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Use regex to replace words before any digit with nothing

I want to replace words and spaces that appear before a digit in a string with nothing. For example, for the string = 'Juice of 1/2', I want to return '1/2'. I tried the following, but it did not work.
string = "Juice of 1/2"
new = string.replace(r"^.+?(?=\d)", "")
Also I am trying to perform this on every cell of a list of columns using the following code. How would I incorporate the new regex pattern into the existing pattern of r"(|)|?
df[pd.Index(cols2) + "_clean"] = (
df[cols2]
.apply(lambda col: col.str.replace(r"\(|\)|,", "", regex=True))
)

You might be able to phrase this using str.extract:
df["col2"] = df["col2"].str.extract(r'([0-9/-]+)')

.+? will match anything, including other digits. It will also match the / in 1/2. Since you only want to replace letters and spaces, use [a-z\s]+.
You also have to use re.sub(), not string.replace() (in Pandas, .str.replace() processes regular expressions by default).
new = re.sub(r'[a-z\s]+(?=\d)', '', string, flags=re.I)

May be something like this might work.
# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility
import re
regex = r"[A-Za-z\s]+"
test_str = "Juice of 1/2 hede"
subst = ""
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0)
if result:
print (result)
# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

Python Regex in Zapier issue

I am trying to take a field from salesforce that has line breaks and pull out the words and punctuation with a python step in zapier. Here is my code but it returns and empty string. If there is a better/easier way let me know, I am super new to any code and Frankensteined this together from googling.
import re
string = input_data['ac']
regex = r"^[a-z,A-Z].*[?.!]$"
cleaned = re.findall(regex, string)
return [{'cleaned': cleaned}]
Here are 2 pictures of, original comment and the current result, I have it working but would like to keep the punctuation by updating code.
Original Comment
Current Result
JSON parser error

The following just finds sentences by looking for a letter and then scanning until it finds one of the sentence-terminating characters.
import re
s = input_data['ac']
# remove multiple, consecutive carriage returns and/or newlines
s = re.sub(r'[\r\n]+', '', s)
regex = r"""(?x) # verbose flag
[A-Za-z] # a letter
[^?.!]* # one or more non-sentence-ending characters or .*? (non-greedy)
[?.!] # a sentence-ending character
"""
cleaned = re.findall(regex, s)
result = [{'cleaned': cleaned}]
#return result # only legal in a function
Regex Demo

How can I remove a specific character from multi line string using regex in python

I have a multiline string which looks like this:
st = '''emp:firstinfo\n
:secondinfo\n
thirdinfo
'''
print(st)
What I am trying to do is to skip the second ':' from my string, and get an output which looks like this:
'''emp:firstinfo\n
secondinfo\n
thirdinfo
'''
simply put if it starts with a ':' I'm trying to ignore it.
Here's what I've done:
mat_obj = re.match(r'(.*)\n*([^:](.*))\n*(.*)' , st)
print(mat_obj.group())
Clearly, I don't see my mistake but could anyone please help me telling where I am getting it wrong?

You may use re.sub with this regex:
>>> print (re.sub(r'([^:\n]*:[^:\n]*\n)\s*:(.+)', r'\1\2', st))
emp:firstinfo
secondinfo
thirdinfo
RegEx Demo
RegEx Details:
(: Start 1st capture group
[^:\n]*: Match 0 or more of any character that is not : and newline
:: Match a colon
[^:\n]*: Match 0 or more of any character that is not : and newline
\n: Match a new line
): End 1st capture group
\s*: Match 0 or more whitespaces
:: Match a colon
(.+): Match 1 or more of any characters (except newlines) in 2nd capture group
\1\2: Is used in replacement to put back substring captured in groups 1 and 2.

