I have been writing a program recently and a part of it requires me to get information form inside a string. I need to find where there is 1 letter immediately followed by 2 numbers (e.g. S07) and I can't work out the RegEx for it.
def get_season(filenames):
pattern = "^[a-zA-z]{1}[\d]{2}$"
found = re.search(filenames[0], pattern)
season_name = found.string
season = season_name[1:3]
print(season)
I know that this information is in the string but it keeps giving me "None" in response
(I'm not too sure if the code section has formatted correctly, in the preview it shows as on the same line, but the indentation in my program is correct)
You swapped the arguments to re.search(). The first argument is the pattern, not the string to match:
found = re.search(pattern, filenames[0])
Your pattern is also overly wide; A-z matches everything between Z (uppercase) and a (lowercase) too. The correct pattern is:
pattern = "^[a-zA-Z]\d{2}$"
where {1} is the default, so I omitted that.
If you are matching this against filenames, you probably do not want to use the start or end anchors, that would limit matches to exact strings only:
>>> re.search("^[a-zA-Z]\d{2}$", "S07").string
'S20'
>>> re.search("^[a-zA-Z]\d{2}$", "S07E01 - Meet the New Boss.avi") is None
True
>>> re.search("^[a-zA-Z]\d{2}$", "S07E01 - Meet the New Boss.avi") is None
True
>>> re.search("[a-zA-Z]\d{2}", "S07E01 - Meet the New Boss.avi").string
'S07E01 - Meet the New Boss.avi'
And you want to use .group() to get the matched portion, not string (which is the original input string):
>>> re.search("[a-zA-Z]\d{2}", "S07E01 - Meet the New Boss.avi").group()
'S07'
If you only wanted the numbers, you need to add a group, and pick that. You create a capturing group with parenthesis:
>>> re.search("[a-zA-Z](\d{2})", "S07E01 - Meet the New Boss.avi").group(1)
'07'
This selects the first group (.group(1)), which is the parenthesis around the 2 digits portion.
Your regex will catch only the string which consits only of one letter and two digits, to check whole string for multiple occurences use these:
Try this regex:
[a-zA-Z]\d{2}
INPUT
asdasdasS01asfasfsa
OUTPUT
S01
If you want to find a word wich consists only of a letter followed by two digits use this regex:
\b[a-zA-Z]\d{2}\b
Only numebers capture regex:
[a-zA-Z](\d{2})
INPUT
asdasdasS01asfasfsa
OUTPUT
01
Also swap the arguments in serach method.
Related
I'm learning about regular expressions and I to want extract a string from a text that has the following characteristic:
It always begins with the letter C, in either lowercase or
uppercase, which is then followed by a number of hexadecimal
characters (meaning it can contain the letters A to F and numbers
from 1 to 9, with no zeros included).
After those hexadecimal
characters comes a letter P, also either in lowercase or uppercase
And then some more hexadecimal characters (again, excluding 0).
Meaning I want to capture the strings that come in between the letters C and P as well as the string that comes after the letter P and concatenate them into a single string, while discarding the letters C and P
Examples of valid strings would be:
c45AFP2
CAPF
c56Bp26
CA6C22pAAA
For the above examples what I want would be to extract the following, in the same order:
45AF2 # Original string: c45AFP2
AF # Original string: CAPF
56B26 # Original string: c56Bp26
A6C22AAA # Original string: CA6C22pAAA
Examples of invalid strings would be:
BCA6C22pAAA # It doesn't begin with C
c56Bp # There aren't any characters after P
c45AF0P2 # Contains a zero
I'm using python and I want a regex to extract the two strings that come both in between the characters C and P as well as after P
So far I've come up with this:
(?<=\A[cC])[a-fA-F1-9]*(?<=[pP])[a-fA-F1-9]*
A breakdown would be:
(?<=\A[cC]) Positive lookbehind assertion. Asserts that what comes before the regex parser’s current position must match [cC] and that [cC] must be at the beginning of the string
[a-fA-F1-9]* Matches a single character in the list between zero and unlimited times
(?<=[pP]) Positive lookbehind assertion. Asserts that what comes before the regex parser’s current position must match [pP]
[a-fA-F1-9]* Matches a single character in the list between zero and unlimited times
But with the above regex I can't match any of the strings!
