I am new to regular expressions and I am trying to write a pattern of phone numbers, in order to identify them and be able to extract them. My doubt can be summarized to the following simple example:
I try first to identify whether in the string is there something like (+34) which should be optional:
prefixsrch = re.compile(r'(\(?\+34\)?)?')
that I test in the following string in the following way:
line0 = "(+34)"
print prefixsrch.findall(line0)
which yields the result:
['(+34)','']
My first question is: why does it find two occurrences of the pattern? I guess that this is related to the fact that the prefix thing is optional but I do not completely understand it. Anyway, now for my big doubt
If we do a similar thing searching for a pattern of 9 digits we get the same:
numsrch = re.compile(r'\d{9}')
line1 = "971756754"
print numsrch.findall(line1)
yields something like:
['971756754']
which is fine. Now what I want to do is identify a 9 digits number, preceded or not, by (+34). So to my understanding I should do something like:
phonesrch = re.compile(r'(\(?\+34\)?)?\d{9}')
If I test it in the following strings...
line0 = "(+34)971756754"
line1 = "971756754"
print phonesrch.findall(line0)
print phonesrch.findall(line1)
this is, to my surprise, what I get:
['(+34)']
['']
What I was expecting to get is ['(+34)971756754'] and ['971756754']. Does anybody has the insight of this? thank you very much in advance.
Your capturing group is wrong. Make the country code within a non-capturing group and the entire expression in the capturing group
>>> line0 = "(+34)971756754"
>>> line1 = "971756754"
>>> re.findall(r'((?:\(?\+34\)?)?\d{9})',line0)
['(+34)971756754']
>>> re.findall(r'((?:\(?\+34\)?)?\d{9})',line1)
['971756754']
My first question is: why does it find two occurrences of the pattern?
This is because, ? which means it match 0 or 1 repetitions, so an empty string is also a valid match
Related
This question already has answers here:
Python extract pattern matches
(10 answers)
Closed 2 years ago.
I am new to regex's with python
I have a string which has got a sub-string which I would like to extract from
I have the following pattern:
r = re.compile("(flag{.+[^}]})")
and the string is
Something has gone horribly wrong\n\nflag{Hi!}
I would like to get hold of just flag{Hi!}
I have tried it with:
a = re.search(r,string)
a = re.split(r,string)
None of the approaches work, if I print a I get None
How can I get hold of the desired flag.
Thanks in advance
import re
str="Something has gone horribly wrong\n\nflag{Hi!}"
r = re.compile("(flag{.+[^}]})")
a = re.search(r,str)
print(a.group())
This worked.
Firstly, as mentioned in the comments, your output is not None. You do get a match, the match you were looking for. You actually get a Match object that spans from position 35 -> 44 and matches flag{Hi!}. You can use group() to get the match represented as a string:
>>> a = re.search(r, string)
>>> print(a.group())
"flag{Hi!}"
You can also shorten your regex a little bit. There really isn't a need to use .+ because it becomes redundant when you add [^}], which matches all characters that aren't a closing curly bracket (}):
"(flag{[^}]+})"
You can replace the +, which matches one or more with * which matches zero or more if you want to match things like flag{} where there are no characters inside the curly brackets.
We can directly search the string for matching string.
import re
line = 'Something has gone horribly wrong\n\nflag{Hi!}'
r = re.search("(flag{[^}]*})", line)
print(r.group())
Output:-
flag{Hi!}
I'm not sure if this is a problem in my understanding of regex modules, or a silly mistake I'm making in my for loop.
I have a list of numbers that look like this:
4; 94
3; 92
1; 53
etc.
I made a regex pattern to match just the last two digits of the string:
'^.*\s([0-9]+)$'
This works when I take each element of the list 1 at a time.
However when I try and make a for loop
for i in xData:
if re.findall('^.*\s([0-9]+)$', i)
print i
The output is simply the entire string instead of just the last two digits.
I'm sure I'm missing something very simple here but if someone could point me in the right direction that would be great. Thanks.
You are printing the whole string, i. If you wanted to print the output of re.findall(), then store the result and print that result:
for i in xData:
results = re.findall('^.*\s([0-9]+)$', i)
if results:
print results
I don't think that re.findall() is the right method here, since your lines contain just the one set of digits. Use re.search() to get a match object, and if the match object is not None, take the first group data:
for i in xData:
match = re.search('^.*\s([0-9]+)$', i)
if match:
print match.group(1)
I might be missing something here, but if all you're looking to do is get the last 2 characters, could you use the below?
for i in xData:
print(i[-2:])
There are probably several ways to solve this problem, so I'm open to any ideas.
I have a file, within that file is the string "D133330593" Note: I do have the exact position within the file this string exists, but I don't know if that helps.
Following this string, there are 6 digits, I need to replace these 6 digits with 6 other digits.
