I want to select select all character strings that begin with 0
x= '1,1,1075 1,0,39 2,4,1,22409 0,1,1,755,300 0,1,1,755,50'
I have
re.findall(r'\b0\S*', x)
but this returns
['0,39', '0,1,1,755,300', '0,1,1,755,50']
I want
['0,1,1,755,300', '0,1,1,755,50']
The problem is that \b matches the boundaries between digits and commas too. The simplest way might be not to use a regex at all:
thingies = [thingy for thingy in x.split() if thingy.startswith('0')]
Instead of using the boundary \b which will match between the comma and number (between any word [a-zA-Z0-9_] and non word character), you will want to match on start of string or space like (^|\s).
(^|\s)0\S*
https://regex101.com/r/Mrzs8a/1
Which will match the start of string or a space preceding the target string. But that will also include the space if present so I would suggest either trimming your matched string or wrapping the latter part with parenthesis to make it a group and then just getting group 1 from the matches like:
(?:^|\s)(0\S*)
https://regex101.com/r/Mrzs8a/2
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 am new to python.
I want to write a regex to find a whole string using re.match.
For example,
word1=I
word2=U
str_to_match = word1+"[There could be some space in between]"+ word2[this is the end of string I want to match]"
In this case, it should match I[0 or more spaces in between]U.
It shouldn't match abI[0 or more spaces in between]U, OR abI[0 or more spaces in between]Ucd, OR
I[0 or more spaces in between]Ucd.
I know you could use boundary \b to set the word boundary but since there could be a number of spaces in between word1 and word2, the whole string is like a variable, not a fixed string, it won't work for me:
ret=re.match(r"\b"+word1+r"\s+" +word2+r"\b")
not working
Does anyone know how could I find the correct way to match this case?
Thanks!
match only matches at the beginning of the string. Which since you are using \b should not be what you want. findall will find all matching strings.
If you want to find the first occurrence and just test if it exists in the string, and not find all matches you can use search:
word1 = 'I'
word2 = 'U'
ret = re.findall(rf"\b{word1}\s*{word2}\b"," I U IU I U aI U I Ub")
print(ret)
ret = re.search(rf"\b{word1}\s*{word2}\b"," IUb .I U ")
print(ret)
I want to match strings in which every second character is same.
for example 'abababababab'
I have tried this : '''(([a-z])[^/2])*'''
The output should return the complete string as it is like 'abababababab'
This is actually impossible to do in a real regular expression with an amount of states polynomial to the alphabet size, because the expression is not a Chomsky level-0 grammar.
However, Python's regexes are not actually regular expressions, and can handle much more complex grammars than that. In particular, you could put your grammar as the following.
(..)\1*
(..) is a sequence of 2 characters. \1* matches the exact pair of characters an arbitrary (possibly null) number of times.
I interpreted your question as wanting every other character to be equal (ababab works, but abcbdb fails). If you needed only the 2nd, 4th, ... characters to be equal you can use a similar one.
.(.)(.\1)*
You could match the first [a-z] followed by capturing ([a-z]) in a group. Then repeat 0+ times matching again a-z and a backreference to group 1 to keep every second character the same.
^[a-z]([a-z])(?:[a-z]\1)*$
Explanation
^ Start of the string
[a-z]([a-z]) Match a-z and capture in group 1 matching a-z
)(?:[a-z]\1)* Repeat 0+ times matching a-z followed by a backreference to group 1
$ End of string
Regex demo
Though not a regex answer, you could do something like this:
def all_same(string):
return all(c == string[1] for c in string[1::2])
string = 'abababababab'
print('All the same {}'.format(all_same(string)))
string = 'ababacababab'
print('All the same {}'.format(all_same(string)))
the string[1::2] says start at the 2nd character (1) and then pull out every second character (the 2 part).
This returns:
All the same True
All the same False
This is a bit complicated expression, maybe we would start with:
^(?=^[a-z]([a-z]))([a-z]\1)+$
if I understand the problem right.
Demo
I expect to fetch all alphanumeric characters after "-"
For an example:
>>> str1 = "12 - mystr"
>>> re.findall(r'[-.\:alnum:](.*)', str1)
[' mystr']
First, it's strange that white space is considered alphanumeric, while I expected to get ['mystr'].
Second, I cannot understand why this can be fetched, if there is no "-":
>>> str2 = "qwertyuio"
>>> re.findall(r'[-.\:alnum:](.*)', str2)
['io']
First of all, Python re does not support POSIX character classes.
