My string is of the form my_str = "2a1^ath67e22^yz2p0". I would like to split based on the pattern '^(any characters) and get ["2a1", "67e22", "2p0"]. The pattern could also appear in the front or the back part of the string, such as ^abc27e4 or 27c2^abc. I tried re.split("(.*)\^[a-z]{1,100}(.*)", my_str) but it only splits one of those patterns. I am assuming here that the number of characters appearing after ^ will not be larger than 100.
you don't need regex for simple string operations, you can use
my_list = my_str.split('^')
EDIT: sorry, I just saw that you don't want to split just on the ^ character but also on strings following. Therefore you will need regex.
my_list = re.split('\^[a-z]+', my_str)
If the pattern is at the front or the end of the string, this will create an empty list element. you can remove them with
my_list = list(filter(None, my_list))
if you want to use regex library, you can just split by '\^'
re.split('\^', my_str)
# output : ['2a1', 'ath67e22', 'yz2p0']
Related
I do have got the below string and I am looking for a way to split it in order to consistently end up with the following output
'1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
['1GB 02060250396L1.060,70',
'2BE 129517720L2.639,40',
'3NL 134187650L4.024,23',
'4DE 165893440L8.111,00',
'5PL 65775644897L3.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L8.0221,30']
My current approach
re.split("([0-9][0-9][0-9][A-Z][A-Z])", input) however is also splitting my delimiter which gives and there is no other split possible than the one I am currently using in order to remain consistent. Is it possible to split my delimiter as well and assign a part of it "70" to the string in front and a part "2BE" to the following string?
Use re.findall() instead of re.split().
You want to match
a number \d, followed by
two letters [A-Z]{2}, followed by
a space \s, followed by
a bunch of characters until you encounter a comma [^,]+, followed by
two digits \d{2}
Try it at regex101
So do:
input_str = '1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
re.findall(r"\d[A-Z]{2}\s[^,]+,\d{2}", input_str)
Which gives
['1GB 02060250396L7.067,70',
'2BE 129517720L6.633,40',
'3NL 134187650L3.824,23',
'4DE 165893440L3.111,00',
'5PL 65775644897L1.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L3.001,30']
Alternatively, if you don't want to be so specific with your pattern, you could simply use the regex
[^,]+,\d{2} Try it at regex101
This will match as many of any character except a comma, then a single comma, then two digits.
re.findall(r"[^,]+,\d{2}", input_str)
# Output:
['1GB 02060250396L7.067,70',
'2BE 129517720L6.633,40',
'3NL 134187650L3.824,23',
'4DE 165893440L3.111,00',
'5PL 65775644897L1.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L3.001,30']
Is it possible to split my delimiter as well and assign a part of it "70" to the string in front and a part "2BE" to the following string?
If you must use re.split AT ANY PRICE then you might exploit zero-length assertion for this task following way
import re
text = '1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
parts = re.split(r'(?<=,[0-9][0-9])', text)
print(parts)
output
['1GB 02060250396L7.067,70', '2BE 129517720L6.633,40', '3NL 134187650L3.824,23', '4DE 165893440L3.111,00', '5PL 65775644897L1.010,00', '6DE 811506926L3.547,40', '7AT U16235008L-830,00', '8SE U57469158L3.001,30', '']
Explanation: This particular one is positive lookbehind, it does find zero-length substring preceded by , digit digit. Note that parts has superfluous empty str at end.
I want to split strings only by suffixes. For example, I would like to be able to split dord word to [dor,wor].
I though that \wd would search for words that end with d. However this does not produce the expected results
import re
re.split(r'\wd',"dord word")
['do', ' wo', '']
How can I split by suffixes?
x='dord word'
import re
print re.split(r"d\b",x)
or
print [i for i in re.split(r"d\b",x) if i] #if you dont want null strings.
Try this.
As a better way you can use re.findall and use r'\b(\w+)d\b' as your regex to find the rest of word before d:
>>> re.findall(r'\b(\w+)d\b',s)
['dor', 'wor']
Since \w also captures digits and underscore, I would define a word consisting of just letters with a [a-zA-Z] character class:
print [x.group(1) for x in re.finditer(r"\b([a-zA-Z]+)d\b","dord word")]
See demo
If you're wondering why your original approach didn't work,
re.split(r'\wd',"dord word")
It finds all instances of a letter/number/underscore before a "d" and splits on what it finds. So it did this:
do[rd] wo[rd]
and split on the strings in brackets, removing them.
Also note that this could split in the middle of words, so:
re.split(r'\wd', "said tendentious")
would split the second word in two.
So I have a string in the format of ABCD-EFGH-IJ where A through J are numbers 0-9 in a list of a ton of other strings. I have a regular expression identifying it, but how do I get it to also replace it with the format IJABCDEFGH?
You can use the following regular expression with substitution:
import re
s = '1234-5678-90'
print re.sub(r'(\d{4})-(\d{4})-(\d{2})', r'\3\1\2', s)
Result:
9012345678
\3 matches the content of what inside the third pair of parentheses. So \3\1\2 means to replace with the third group of your numbers, followed by the first followed by the second.