This is what I have so far:
def editfile():
f = open(filein,'r')
filedata = f.read()
f.close()
#This is the line that needs help
newdata = filedata.replace( -TOREPLACE- ,-REPLACER-)
#Basically what I need is something that lets me say "D133330593******"
#->"D133330593123456" Note: The following 6 digits don't need to be
#anything specific, just different from the original 6
f = open(filein,'w')
f.write(newdata)
f.close()
Use the re module to define your pattern and then use the sub() function to substitute occurrence of that pattern with your own string.
import re
...
pat = re.compile(r"D133330593\d{6}")
re.sub(pat, "D133330593abcdef", filedata)
The above defines a pattern as -- your string ("D133330593") followed by six decimal digits. Then the next line replaces ALL occurrences of this pattern with your replacement string ("abcdef" in this case), if that is what you want.
If you want a unique replacement string for each occurrence of pattern, then you could use the count keyword argument in the sub() function, which allows you to specify the number of times the replacement must be done.
Check out this library for more info - https://docs.python.org/3.6/library/re.html
Let's simplify your problem to you having a string:
s = "zshisjD133330593090909fdjgsl"
and you wanting to replace the 6 characters after "D133330593" with "123456" to produce:
"zshisjD133330594123456fdjgsl"
To achieve this, we can first need to find the index of "D133330593". This is done by just using str.index:
i = s.index("D133330593")
Then replace the next 6 characters, but for this, we should first calculate the length of our string that we want to replace:
l = len("D133330593")
then do the replace:
s[:i+l] + "123456" + s[i+l+6:]
which gives us the desired result of:
'zshisjD133330593123456fdjgsl'
I am sure that you can now integrate this into your code to work with a file, but this is how you can do the heart of your problem .
Note that using variables as above is the right thing to do as it is the most efficient compared to calculating them on the go. Nevertheless, if your file isn't too long (i.e. efficiency isn't too much of a big deal) you can do the whole process outlined above in one line:
s[:s.index("D133330593")+len("D133330593")] + "123456" + s[s.index("D133330593")+len("D133330593")+6:]
which gives the same result.
Let's say I have a number with a recurring pattern, i.e. there exists a string of digits that repeat themselves in order to make the number in question. For example, such a number might be 1234123412341234, created by repeating the digits 1234.
What I would like to do, is find the pattern that repeats itself to create the number. Therefore, given 1234123412341234, I would like to compute 1234 (and maybe 4, to indicate that 1234 is repeated 4 times to create 1234123412341234)
I know that I could do this:
def findPattern(num):
num = str(num)
for i in range(len(num)):
patt = num[:i]
if (len(num)/len(patt))%1:
continue
if pat*(len(num)//len(patt)):
return patt, len(num)//len(patt)
However, this seems a little too hacky. I figured I could use itertools.cycle to compare two cycles for equality, which doesn't really pan out:
In [25]: c1 = itertools.cycle(list(range(4)))
In [26]: c2 = itertools.cycle(list(range(4)))
In [27]: c1==c2
Out[27]: False
Is there a better way to compute this? (I'd be open to a regex, but I have no idea how to apply it there, which is why I didn't include it in my attempts)
EDIT:
I don't necessarily know that the number has a repeating pattern, so I have to return None if there isn't one.
Right now, I'm only concerned with detecting numbers/strings that are made up entirely of a repeating pattern. However, later on, I'll likely also be interested in finding patterns that start after a few characters:
magic_function(78961234123412341234)
would return 1234 as the pattern, 4 as the number of times it is repeated, and 4 as the first index in the input where the pattern first presents itself
(.+?)\1+
Try this. Grab the capture. See demo.
import re
p = re.compile(ur'(.+?)\1+')
test_str = u"1234123412341234"
re.findall(p, test_str)
Add anchors and flag Multiline if you want the regex to fail on 12341234123123, which should return None.
^(.+?)\1+$
See demo.
One way to find a recurring pattern and number of times repeated is to use this pattern:
(.+?)(?=\1+$|$)
w/ g option.
It will return the repeated pattern and number of matches (times repeated)
Non-repeated patterns (fails) will return only "1" match
Repeated patterns will return 2 or more matches (number of times repeated).
Demo
I have a regex something like
(\d\d\d)(\d\d\d)(\.\d\d){0,1}
when it matches I can easily get first two groups, but how do I check if third occurred 0 or 1 times.
Also another minor question: in the (\.\d\d) I only care about \d\d part, any other way to tell regex that \.\d\d needs to appear 0 or 1 times, but that I want to capture only \d\d part ?
This was based on a problem of parsing a
hhmmss
string that has optional decimal part for seconds( so it becomes
hhmmss.ss
)... I put \d\d\d in the question so it is clear about what \d\d Im talking about.
import re
value = "123456.33"
regex = re.search("^(\d\d\d)(\d\d\d)(?:\.(\d\d)){0,1}$", value)
if regex:
print regex.group(1)
print regex.group(2)
if regex.group(3) is not None:
print regex.group(3)
else:
print "3rd group not found"
else:
print "value don't match regex"