The white space is not considered alphanumeric, your first pattern matches - with [-.\:alnum:] and then (.*) captures into Group 1 all 0 or more chars other than a newline. The [-.\:alnum:] pattern matches one char that is either -, ., :, a, l, n, u or m. Thus, when run against the qwertyuio, u is matched and io is captured into Group 1.
Alphanumeric chars can be matched with the [^\W_] pattern. So, to capture all alphanumeric chars after - that is followed with 0+ whitespaces you may use
re.findall(r'-\s*([^\W_]+)', s)
See the regex demo
Details
- - a hyphen
\s* - 0+ whitespaces
([^\W_]+) - Capturing group 1: one or more (+) chars that are letters or digits.
Python demo:
print(re.findall(r'-\s*([^\W_]+)', '12 - mystr')) # => ['mystr']
print(re.findall(r'-\s*([^\W_]+)', 'qwertyuio')) # => []
Your regex says: "Find any one of the characters -.:alnum, then capture any amount of any characters into the first capture group".
In the first test, it found - for the first character, then captured mystr in the first capture group. If any groups are in the regex, findall returns list of found groups, not the matches, so the matched - is not included.
Your second test found u as one of the -.:alnum characters (as none of qwerty matched any), then captured and returned the rest after it, io.
As #revo notes in comments, [....] is a character class - matching any one character in it. In order to include a POSIX character class (like [:alnum:]) inside it, you need two sets of brackets. Also, there is no order in a character class; the fact that you included - inside it just means it would be one of the matched characters, not that alphanumeric characters would be matched without it. Finally, if you want to match any number of alphanumerics, you have your quantifier * on the wrong thing.
Thus, "match -, then any number of alphanumeric characters" would be -([[:alnum:]]*), except... Python does not support POSIX character classes. So you have to write your own: -([A-Za-z0-9]*).
However, that will not match your string because the intervening space is, as you note, not an alphanumeric character. In order to account for that, -\s*([A-Za-z0-9]*).
Not quite sure what you want to match. I'll assume you don't want to include '-' in any matches.
If you want to get all alphanumeric chars after the first '-' and skip all other characters you can do something like this.
re.match('.*?(?<=-)(((?<=\s+)?[a-zA-Z\d]+(?=\s+)?)+)', inputString)
If you want to find each string of alphanumerics after a each '-' then you can do this.
re.findall('(?<=-)[a-zA-Z\d]+')
I defined
s='f(x) has an occ of x but no y'
def italicize_math(line):
p="(\W|^)(x|y|z|f|g|h)(\W|$)"
repl=r"\1<i>\2</i>\3"
return re.sub(p,repl,line)
and made the following call:
print(italicize_math(s)
The result is
'<i>f</i>(x) has an occ of <i>x</i> but no <i>y</i>'
which is not what I expected. I wanted this instead:
'<i>f</i>(<i>x</i>) has an occ of <i>x</i> but no <i>y</i>'
Can anyone tell me why the first occurence of x was not enclosed in inside the "i" tags?
You seem to be trying to match non-alphanumeric characters (\W) when you really want a word boundary (\b):
>>> p=r"(\b)(x|y|z|f|g|h)(\b)"
>>> re.sub(p,repl,s)
'<i>f</i>(<i>x</i>) has an occ of <i>x</i> but no <i>y</i>'
Of course, ( is non alpha-numeric -- The reason your inner content doesn't match is because \W consumes a character in the match. so with a string like 'f(x)', you match the ( when you match f. Since ( was already matched, it won't match again when you try to match x. By contrast, word boundaries don't consume any characters.
Because the group construct is matching the position at the beginning of the string first and x would overlap the previous match. Also, the first and third groups are redundant since they can be replaced by word boundaries; and you can make use of a character class to combine letters.
p = r'\b([fghxyz])\b'
repl = r'<i>\1</i>'
Like previous answer mention, its because the ( char being consume when matching f thus cause subsequent x to fail the match.
beside replace with word boundary \b, you could also use lookahead regex which just do a peek and won't consume anything match inside the lookahead. Since it didn't consume anything, you don't need the \3 either
p=r"(\W|^)(x|y|z|f|g|h)(?=\W|$)"
repl=r"\1<i>\2</i>"
re.sub(p,repl,line)