When I insert a | in between (?<=[cC])[a-fA-F1-9]* and (?<=[pP])[a-fA-F1-9]* it works.
Meaning the below regex works:
(?<=[cC])[a-fA-F1-9]*|(?<=[pP])[a-fA-F1-9]*
I know that | means that it should match at most one of the specified regex expressions. But it's non greedy and it returns the first match that it finds. The remaining expressions aren’t tested, right?
But using | means the string BCA6C22pAAA is a partial match to AAA since it comes after P, even though the first assertion isn't true, since it doesn't begin with a C.
That shouldn't be the case. I want it to only match if all conditions explained in the beginning are true.
Could someone explain to me why my first attempt doesn't produces the result I want? Also, how can I improve my regex?
I still need it to:
Not be a match if the string contains the number 0
Only be a match if ALL conditions are met
Thank you
To match both groups before and after P or p
(?<=^[Cc])[1-9a-fA-F]+(?=[Pp]([1-9a-fA-F]+$))
(?<=^[Cc]) - Positive Lookbehind. Must match a case insensitive C or c at the start of the line
[1-9a-fA-F]+ - Matches hexadecimal characters one or more times
(?=[Pp] - Positive Lookahead for case insensitive p or P
([1-9a-fA-F]+$) - Cature group for one or more hexadecimal characters following the pP
View Demo
Your main problem is you're using a look behind (?<=[pP]) for something ahead, which will never work: You need a look ahead (?=...).
Also, the final quantifier should be + not * because you require at least one trailing character after the p.
The final mistake is that you're not capturing anything, you're only matching, so put what you want to capture inside brackets, which also means you can remove all look arounds.
If you use the case insensitive flag, it makes the regex much smaller and easier to read.
A working regex that captures the 2 hex parts in groups 1 and 2 is:
(?i)^c([a-f1-9]*)p([a-f1-9]+)
See live demo.
Unless you need to use \A, prefer ^ (start of input) over \A (start of all input in multi line scenario) because ^ easier to read and \A won't match every line, which is what many situations and tools expect. I've used ^.
I have a list of 4000 strings. The naming convention needs to be changed for each string and I do not want to go through and edit each one individually.
The list looks like this:
data = list()
data = ['V2-FG2110-EMA-COMPRESSION',
'V2-FG2110-SA-COMPRESSION',
'V2-FG2110-UMA-COMPRESSION',
'V2-FG2120-EMA-DISTRIBUTION',
'V2-FG2120-SA-DISTRIBUTION',
'V2-FG2120-UMA-DISTRIBUTION',
'V2-FG2140-EMA-HEATING',
'V2-FG2140-SA-HEATING',
'V2-FG2140-UMA-HEATING',
'V2-FG2150-EMA-COOLING',
'V2-FG2150-SA-COOLING',
'V2-FG2150-UMA-COOLING',
'V2-FG2160-EMA-TEMPERATURE CONTROL']
I need all each 'SA' 'UMA' and 'EMA' to be moved to before the -FG.
Desired output is:
V2-EMA-FG2110-Compression
V2-SA-FG2110-Compression
V2-UMA-FG2110-Compression
...
The V2-FG2 does not change throughout the list so I have started there and I tried re.sub and re.search but I am pretty new to python so I have gotten a mess of different results. Any help is appreciated.
You can rearrange the strings.
new_list = []
for word in data:
arr = word.split('-')
new_word = '%s-%s-%s-%s'% (arr[0], arr[2], arr[1], arr[3])
new_list.append(new_word)
You can replace matches of the following regular expression with the contents of capture group 1:
(?<=^[A-Z]\d)(?=.*(-(?:EMA|SA|UMA))(?=-))|-(?:EMA|SA|UMA)(?=-)
Demo
The regular expression can be broken down as follows.