You can use sub instead, just don't capture the undesired part.
(.*\n)[^:]*:(.*\n)(.*)
Replace by
\1\2\3
Regex Demo
import re
regex = r"(.*\n)[^:]*:(.*\n)(.*)"
test_str = ("emp:firstinfo\\n\n"
" :secondinfo\\n\n"
" thirdinfo")
subst = "\\1\\2\\3"
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE)

#import regex library
import re
#remove character in a String and replace with empty string.
text = "The film Pulp Fiction was released in year 1994"
result = re.sub(r"[a-z]", "", text)
print(result)

Python regex to match after the text and the dot [duplicate]

I am using Python and would like to match all the words after test till a period (full-stop) or space is encountered.
text = "test : match this."
At the moment, I am using :
import re
re.match('(?<=test :).*',text)
The above code doesn't match anything. I need match this as my output.

Everything after test, including test
test.*
Everything after test, without test
(?<=test).*
Example here on regexr.com

You need to use re.search since re.match tries to match from the beging of the string. To match until a space or period is encountered.
re.search(r'(?<=test :)[^.\s]*',text)
To match all the chars until a period is encountered,
re.search(r'(?<=test :)[^.]*',text)

In a general case, as the title mentions, you may capture with (.*) pattern any 0 or more chars other than newline after any pattern(s) you want:
import re
p = re.compile(r'test\s*:\s*(.*)')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want . to match across multiple lines, compile the regex with re.DOTALL or re.S flag (or add (?s) before the pattern):
p = re.compile(r'test\s*:\s*(.*)', re.DOTALL)
p = re.compile(r'(?s)test\s*:\s*(.*)')
However, it will retrun match this.. See also a regex demo.
You can add \. pattern after (.*) to make the regex engine stop before the last . on that line:
test\s*:\s*(.*)\.
Watch out for re.match() since it will only look for a match at the beginning of the string (Avinash aleady pointed that out, but it is a very important note!)
See the regex demo and a sample Python code snippet:
import re
p = re.compile(r'test\s*:\s*(.*)\.')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want to make sure test is matched as a whole word, add \b before it (do not remove the r prefix from the string literal, or '\b' will match a BACKSPACE char!) - r'\btest\s*:\s*(.*)\.'.

I don't see why you want to use regex if you're just getting a subset from a string.
This works the same way:
if line.startswith('test:'):
print(line[5:line.find('.')])
example:
>>> line = "test: match this."
>>> print(line[5:line.find('.')])
match this
Regex is slow, it is awkward to design, and difficult to debug. There are definitely occassions to use it, but if you just want to extract the text between test: and ., then I don't think is one of those occasions.
See: https://softwareengineering.stackexchange.com/questions/113237/when-you-should-not-use-regular-expressions
For more flexibility (for example if you are looping through a list of strings you want to find at the beginning of a string and then index out) replace 5 (the length of 'test:') in the index with len(str_you_looked_for).

Regular expression to replace with XML node

I'm using Python to write a regular expression for replacing parts of the string with a XML node.
The source string looks like:
Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace
And the result string should be like:
Hello
<replace name="str1"> this is to replace </replace>
<replace name="str2"> this is to replace </replace>
Can anyone help me?

