How to Extract Everything Until First Occurrence of Pattern

How to extract everything until first occurrence of pattern

To get L0, you may use

> library(stringr)
> str_extract("L0_123_abc", "[^_]+")
[1] "L0"

The [^_]+ matches 1 or more chars other than _.

Also, you may split the string with _:

x <- str_split("L0_123_abc", fixed("_"))
> x
[[1]]
[1] "L0"  "123" "abc"

This way, you will have all the substrings you need.

The same can be achieved with

> str_extract_all("L0_123_abc", "[^_]+")
[[1]]
[1] "L0"  "123" "abc"

Regex: matching up to the first occurrence of a character

You need

/^[^;]*/

The [^;] is a character class, it matches everything but a semicolon.

^ (start of line anchor) is added to the beginning of the regex so only the first match on each line is captured. This may or may not be required, depending on whether possible subsequent matches are desired.

To cite the perlre manpage:

You can specify a character class, by enclosing a list of characters in [] , which will match any character from the list. If the first character after the "[" is "^", the class matches any character not in the list.

This should work in most regex dialects.

Remove everything until the first occurrence of a bracket ( in R

You can use sub with *\\(.* to remove everything after the first ( and also spaces before.

example$LGA <- sub(" *\\(.*", "", example$LGA_formal)
identical(example, example_desired) #test if desired is reached
#[1] TRUE

How can I extract everything in between _ characters starting with nth occurrence of said character (from the end of the string)?

Here are four suitable regular expressions utilizing positive lookarounds. Let me know if they work:

"(?<=\_)[^_]+(?=_[^_]+_[^_]+\.)"
"(?<=\_)[^_]+(?=_[^_]+\.)"
"(?<=\_)[^_]+(?=\.)"
"(?<=\.).*$"

As Google Data Studio cannot implement lookaround, here is an alternative workaround with multiple steps, which is written in R but can be translated to your language of choice:

text1 <- "Parameter1_Parameter3_Parameter4_ParamaterA_ParameterB_ParamaterC.mp4"

last_three <- str_extract(text1, "[^_]+_[^_]+_[^_]+\\..+")

str_extract(last_three, "^[^_]+")

str_replace_all(str_extract(last_three, "_[^_\\.]+_"), "_", "")

str_replace(str_extract(last_three, "[^_\\.]+\\."), "\\.", "")

str_replace(str_extract(last_three, "\\..+$"), "\\.", "")

https://support.google.com/datastudio/table/6379764?hl=en

Google Data Studio has the required commands for this: REGEXP_EXTRACT and REGEXP_REPLACE.

Regex match until first instance of certain character

You added the " into the consuming part of the pattern, remove it.

^.+?(?=\")

Or, if you need to match any chars including line breaks, use either

(?s)^.+?(?=\")
^[\w\W]+?(?=\")

See demo. Here, ^ matches start of string, .+? matches any 1+ chars, as few as possible, up to the first " excluding it from the match because the "` is a part of the lookahead (a zero-width assertion).

In the two other regexps, (?s) makes the dot match across lines, and [\w\W] is a work-around construct that matches any char if the (s) (or its /s form) is not supported.

Best is to use a negated character class:

^[^"]+

See another demo. Here, ^[^"]+ matches 1+ chars other than " (see [^"]+) from the start of a string (^).

extracting data before a sign in R

Using sub does the job:

sub("(.*)-.*", "\\1", c(text1, text2, text3))
# [1] "Médicos"   "Disturbio" "Accidente"

Here we split each character into: what goes before the dash ((.*)), the dash itself, and what goes after the dash (.*). Each character then is replaced by the first part (\\1).

Analogously you may extract the second half:

sub(".*-(.*)", "\\1", c(text1, text2, text3))
# [1] "Otros"     "Escándalo" "Choque"   

R/Stringr Extract String after nth occurrence of _ and end with first occurrence of _

We could create a pattern based on the 'n'

n <- 2
pat <- sprintf('([^_]+_){%d}([^_]+)_.*', n)
sub(pat, '\\2', df)
#[1] "HERE" "THIS"

Details -

Capture one or more characters that are not a _ ([^_]+) followed by a _ that is repeated 'n' times (2), followed by the next set of characters that are not a _ (([^_]+)) followed by a _ and other characters. In the replacement, specify the backreference of the second captured group

How can I match anything up until this sequence of characters in a regular expression?

You didn't specify which flavor of regex you're using, but this will
work in any of the most popular ones that can be considered "complete".

/.+?(?=abc)/

How it works

The .+? part is the un-greedy version of .+ (one or more of
anything). When we use .+, the engine will basically match everything.
Then, if there is something else in the regex it will go back in steps
trying to match the following part. This is the greedy behavior,
meaning as much as possible to satisfy.

When using .+?, instead of matching all at once and going back for
other conditions (if any), the engine will match the next characters by
step until the subsequent part of the regex is matched (again if any).
This is the un-greedy, meaning match the fewest possible to
satisfy.

/.+X/  ~ "abcXabcXabcX"        /.+/  ~ "abcXabcXabcX"
          ^^^^^^^^^^^^                  ^^^^^^^^^^^^

/.+?X/ ~ "abcXabcXabcX"        /.+?/ ~ "abcXabcXabcX"
          ^^^^                          ^

Following that we have (?={contents}), a zero width
assertion, a look around. This grouped construction matches its
contents, but does not count as characters matched (zero width). It
only returns if it is a match or not (assertion).

Thus, in other terms the regex /.+?(?=abc)/ means:

Match any characters as few as possible until a "abc" is found,
without counting the "abc".

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