for. For what you ask you can use the following code example to help:
testStrings <- c("KEEPB15", "KEEPB15A", "KEEPB15ABCDE")
gsub("B15.{2}", "", testStrings)
gsub is the base R function to replace a pattern with something else in one or a series of inputs. To test our regex I created the testStrings vector for different examples.
Breaking down the regex code, "B15" is the pattern you're specifically looking for. The "." means any character and the "{2}" is saying what range of any character we want to grab after "B15". You can change it as you need. If you want to remove everything after "B15". replace the pattern with "B15.". the "" means everything till the end.
edit: If you want to specify that "B15" must be at the start of the string, you can add "^" to the start of the pattern as so: "^B15.{2}"
https://www.rstudio.com/wp-content/uploads/2016/09/RegExCheatsheet.pdf has a info on different regex's you can make to be more particular.
Trying to code up a Regex in R to match everything before the first occurrence of a colon.
Let's say I have:
time = "12:05:41"
I'm trying to extract just the 12. My strategy was to do something like this:
grep(".+?(?=:)", time, value = TRUE)
But I'm getting the error that it's an invalid Regex. Thoughts?
Your regex seems fine in my opinion, I don't think you should use grep, also you are missing perl=TRUE that is why you are getting the error.
I would recommend using :
stringr::str_extract( time, "\\d+?(?=:)")
grep is little different than it is being used here, its good for matching separate values and filtering out those which has similar pattern, but you can't pluck out values within a string using grep.
If you want to use Base R you can also go for sub:
sub("^(\\d+?)(?=:)(.*)$","\\1",time, perl=TRUE)
Also, you may split the string using strsplit and filter out the first string like below:
strsplit(time, ":")[[1]][1]
In R, I currently have a long vector of dates and times saved as a string. So depending on the given date, the string can be 16 or 17 or 18 characters long and so I cannot just subset the first the 8 or 10 characters in the string, since that would not work for every date. But since there is a space between the date and time values, I am wondering how can I subset this string so that I only get the characters before the space?
Just to show how the string looks like now, here are a couple of examples:
"4/18/1950 0:00:00"
"6/8/1951 0:00:00"
"11/15/1951 0:00:00"
I'm not sure if you are familiar with regular expressions, if not you should learn as they are extremely useful:
tutorial
As akrun pointed out you can use the "sub" command to remove the space and everything after it like this:
sub(" .*","",stringVar)
First argument is the regular expression code which matches the space and everything that follows.
Second argument is what you want to replace the match with, in this case nothing
Third argument is the input string
Alternatively, you can just split the string at the space and select the first half using "strsplit"
strsplit(stringVar," ")[1]
Used to run R with numbers and matrix, when it comes to play with strings and characters I am lost. I want to analyze some data where the time is read into R as follow:
>my.time.char[1]
[1] "\"2011-10-05 15:55:00\""
I want to end up with a string containing only:
"2011-10-05 15:55:00"
Using the function sub() (that i barely understand...), I got the following result:
> sub("(\")","",my.time.char[1])
[1] "2011-10-05 15:55:00\""
This is closer to the format i am looking for, but I still need to get rid of the two last characters (\").
The second line from ?sub explains:
sub and gsub perform replacement of the first and all matches respectively.
which should tell you to use gsub instead.