(?<=^[A-Z]\d) # current string position must be preceded by a capital
# letter followed by a digit at the start of the string
(?= # begin a positive lookahead
.* # match >= 0 chars other than a line terminator
(-(?:EMA|SA|UMA)) # match a hyphen followed by one of the three strings
# and save to capture group 1
(?=-) # the next char must be a hyphen
) # end positive lookahead
| # or
-(?:EMA|SA|UMA) # match a hyphen followed by one of the three strings
(?=-) # the next character must be a hyphen
(?=-) is a positive lookahead.
Evidently this may not work for versions of Python prior to 3.5, because the match in the second part of the alternation does not assign a value to capture group 1: "Before Python 3.5, backreferences to failed capture groups in Python re.sub were not populated with an empty string.. This quote is from
#WiktorStribiżew 's answer at the link. For what it's worth I confirmed that Ruby has the same behaviour ("V2-FG2110-EMA-COMPRESSION".gsub(rgx,'\1') #=> "V2-EMA-FG2110-COMPRESSION").
One could of course instead replace matches of (?<=^[A-Z]\d)(-[A-Z]{2}\d{4})(-(?:EMA|SA|UMA))(?=-)) with $2 + $1. That's probably more sensible even if it's less interesting.
I have two kinds of documents to parse:
1545994641 INFO: ...
and
'{"deliveryDate":"1545994641","error"..."}'
I want to extract the timestamp 1545994641 from each of them.
So, I decided to write a regex to match both cases:
(\d{10}\s|\"\d{10}\")
In the 1st kind of document, it matches the timestamp and groups it, using the first expression in the "or" above (\d{10}\s):
>>> regex = re.compile("(\d{10}\s|\"\d{10}\")")
>>> msg="1545994641 INFO: ..."
>>> regex.search(msg).group(0)
'1545994641 '
(So far so good.)
However, in the 2nd kind, using the second expression in the "or" (\"\d{10}\") it matches the timestamp and quotation marks, grouping them. But I just want the timestamp, not the "":
>>> regex = re.compile("(\d{10}\s|\"\d{10}\")")
>>> msg='{"deliveryDate":"1545994641","error"..."}'
>>> regex.search(msg).group(0)
'"1545994641"'
What I tried:
I decided to use a non-capturing group for the quotation marks:
(\d{10}\s|(?:\")\d{10}(?:\"))
but it doesn't work as the outer group catches them.
I also removed the outer group, but the result is the same.
Unwanted ways to solve:
I can surpass this by creating a group for each expression in the or,
but I just want it to output a single group (to abstract the code
from the regex).
I could also use a 2nd step of regex to capture the timestamp from
the group that has the quotation marks, but again that would break
the code abstraction.
I could omit the "" in the regex but that would match a timestamp in the middle of the message , as I want it to be objective to capture the timestamp as a value of a key or in the beginning of the document, followed by a space.
Is there a way I can match both cases above but, in the case it matches the second case, return only the timestamp? Or is it impossible?
EDIT:
As noticed by #Amit Bhardwaj, the first case also returns a space after the timestamp. It's another problem (I didn't figure out) with the same solution, probably!
You may use lookarounds if your code can only access the whole match:
^\d{10}(?=\s)|(?<=")\d{10}(?=")
See the regex demo.
In Python, declare it as
rx = r'^\d{10}(?=\s)|(?<=")\d{10}(?=")'
Pattern details
^\d{10}(?=\s):
^ - string start
\d{10} - ten digits
(?=\s) - a positive lookahead that requires a whitespace char immediately to the right of the current location
| - or
(?<=")\d{10}(?="):
(?<=") - a " char
\d{10} - ten digits
(?=") - a positive lookahead that requires a double quotation mark immediately to the right of the current location.
You could use lookarounds, but I think this solution is simpler, if you can just get the group:
"?(\d{10})(?:\"|\s)
EDIT:
Considering if there is a first " there must be a ", try this:
(^\d{10}\s|(?<=\")\d{10}(?=\"))
EDIT 2:
To also remove the trailing space in the end, use a lookahead too:
(^\d{10}(?=\s)|(?<=\")\d{10}(?=\"))
This is my first post and I am a newbie to Python. I am trying to get this to work.
string 1 = [1/0/1, 1/0/2]
string 2 = [1/1, 1/2]
Trying to check the string if I see two / then I just need to replace the 0 with 1 so it becomes 1/1/1 and 1/1/2.
