return poker_hand(list_of_five_cards) returns a string similar to this:
**4-Diamonds/2-Clubs/5-Hearts/4-Spades/King-Spades (One pair.)
and I have created a string out of it I want the information inside the brackets. in this vein I have tried:
s = str(poker_hand(one_man))
print s
the_search = re.search(r"\((\w+)\)", s)
and this returns None when you type print the_search. I have also tried
s[s.find("(")+1:s.find(')')]
print s
which returns the whole string. Does anyone know what I am doing wrong?
EDIT sorry for the confusion I should be better,
input is 7-Spades/4-Clubs/3-Diamonds/3-Hearts/8-Spades (One pair.)
desired output is One pair
re the assigning... trying to assign it now, will post the results
the pattern you are using to find the item in brackets is not right.
you can try to test your regex in http://regexr.com/
import re
s = '**4-Diamonds/2-Clubs/5-Hearts/4-Spades/King-Spades (One pair.)'
pattern = r'\(.+\.\)'
for item in re.findall(pattern,s):
print item.strip('().')
output:
One pair
IIUC at the end of your string you always have the closed brackets. Then try this:
'**4-Diamonds/2-Clubs/5-Hearts/4-Spades/King-Spades (One pair.)'.split('(')[1][:-1]
Out[1]: 'One pair.'
The idea is to split by the opening brackets, taking what's after, and deleting the closing brackets.
input is 7-Spades/4-Clubs/3-Diamonds/3-Hearts/8-Spades (One pair.)
desired output is One pair
You can use something like:
import re
string = "7-Spades/4-Clubs/3-Diamonds/3-Hearts/8-Spades (One pair.)"
result = re.findall(r"\((.*?)\.?\)", string )
print result[0]
Ideone Demo
Regex Explanation:
\((.*?)\.?\)
Match the character “(” literally «\(»
Match the regex below and capture its match into backreference number 1 «(.*?)»
Match any single character that is NOT a line break character (line feed) «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “.” literally «\.?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match the character “)” literally «\)»
Use the groups:
import re
s = '**4-Diamonds/2-Clubs/5-Hearts/4-Spades/King-Spades (One pair.)'
print (s)
m = re.search(r'\(([\s\S]+)\.\)', s)
print(m.group(1))
Related
I currently investigate a problem that I want to replace something in a string.
For example. I have the following string:
'123.49, 19.30, 02\n'
I only want the first two numbers like '123.49, 19.30'. The split function is not possible, because a I have a lot of data and some with and some without the last number.
I tried something like this:
import re as regex
#result = regex.match(', (.*)\n', string)
result = re.search(', (.*)\\n', string)
print(result.group(1))
This is not working finde. Can someone help me?
Thanks in advance
You could do something like this:
reg=r'(\d+\.\d+), (\d+\.\d+).*'
if(re.search(reg, your_text)):
match = re.search(reg, your_text)
first_num = match.group(1)
second_num = match.group(2)
Alternatively, also adding the ^ sign at the beginning, making sure to always only take the first two.
import re
string = '123.49, 19.30, 02\n'
pattern = re.compile('^(\d*.?\d*), (\d*.?\d*)')
result = re.findall(pattern, string)
result
Output:
[('123.49', '19.30')]
In the code you are using import re as regex. If you do that, you would have to use regex.search instead or re.search.
But in this case you can just use re.
If you use , (.*) you would capture all after the first occurrence of , and you are not taking digits into account.
If you want the first 2 numbers as stated in the question '123.49, 19.30' separated by comma's you can match them without using capture groups:
\b\d+\.\d+,\s*\d+\.\d+\b
Or matching 1 or more repetitions preceded by a comma:
\b\d+\.\d+(?:,\s*\d+\.\d+)+\b
regex demo | Python demo
As re.search can also return None, you can first check if there is a result (no need to run re.search twice)
import re
regex = r"\b\d+\.\d+(?:,\s*\d+\.\d+)+\b"
s = "123.49, 19.30, 02"
match = re.search(regex, s)
if match:
print(match.group())
Output
123.49, 19.30
I am trying to find all occurrences of either "_"+digit or "^"+digit, using the regex ((_\^)[1-9])
The groups I'd expect back eg for "X_2ZZZY^5" would be [('_2'), ('^5')] but instead I am getting [('_2', '_'), ('^5', '^')]
Is my regex incorrect? Or is my expectation of what gets returned incorrect?
