I have a String from which I want to take the values within the parenthesis. Then, get the values that are separated from a comma.
Example: x(142,1,23ERWA31)
I would like to get:
142
1
23ERWA31
Is it possible to get everything with one regex?
I have found a method to do so, but it is ugly.
This is how I did it in python:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
secondResult = re.search("(?<=\()(.*?)(?=\))", firstResult.group(0))
finalResult = [x.strip() for x in secondResult.group(0).split(',')]
for i in finalResult:
print(i)
142
1
23ERWA31
This works for your example string:
import re
string = "x(142,1,23ERWA31)"
l = re.findall (r'([^(,)]+)(?!.*\()', string)
print (l)
Result: a plain list
['142', '1', '23ERWA31']
The expression matches a sequence of characters not in (,,,) and – to prevent the first x being picked up – may not be followed by a ( anywhere further in the string. This makes it also work if your preamble x consists of more than a single character.
findall rather than search makes sure all items are found, and as a bonus it returns a plain list of the results.
You can make this a lot simpler. You are running your first Regex but then not taking the result. You want .group(1) (inside the brackets), not .group(0) (the whole match). Once you have that you can just split it on ,:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
for e in firstResult.group(1).split(','):
print(e)
A little wonky looking, and also assuming there's always going to be a grouping of 3 values in the parenthesis - but try this regex
\((.*?),(.*?),(.*?)\)
To extract all the group matches to a single object - your code would then look like
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?),(.*?),(.*?)\)", string).groups()
You can then call the firstResult object like a list
>> print(firstResult[2])
23ERWA31
Related
So I have the following strings and I have been trying to figure out how to manipulate them in such a way that I get a specific format.
string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo
I want to be able to get rid of any of the last string so I am just left with the month and year, like below:
string1-itd_jan2021
string2itd_mar2021
string3itd_feb2021
string4-itd_mar2021
string5itd_jun2021
string6-itd_feb2021
I thought about using string.split on the - but then realized that for some strings this wouldn't work. I also thought about getting rid of a set amount of characters by putting it into a list and slicing but the end is varying characters length?
Is there anything I can do it with regex or any other python module?
Use str.rsplit with the appropriate maxsplit parameter:
s = s.rsplit("-", 1)[0]
You could also use str.split (even though this is clearly the worse choice):
s = "-".join(s.split("-")[:-1])
Or using regular expressions:
s = re.sub(r'-[^-]*$', '', s)
# "-[^-]*" a "-" followed by any number of non-"-"
With a regex:
import re
re.sub(r'([0-9]{4}).*$', r'\1', s)
Use re.sub like so:
import re
lines = '''string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo'''
for old in lines.split('\n'):
new = re.sub(r'[-][^-]+$', '', old)
print('\t'.join([old, new]))
Prints:
string1-itd_jan2021-internal string1-itd_jan2021
string2itd_mar2021-space string2itd_mar2021
string3itd_feb2021-internal string3itd_feb2021
string4-itd_mar2021-moon string4-itd_mar2021
string5itd_jun2021-internal string5itd_jun2021
string6-itd_feb2021-apollo string6-itd_feb2021
Explanation:
r'[-][^-]+$' : Literal dash (-), followed by any character other than a dash ([^-]) repeated 1 or more times, followed by the end of the string ($).
I am handed a bunch of data and trying to get rid of certain characters. The data contains multiple instances of "^{number}" → "^0", "^1", "^2", etc.
I am trying to set all of these instances to an empty string, "", is there a better way to do this than
string.replace("^0", "").replace("^1", "").replace("^2", "")
I understand you can use a dictionary, but it seems a little overkill considering each item will be replaced with "".
I understand that the digits are always at the end of the string, have a look at the solutions below.
with regex:
import re
text = 'xyz125'
s = re.sub("\d+$",'', text)
print(s)
it should print:
'xyz'
without regex, keep in mind that this solution removes all digits and not only the ones at the end of a string:
text = 'xyz125'
result = ''.join(i for i in text if not i.isdigit())
print(result)
it should print:
'xyz'
In python I have long string like (of which I removed all breaks)
stringA = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
What I want to do is to search this string for all occurrences of "key:", then extract the "values" following "key:".
One further complication for me is that I don't know how long these values belonging to key are (e.g. key:12/eas9 and key:43/e3). All I do know is that they do have to end with a digit whereas the rest of the string does not contain any digits.
This is why my idea was to slice from the indices of key plus the next say 10 characters (e.g. key:12/eas9g) and then work backward until isdigit() is false.
I tried to split my initial string (that did contain breaks):
stringA_split = re.split("\n", stringA)
for linex in stringA_split:
index_start = linex.rfind("key:")
index_end = index_start + 8
print(linex[index_start:index_end]
#then work backward
However, inserting line breaks does not help in any way as they are meaningless from a pdf-to-txt conversion.
How would I then solve this (e.g. as a start with getting all indices of '"key:"' and slice this to a list)?
import re
>>> re.findall('key:(\d+[^\d]+[\d])', stringA)
['12/eas9', '43/e3']
\d+ # One or more digits.
[^\d]+ # Everything except a digit (equivalent to [\D]).