If I don't have two / then I need to add one in along with a 1 and change it to the format 1/1/1 and 1/1/2 so string 2 becomes [1/1/1,1/1/2]
Ultimate goal is to get all strings match the pattern x/1/x. Thanks for all the Input on this.I tried this and it seems to work
for a in Port:
if re.search(r'././', a):
z.append(a.replace('/0/','/1/') )
else:
t1= a.split('/')
if len(t1)>1 :
t2= t1[0] + "/1/" + t1[1]
z.append(t2)
few lines are there to take care of some exceptions but seems to do the job.
The regex pattern for identifying a / is just \/
This could be solved rather simply using the built in string functions without having to add all of the overhead and additional computational time caused by using the RegEx engine.
For example:
# The string to test:
sTest = '1/0/2'
# Test the string:
if(sTest.count('/') == 2):
# There are two forward slashes in the string
# If the middle number is a 0, we'll replace it with a one:
sTest = sTest.replace('/0/', '/1/')
elif(sTest.count('/') == 1):
# One forward slash in string
# Insert a 1 between first portion and the last portion:
sTest = sTest.replace('/', '/1/')
else:
print('Error: Test string is of an unknown format.')
# End If
If you really want to use RegEx, though, you could simply match the string against these two patterns: \d+/0/\d+ and \d+/\d+(?!/) If matching against the first pattern fails, then attempt to match against the second pattern. Then, you can use a either grouping, splitting, or simply calling .replace() (like I'm doing above) to format the string as you need.
EDIT: for clarification, I'll explain the two patterns:
Pattern 1: \d+/0/\d+ could essentially be read as "match any number (consisting of one (1) or more digits) followed by a forward slash, a zero (0), another forward slash and then followed by any number (consisting of one (1) or more digits).
Pattern 2: \d+/\d+(?!/) could be read as "match any number (consisting of one (1) or more digits) followed by a forward slash and any other number (consisting of one (1) or more digits) which is then NOT followed by another forward slash." The last part in this pattern could be a little confusing because it uses the negative lookahead abilities of the RegEx engine.
If you wanted to add stricter rules to these patterns to make sure there are not any leading or trailing non-digit characters, you could add ^ to the start of the patterns and $ to the end, to signify the start of the string and the end of the string respectively. This would also allow you to remove the lookahead expression from the second pattern ((?!/)). As such, you would end up with the following patterns: ^\d+/0/\d+$ and ^\d+/\d+$.
https://regex101.com/r/rE6oN2/1
Click code generator on the left side. You get:
import re
p = re.compile(ur'\d/1/\d')
test_str = u"1/1/2"
re.search(p, test_str)
I am trying to parse a chemical formula that is given to me in unicode in the format C7H19N3
I wish to isolate the position of the first number after the letter, I.e 7 is at index 1 and 1 is at index 3. With is this i want to insert "sub" infront of the digits
My first couple attempts had me looping though trying to isolate the position of only the first numbers but to no avail.
I think that Regular expressions can accomplish this, though im quite lost in it.
My end goal is to output the formula Csub7Hsub19Nsub3 so that my text editor can properly format it.
How about this?
>>> re.sub('(\d+)', 'sub\g<1>', "C7H19N3")
'Csub7Hsub19Nsub3'
(\d+) is a capturing group that matches 1 or more digits. \g<1> is a way of referring to the saved group in the substitute string.
Something like this with lookahead and lookbehind:
>>> strs = 'C7H19N3'
>>> re.sub(r'(?<!\d)(?=\d)','sub',strs)
'Csub7Hsub19Nsub3'
This matches the following positions in the string:
C^7H^19N^3 # ^ represents the positions matched by the regex.
Here is one which literally matches the first digit after a letter:
>>> re.sub(r'([A-Z])(\d)', r'\1sub\2', "C7H19N3")
'Csub7Hsub19Nsub3'
It's functionally equivalent but perhaps more expressive of the intent? \1 is a shorter version of \g<1>, and I also used raw string literals (r'\1sub\2' instead of '\1sub\2').