Easiest way to explain this is an example:
I have this string: 'Docs/src/Scripts/temp'
Which I know how to split two different ways:
re.split('/', 'Docs/src/Scripts/temp') -> ['Docs', 'src', 'Scripts', 'temp']
re.split('(/)', 'Docs/src/Scripts/temp') -> ['Docs', '/', 'src', '/', 'Scripts', '/', 'temp']
Is there a way to split by the forward slash, but keep the slash part of the words?
For example, I want the above string to look like this:
['Docs/', '/src/', '/Scripts/', '/temp']
Any help would be appreciated!
Interesting question, I would suggest doing something like this:
>>> 'Docs/src/Scripts/temp'.replace('/', '/\x00/').split('\x00')
['Docs/', '/src/', '/Scripts/', '/temp']
The idea here is to first replace all / characters by two / characters separated by a special character that would not be a part of the original string. I used a null byte ('\x00'), but you could change this to something else, then finally split on that special character.
Regex isn't actually great here because you cannot split on zero-length matches, and re.findall() does not find overlapping matches, so you would potentially need to do several passes over the string.
Also, re.split('/', s) will do the same thing as s.split('/'), but the second is more efficient.
A solution without split() but with lookaheads:
>>> s = 'Docs/src/Scripts/temp'
>>> r = re.compile(r"(?=((?:^|/)[^/]*/?))")
>>> r.findall(s)
['Docs/', '/src/', '/Scripts/', '/temp']
Explanation:
(?= # Assert that it's possible to match...
( # and capture...
(?:^|/) # the start of the string or a slash
[^/]* # any number of non-slash characters
/? # and (optionally) an ending slash.
) # End of capturing group
) # End of lookahead
Since a lookahead assertion is tried at every position in the string and doesn't consume any characters, it doesn't have a problem with overlapping matches.
1) You do not need regular expressions to split on a single fixed character:
>>> 'Docs/src/Scripts/temp'.split('/')
['Docs', 'src', 'Scripts', 'temp']
2) Consider using this method:
import os.path
def components(path):
start = 0
for end, c in enumerate(path):
if c == os.path.sep:
yield path[start:end+1]
start = end
yield path[start:]
It doesn't rely on clever tricks like split-join-splitting, which makes it much more readable, in my opinion.
If you don't insist on having slashes on both sides, it's actually quite simple:
>>> re.findall(r"([^/]*/)", 'Docs/src/Scripts/temp')
['Docs/', 'src/', 'Scripts/']
Neither re nor split are really cut out for overlapping strings, so if that's what you really want, I'd just add a slash to the start of every result except the first.
Try about this:
re.split(r'(/)', 'Docs/src/Scripts/temp')
From python's documentation
re.split(pattern, string, maxsplit=0, flags=0)
Split string by the
occurrences of pattern. If capturing parentheses are used in pattern,
then the text of all groups in the pattern are also returned as part
of the resulting list. If maxsplit is nonzero, at most maxsplit splits
occur, and the remainder of the string is returned as the final
element of the list. (Incompatibility note: in the original Python 1.5
release, maxsplit was ignored. This has been fixed in later releases.)
I'm not sure there is an easy way to do this. This is the best I could come up with...
import re
lSplit = re.split('/', 'Docs/src/Scripts/temp')
print [lSplit[0]+'/'] + ['/'+x+'/' for x in lSplit][1:-1] + ['/'+lSplit[len(lSplit)-1]]
Kind of a mess, but it does do what you wanted.
Let's say I want to remove all duplicate chars (of a particular char) in a string using regular expressions. This is simple -
import re
re.sub("a*", "a", "aaaa") # gives 'a'
What if I want to replace all duplicate chars (i.e. a,z) with that respective char? How do I do this?
import re
re.sub('[a-z]*', <what_to_put_here>, 'aabb') # should give 'ab'
re.sub('[a-z]*', <what_to_put_here>, 'abbccddeeffgg') # should give 'abcdefg'
NOTE: I know this remove duplicate approach can be better tackled with a hashtable or some O(n^2) algo, but I want to explore this using regexes
>>> import re
>>> re.sub(r'([a-z])\1+', r'\1', 'ffffffbbbbbbbqqq')
'fbq'
The () around the [a-z] specify a capture group, and then the \1 (a backreference) in both the pattern and the replacement refer to the contents of the first capture group.
Thus, the regex reads "find a letter, followed by one or more occurrences of that same letter" and then entire found portion is replaced with a single occurrence of the found letter.
On side note...
Your example code for just a is actually buggy:
>>> re.sub('a*', 'a', 'aaabbbccc')
'abababacacaca'
You really would want to use 'a+' for your regex instead of 'a*', since the * operator matches "0 or more" occurrences, and thus will match empty strings in between two non-a characters, whereas the + operator matches "1 or more".
In case you are also interested in removing duplicates of non-contiguous occurrences you have to wrap things in a loop, e.g. like this
s="ababacbdefefbcdefde"
while re.search(r'([a-z])(.*)\1', s):
s= re.sub(r'([a-z])(.*)\1', r'\1\2', s)
print s # prints 'abcdef'
A solution including all category:
re.sub(r'(.)\1+', r'\1', 'aaaaabbbbbb[[[[[')
gives:
'ab['