Many thanks
** my original re used (_|\^) this was incorrect, and should have been (_\^) -- question has been amended accordingly
You have 2 groups in your regex - so you're getting 2 groups. And you need to match atleast 1 number that follows.
try this:
([_\^][1-9]+)
See it in action here
Demand at least 1 digit (1-9) following the special characters _ or ^, placed inside a single capture group:
import re
text = "X_2ZZZY^5"
pattern = r"([_\^][1-9]{1,})"
regex = re.compile(pattern)
res = re.findall(regex, text)
print(res)
Returning:
['_2', '^5']
I have always 2 numbers in between and I want to extract everything before 3 so Salvatore and everything after 2 Abdulla
For example I have the following:
txt = "Salvatore32Abdulla"
first = re.findall("^\D+", txt)
last = re.search(,txt)
Expected result:
first = 'Salvatore'
last = 'Abdulla'
I can get the first part, but after 2 I can't get the last part
You could also do this in a single line by slightly changing the solution suggested by #ctwheels as follows. I would suggest you to use re.findall as that gets the job done with a single blow.
import re
txt = "Salvatore32Abdulla"
Option-1
Single line extraction of the non-numeric parts.
first, last = re.findall("\D+", txt)
print((first, last))
('Salvatore', 'Abdulla')
Option-2
If you would (for some reason) also want to keep track of the number in between:
first, num, last = re.findall("(\D+)(\d{2})(\D+)", txt)
print((first, num, last))
('Salvatore', '32', 'Abdulla')
Option-3
As an extension of Option-2 and considering the text with a form 'Salvatore####...###Abdulla', where ####...### denotes a continuous block of digits separating the non-numeric parts and you may or may not have any idea of how many digits could be in-between, you could use the following:
first, num, last = re.findall("(\D+)(\d*)(\D+)", txt)
print((first, num, last))
('Salvatore', '32', 'Abdulla')
Why am I not getting the expected results?
You currently have one issue with your regex and one with your code.
Your regex contains ^, which anchors it to the start of the string. This will only allow you to match Salvatore. You're using findall (which is the appropriate choice if you change the regex to simply \D+), but right now it's only getting one result.
The second re.search call is not needed as you can capture first and last with the findall given an appropriate pattern (see below).
How do I fix it?
See code in use here
import re
txt = "Salvatore32Abdulla"
x = re.findall("\D+", txt)
print(x)
Result:
['Salvatore', 'Abdulla']
You could use a regex like this:
txt = "Salvatore32Abdulla"
regex = r"(\D+)\d\d(\D+)"
match = re.match(regex, txt)
first = match.group(1)
last = match.group(2)
Part after last digit:
match = re.search(r'\D+$',txt)
if match:
print(match.group())
See Python proof | regex proof.
Results: Abdulla
EXPLANATION
--------------------------------------------------------------------------------
\D+ non-digits (all but 0-9) (1 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
I have a string in which there are certain values that I need to extract from it. For example: "FEFEWFSTARTFFFPENDDCDC". How could I make an expression that would take a slice from "START" all the way to "END"?
I tried doing this previously by creating functions which used a for loop and string.find("START") to locate the beginning and ends, but this didn't appear to work effectively and seemed overly complex. Is there an easier way to do this without using complex loops?
EDIT:
Forgot this part. What if there were different end values? In other words, instead of just ending with "END", the values "DONE" and "NOMORE" would also end it? And in addition to that, there were multiple starts and ends throughout the string. For example: "STARTFFEFFDONEFEWFSTARTFEFFENDDDW".
EDIT2: Sample run: Start value: ATG. End values: TAG,TAA,TGA
"Enter a string": TTATGTTTTAAGGATGGGGCGTTAGTT
TTT
GGGCGT
And
"Enter a string": TGTGTGTATAT
"No string found"
That's a perfect fit for a regular expression:
>>> import re
>>> s = "FEFEWFSTARTFFFPENDDCDCSTARTDOINVOIJHSDFDONEDFOIER"
>>> re.findall("START.*?(?:END|DONE|NOMORE)", s)
['STARTFFFPEND', 'STARTDOINVOIJHSDFDONE']
.* matches any number of characters (except newlines), the additional ? makes the quantifier lazy, telling it to match as few characters as possible. Otherwise, there would be only one match, namely STARTFFFPENDDCDCSTARTDOINVOIJHSDFDONE.
