i'm trying to get the start and end positions of a query in sequence by using re.findall
import re
sequence = 'aaabbbaaacccdddeeefff'
query = 'aaa'
findall = re.findall(query,sequence)
>>> ['aaa','aaa']
how do i get something like findall.start() or findall.end() ?
i would like to get
start = [0,6]
end = [2,8]
i know that
search = re.search(query,sequence)
print search.start(),search.end()
>>> 0,2
would give me only the first instance
Use re.finditer:
>>> import re
>>> sequence = 'aaabbbaaacccdddeeefff'
>>> query = 'aaa'
>>> r = re.compile(query)
>>> [[m.start(),m.end()] for m in r.finditer(sequence)]
[[0, 3], [6, 9]]
From the docs:
Return an iterator yielding MatchObject instances over all
non-overlapping matches for the RE pattern in string. The string is
scanned left-to-right, and matches are returned in the order found.
You can't. findall is a convenience function that, as the docs say, returns "a list of strings". If you want a list of MatchObjects, you can't use findall.
However, you can use finditer. If you're just iterating over the matches for match in re.findall(…):, you can use for match in re.finditer(…) the same way—except you get MatchObject values instead of strings. If you actually need a list, just use matches = list(re.finditer(…)).
Use finditer instead of findall. This gives you back an iterator yielding MatchObject instances and you can get start/end from the MatchObject.
Related
I want to get String before last occurrence of my given sub string.
My String was,
path =
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov
my substring, 1001-1010 which will occurred twice. all i want is get string before its last occurrence.
Note: My substring is dynamic with different padding but only number.
I want,
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v
I have done using regex and slicing,
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> q = re.findall("\d*-\d*",p)
>>> q[-1].join(p.split(q[-1])[:-1])
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v'
>>>
Is their any better way to do by purely using regex?
Please Note I have tried so many eg:
regular expression to match everything until the last occurrence of /
Regex Last occurrence?
I got answer by using regex with slicing but i want to achieve by using regex alone..
Why use regex. Just use built in string methods:
path = "D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov"
index = path.rfind("1001-1010")
print(path[:index])
You can use a simple greedy match and a capture group:
(.*)1001-1010
Your match is in capture group #1
Since .* is greedy by nature, it will match longest match before matching your keyword 1001-1010.
RegEx Demo
As per comments below if keyword is not a static string then you may use this regex:
r'(.*\D)\d+-\d+'
Python Code:
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> print (re.findall(r'(.*\D)\d+-\d+', p))
['D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v']
Thanks #anubhava,
My first regex was,
.*(\d*-\d*)\/
Now i have corrected mine..
.*(\d*-\d*)
or
(.*)(\d*-\d*)
which gives me,
>>> q = re.search('.+(\d*-\d*)', p)
>>> q.group()
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v0001-1001'
>>>
(.*\D)\d+-\d+
this gives me exactly what i want...
>>> q = re.search('(.*\D)\d+-\d+', p)
>>> q.groups()
('D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v',)
>>>
I have a list of strings
['time_10', 'time_23', 'time_345', 'date_10', 'date_23', 'date_345']
I want to use regular expression to get strings that end with a specific number.
As I understand, first I have to combine all strings from the list into large string, then use form some kind of a pattern to use it for regular expression
I would be grateful if you could provide
regex(some_pattern, some_string)
that would return
['time_10', 'date_10']
or just
'time_10, date_10'
str.endswith is enough.
l = ['time_10', 'time_23', 'time_345', 'date_10', 'date_23', 'date_345']
result = [s for s in l if s.endswith('10')]
print(result)
['time_10', 'date_10']
If you insist on using regex,
import re
result = [s for s in l if re.search('10$', s)]
I'm tackling a python challenge problem to find a block of text in the format xXXXxXXXx (lower vs upper case, not all X's) in a chunk like this:
jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn
I have tested the following RegEx and found it correctly matches what I am looking for from this site (http://www.regexr.com/):
'([a-z])([A-Z]){3}([a-z])([A-Z]){3}([a-z])'
However, when I try to match this expression to the block of text, it just returns the entire string:
In [1]: import re
In [2]: example = 'jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn'
In [3]: expression = re.compile(r'([a-z])([A-Z]){3}([a-z])([A-Z]){3}([a-z])')
In [4]: found = expression.search(example)
In [5]: print found.string
jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn
Any ideas? Is my expression incorrect? Also, if there is a simpler way to represent that expression, feel free to let me know. I'm fairly new to RegEx.
You need to return the match group instead of the string attribute.
>>> import re
>>> s = 'jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn'
>>> rgx = re.compile(r'[a-z][A-Z]{3}[a-z][A-Z]{3}[a-z]')
>>> found = rgx.search(s).group()
>>> print found
nJDKoJIWh
The string attribute always returns the string passed as input to the match. This is clearly documented:
string
The string passed to match() or search().
The problem has nothing to do with the matching, you're just grabbing the wrong thing from the match object. Use match.group(0) (or match.group()).
Based on xXXXxXXXx if you want upper letters with len 3 and lower with len 1 between them this is what you want :
([a-z])(([A-Z]){3}([a-z]))+
also you can get your search function with group()
print expression.search(example).group(0)
I have a regex match object in Python. I want to get the text it matched. Say if the pattern is '1.3', and the search string is 'abc123xyz', I want to get '123'. How can I do that?
I know I can use match.string[match.start():match.end()], but I find that to be quite cumbersome (and in some cases wasteful) for such a basic query.
Is there a simpler way?
You can simply use the match object's group function, like:
match = re.search(r"1.3", "abc123xyz")
if match:
doSomethingWith(match.group(0))
to get the entire match. EDIT: as thg435 points out, you can also omit the 0 and just call match.group().
Addtional note: if your pattern contains parentheses, you can even get these submatches, by passing 1, 2 and so on to group().
You need to put the regex inside "()" to be able to get that part
>>> var = 'abc123xyz'
>>> exp = re.compile(".*(1.3).*")
>>> exp.match(var)
<_sre.SRE_Match object at 0x691738>
>>> exp.match(var).groups()
('123',)
>>> exp.match(var).group(0)
'abc123xyz'
>>> exp.match(var).group(1)
'123'
or else it will not return anything:
>>> var = 'abc123xyz'
>>> exp = re.compile("1.3")
>>> print exp.match(var)
None
I need to find the value of "taxid" in a large number of strings similar to one given below. For this particular string, the 'taxid' value is '9606'. I need to discard everything else. The "taxid" may appear anywhere in the text, but will always be followed by a ":" and then number.
score:0.86|taxid:9606(Human)|intact:EBI-999900
How to write regular expression for this in python.
>>> import re
>>> s = 'score:0.86|taxid:9606(Human)|intact:EBI-999900'
>>> re.search(r'taxid:(\d+)', s).group(1)
'9606'
If there are multiple taxids, use re.findall, which returns a list of all matches:
>>> re.findall(r'taxid:(\d+)', s)
['9606']
for line in lines:
match = re.match(".*\|taxid:([^|]+)\|.*",line)
print match.groups()