[\d] # The final digit
(\d+[^\d]+[\d]) # The group of the expression above
'key:(\d+[^\d]+[\d])' # 'key:' followed by the group expression
If you want key: in your result:
>>> re.findall('(key:\d+[^\d]+[\d])', stringA)
['key:12/eas9', 'key:43/e3']
I'm not 100% sure I understand your definition of what defines a value, but I think this will get you what you described
import re
stringA = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
for v in stringA.split('key:'):
ma = re.match(r'(\d+\/.*\d+)', v)
if ma:
print ma.group(1)
This returns:
12/eas9
43/e3
You can apply just one RE that gets all the keys into an array of tuples:
import re
p=re.compile('key\:(\d+)\/([^\d]+\d)')
ret=p.findall(stringA)
After the execution, you have:
ret
[('12', 'eas9'), ('43', 'e3')]
edit: a better answer was posted above. I misread the original question when proposing to reverse here, which really wasn't necessary. Good luck!
If you know that the format is always key:, what if you reversed the string and rex for :yek? You'd isolate all keys and then can reverse them back
import re
# \w is alphanumeric, you may want to add some symbols
rex = re.compile("\w*:yek")
word = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
matches = re.findall(rex, word[::-1])
matches = [match[::-1] for match in matches]
I am trying to do something which I thought would be simple (and probably is), however I am hitting a wall. I have a string that contains document numbers. In most cases the format is ######-#-### however in some cases, where the single digit should be, there are multiple single digits separated by a comma (i.e. ######-#,#,#-###). The number of single digits separated by a comma is variable. Below is an example:
For the string below:
('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
I need to return:
['030421-1-001', '030421-2-001' '030421-1-002', '030421-1-002', '030421-2-002', '030421-3-002' '030421-1-003']
I have only gotten as far as returning the strings that match the ######-#-### pattern:
import re
p = re.compile('\d{6}-\d{1}-\d{3}')
m = p.findall('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
print m
Thanks in advance for any help!
Matt
Perhaps something like this:
>>> import re
>>> s = '030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003'
>>> it = re.finditer(r'(\b\d{6}-)(\d(?:,\d)*)(-\d{3})\b', s)
>>> for m in it:
a, b, c = m.groups()
for x in b.split(','):
print a + x + c
...
030421-1-001
030421-2-001
030421-1-002
030421-1-002
030421-2-002
030421-3-002
030421-1-003
Or using a list comprehension
>>> [a+x+c for a, b, c in (m.groups() for m in it) for x in b.split(',')]
['030421-1-001', '030421-2-001', '030421-1-002', '030421-1-002', '030421-2-002', '030421-3-002', '030421-1-003']
Use '\d{6}-\d(,\d)*-\d{3}'.
* means "as many as you want (0 included)".
It is applied to the previous element, here '(,\d)'.
I wouldn't use a single regular expression to try and parse this. Since it is essentially a list of strings, you might find it easier to replace the "&" with a comma globally in the string and then use split() to put the elements into a list.
Doing a loop of the list will allow you to write a single function to parse and fix the string and then you can push it onto a new list and the display your string.
replace(string, '&', ',')
initialList = string.split(',')
for item in initialList:
newItem = myfunction(item)
newList.append(newItem)
newstring = newlist(join(','))
(\d{6}-)((?:\d,?)+)(-\d{3})
We take 3 capturing groups. We match the first part and last part the easy way. The center part is optionally repeated and optionally contains a ','. Regex will however only match the last one, so ?: won't store it at all. What where left with is the following result:
>>> p = re.compile('(\d{6}-)((?:\d,?)+)(-\d{3})')
>>> m = p.findall('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
>>> m
[('030421-', '1,2', '-001'), ('030421-', '1', '-002'), ('030421-', '1,2,3', '-002'), ('030421-', '1', '-003')]
You'll have to manually process the 2nd term to split them up and join them, but a list comprehension should be able to do that.
I have a lot of long strings - not all of them have the same length and content, so that's why I can't use indices - and I want to extract a string from all of them. This is what I want to extract:
http://www.someDomainName.com/anyNumber
SomeDomainName doesn't contain any numbers and and anyNumber is different in each long string. The code should extract the desired string from any string possible and should take into account spaces and any other weird thing that might appear in the long string - should be possible with regex right? -. Could anybody help me with this? Thank you.
Update: I should have said that www. and .com are always the same. Also someDomainName! But there's another http://www. in the string
import re
results = re.findall(r'\bhttp://www\.someDomainName\.com/\d+\b', long_string)
>>> import re
>>> pattern = re.compile("(http://www\\.)(\\w*)(\\.com/)(\\d+)")
>>> matches = pattern.search("http://www.someDomainName.com/2134")
>>> if matches:
print matches.group(0)
print matches.group(1)
print matches.group(2)
print matches.group(3)
print matches.group(4)
http://www.someDomainName.com/2134
http://www.
someDomainName
.com/
2134
In the above pattern, we have captured 5 groups -
One is the complete string that is matched
Rest are in the order of the brackets you see.. (So, you are looking for the second one..) - (\\w*)
If you want, you can capture only the part of the string you are interested in.. So, you can remove the brackets from rest of the pattern that you don't want and just keep (\w*)
>>> pattern = re.compile("http://www\\.(\\w*)\\.com/\\d+")
>>> matches = patter.search("http://www.someDomainName.com/2134")
>>> if matches:
print matches.group(1)
someDomainName
In the above example, you won't have groups - 2, 3 and 4, as in the previous example, as we have captured only 1 group.. And yes group 0 is always captured.. That is the complete string that matches..
Yeah, your simplest bet is regex. Here's something that will probably get the job done:
import re
matcher = re.compile(r'www.(.+).com\/(.+)
matches = matcher.search(yourstring)
if matches:
str1,str2 = matches.groups()
If you are sure that there are no dots in SomeDomainName you can just take the first occurence of the string ".com/" and take everything from that index on
this will avoid you the use of regex which are harder to maintain
exp = 'http://www.aejlidjaelidjl.com/alieilael'
print exp[exp.find('.com/')+5:]