As #BurhanKhalid noted, if you add a capturing group, only the substring matched by that part of the regex will be captured:
>>> re.findall("START(.*?)(?:END|DONE|NOMORE)", s)
['FFFP', 'DOINVOIJHSDF']
Explanation:
START # Match "START"
( # Match and capture in group number 1:
.*? # Any character, any number of times, as few as possible
) # End of capturing group 1
(?: # Start a non-capturing group that matches...
END # "END"
| # or
DONE # "DONE"
| # or
NOMORE # "NOMORE"
) # End of non-capturing group
And if your real goal is to match gene sequences, you need to make sure that you always match triplets:
re.findall("ATG(?:.{3})*?(?:TA[AG]|TGA)", s)
a="FEFEWFSTARTFFFPENDDCDC"
a[a.find('START'):]
'STARTFFFPENDDCDC'
The simple way (no loop, no regex):
s = "FEFEWFSTARTFFFPENDDCDC"
tmp = s[s.find("START") + len("START"):]
result = tmp[:tmp.find("END")]
yourString = 'FEFEWFSTARTFFFPENDDCDC'
substring = yourString[yourString.find("START") + len("START") : yourString.find("END")]
Not that efficient but does work.
>>> s = "FEFEWFSTARTFFFPENDDCDC"
>>> s[s.index('START'):s.index('END')+len('END')]
'STARTFFFPEND'
I am quite new to python and regex (regex newbie here), and I have the following simple string:
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
I would like to extract only the last digits in the above string i.e 767980716 and I was wondering how I could achieve this using python regex.
I wanted to do something similar along the lines of:
re.compile(r"""-(.*?)""").search(str(s)).group(1)
indicating that I want to find the stuff in between (.*?) which starts with a "-" and ends at the end of string - but this returns nothing..
I was wondering if anyone could point me in the right direction..
Thanks.
You can use re.match to find only the characters:
>>> import re
>>> s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
>>> re.match('.*?([0-9]+)$', s).group(1)
'767980716'
Alternatively, re.finditer works just as well:
>>> next(re.finditer(r'\d+$', s)).group(0)
'767980716'
Explanation of all regexp components:
.*? is a non-greedy match and consumes only as much as possible (a greedy match would consume everything except for the last digit).
[0-9] and \d are two different ways of capturing digits. Note that the latter also matches digits in other writing schemes, like ୪ or ൨.
Parentheses (()) make the content of the expression a group, which can be retrieved with group(1) (or 2 for the second group, 0 for the whole match).
+ means multiple entries (at least one number at the end).
$ matches only the end of the input.
Nice and simple with findall:
import re
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
print re.findall('^.*-([0-9]+)$',s)
>>> ['767980716']
Regex Explanation:
^ # Match the start of the string
.* # Followed by anthing
- # Upto the last hyphen
([0-9]+) # Capture the digits after the hyphen
$ # Upto the end of the string
Or more simply just match the digits followed at the end of the string '([0-9]+)$'
Your Regex should be (\d+)$.
\d+ is used to match digit (one or more)
$ is used to match at the end of string.
So, your code should be: -
>>> s = "99-my-name-is-John-Smith-6376827-%^-1-2-767980716"
>>> import re
>>> re.compile(r'(\d+)$').search(s).group(1)
'767980716'
And you don't need to use str function here, as s is already a string.
Use the below regex
\d+$
$ depicts the end of string..
\d is a digit
+ matches the preceding character 1 to many times
Save the regular expressions for something that requires more heavy lifting.
>>> def parse_last_digits(line): return line.split('-')[-1]
>>> s = parse_last_digits(r"99-my-name-is-John-Smith-6376827-%^-1-2-767980716")
>>> s
'767980716'
I have been playing around with several of these solutions, but many seem to fail if there are no numeric digits at the end of the string. The following code should work.
import re
W = input("Enter a string:")
if re.match('.*?([0-9]+)$', W)== None:
last_digits = "None"
else:
last_digits = re.match('.*?([0-9]+)$', W).group(1)
print("Last digits of "+W+" are "+last_digits)
Try using \d+$ instead. That matches one or more numeric characters followed by the end of the string.