What makes your problem a little bit tricky is that you want to match inside of a multiline string. You need to use the re.MULTILINE flag to make that work.
Then, you need to match some groups inside your source string, and use those groups in the final output. Here is code that works to solve your problem:
import re
s_pat = "^\s*REPLACE\(([^)]+)\)(.*)$"
pat = re.compile(s_pat, re.MULTILINE)
s_input = """\
Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace"""
def mksub(m):
return '<replace name="%s">%s</replace>' % m.groups()
s_output = re.sub(pat, mksub, s_input)
The only tricky part is the regular expression pattern. Let's look at it in detail.
^ matches the start of a string. With re.MULTILINE, this matches the start of a line within a multiline string; in other words, it matches right after a newline in the string.
\s* matches optional whitespace.
REPLACE matches the literal string "REPLACE".
\( matches the literal string "(".
( begins a "match group".
[^)] means "match any character but a ")".
+ means "match one or more of the preceding pattern.
) closes a "match group".
\) matches the literal string ")"
(.*) is another match group containing ".*".
$ matches the end of a string. With re.MULTILINE, this matches the end of a line within a multiline string; in other words, it matches a newline character in the string.
. matches any character, and * means to match zero or more of the preceding pattern. Thus .* matches anything, up to the end of the line.
So, our pattern has two "match groups". When you run re.sub() it will make a "match object" which will be passed to mksub(). The match object has a method, .groups(), that returns the matched substrings as a tuple, and that gets substituted in to make the replacement text.
EDIT: You actually don't need to use a replacement function. You can put the special string \1 inside the replacement text, and it will be replaced by the contents of match group 1. (Match groups count from 1; the special match group 0 corresponds the the entire string matched by the pattern.) The only tricky part of the \1 string is that \ is special in strings. In a normal string, to get a \, you need to put two backslashes in a row, like so: "\\1" But you can use a Python "raw string" to conveniently write the replacement pattern. Doing so you get this:
import re
s_pat = "^\s*REPLACE\(([^)]+)\)(.*)$"
pat = re.compile(s_pat, re.MULTILINE)
s_repl = r'<replace name="\1">\2</replace>'
s_input = """\
Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace"""
s_output = re.sub(pat, s_repl, s_input)

Here is an excellent tutorial on how to write regular expressions in Python.

Here is a solution using pyparsing. I know you specifically asked about a regex solution, but if your requirements change, you might find it easier to expand a pyparsing parser. Or a pyparsing prototype solution might give you a little more insight into the problem leading toward a regex or other final implementation.
src = """\
Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace
"""
from pyparsing import Suppress, Word, alphas, alphanums, restOfLine
LPAR,RPAR = map(Suppress,"()")
ident = Word(alphas, alphanums)
replExpr = "REPLACE" + LPAR + ident("name") + RPAR + restOfLine("body")
replExpr.setParseAction(
lambda toks : '<replace name="%(name)s">%(body)s </replace>' % toks
)
print replExpr.transformString(src)
In this case, you create the expression to be matched with pyparsing, define a parse action to do the text conversion, and then call transformString to scan through the input source to find all the matches, apply the parse action to each match, and return the resulting output. The parse action serves a similar function to mksub in #steveha's solution.
In addition to the parse action, pyparsing also supports naming individual elements of the expression - I used "name" and "body" to label the two parts of interest, which are represented in the re solution as groups 1 and 2. You can name groups in an re, the corresponding re would look like:
s_pat = "^\s*REPLACE\((?P<name>[^)]+)\)(?P<body>.*)$"
Unfortunately, to access these groups by name, you have to invoke the group() method on the re match object, you can't directly do the named string interpolation as in my lambda parse action. But this is Python, right? We can wrap that callable with a class that will give us dict-like access to the groups by name:
class CallableDict(object):
def __init__(self,fn):
self.fn = fn
def __getitem__(self,name):
return self.fn(name)
def mksub(m):
return '<replace name="%(name)s">%(body)s</replace>' % CallableDict(m.group)
s_output = re.sub(pat, mksub, s_input)
Using CallableDict, the string interpolation in mksub can now call m.group for each field, by making it look like we are retrieving the ['name'] and ['body'] elements of a dict.

Maybe like this ?
import re
mystr = """Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace"""
prog = re.compile(r'REPLACE\((.*?)\)\s(.*)')
for line in mystr.split("\n"):
print prog.sub(r'< replace name="\1" > \2',line)

Something like this should work:
import re,sys
f = open( sys.argv[1], 'r' )
for i in f:
g = re.match( r'REPLACE\((.*)\)(.*)', i )
if g is None:
print i
else:
print '<replace name=\"%s\">%s</replace>' % (g.group(1),g.group(2))
f.close()

import re
a="""Hello
REPLACE(str1) this is to replace
REPLACE(str2) this is to replace"""
regex = re.compile(r"^REPLACE\(([^)]+)\)\s+(.*)$", re.MULTILINE)
b=re.sub(regex, r'< replace name="\1" > \2 < /replace >', a)
print b
will do the replace in one line.