Find "one letter that appears twice" in a string
Question:
I’m trying to catch if one letter that appears twice in a string using RegEx (or maybe there’s some better ways?), for example my string is:
ugknbfddgicrmopn
The output would be:
dd
However, I’ve tried something like:
re.findall('[a-z]{2}', 'ugknbfddgicrmopn')
but in this case, it returns:
['ug', 'kn', 'bf', 'dd', 'gi', 'cr', 'mo', 'pn'] # the except output is `['dd']`
I also have a way to get the expect output:
>>> l = []
>>> tmp = None
>>> for i in 'ugknbfddgicrmopn':
... if tmp != i:
... tmp = i
... continue
... l.append(i*2)
...
...
>>> l
['dd']
>>>
But that’s too complex…
If it’s 'abbbcppq', then only catch:
abbbcppq
^^ ^^
So the output is:
['bb', 'pp']
Then, if it’s 'abbbbcppq', catch bb twice:
abbbbcppq
^^^^ ^^
So the output is:
['bb', 'bb', 'pp']
Answers:
As a Pythonic way You can use zip function within a list comprehension:
>>> s = 'abbbcppq'
>>>
>>> [i+j for i,j in zip(s,s[1:]) if i==j]
['bb', 'bb', 'pp']
If you are dealing with large string you can use iter() function to convert the string to an iterator and use itertols.tee() to create two independent iterator, then by calling the next function on second iterator consume the first item and use call the zip class (in Python 2.X use itertools.izip() which returns an iterator) with this iterators.
>>> from itertools import tee
>>> first = iter(s)
>>> second, first = tee(first)
>>> next(second)
'a'
>>> [i+j for i,j in zip(first,second) if i==j]
['bb', 'bb', 'pp']
Benchmark with RegEx recipe:
# ZIP
~ $ python -m timeit --setup "s='abbbcppq'" "[i+j for i,j in zip(s,s[1:]) if i==j]"
1000000 loops, best of 3: 1.56 usec per loop
# REGEX
~ $ python -m timeit --setup "s='abbbcppq';import re" "[i[0] for i in re.findall(r'(([a-z])2)', 'abbbbcppq')]"
100000 loops, best of 3: 3.21 usec per loop
After your last edit as mentioned in comment if you want to only match one pair of b in strings like "abbbcppq" you can use finditer() which returns an iterator of matched objects, and extract the result with group() method:
>>> import re
>>>
>>> s = "abbbcppq"
>>> [item.group(0) for item in re.finditer(r'([a-z])1',s,re.I)]
['bb', 'pp']
Note that re.I is the IGNORECASE flag which makes the RegEx match the uppercase letters too.
Using back reference, it is very easy:
import re
p = re.compile(ur'([a-z])1{1,}')
re.findall(p, u"ugknbfddgicrmopn")
#output: [u'd']
re.findall(p,"abbbcppq")
#output: ['b', 'p']
For more details, you can refer to a similar question in perl: Regular expression to match any character being repeated more than 10 times
>>> l = ['ug', 'kn', 'bf', 'dd', 'gi', 'cr', 'mo', 'pn']
>>> import re
>>> newList = [item for item in l if re.search(r"([a-z]{1})1", item)]
>>> newList
['dd']
You need use capturing group based regex and define your regex as raw string.
>>> re.search(r'([a-z])1', 'ugknbfddgicrmopn').group()
'dd'
>>> [i+i for i in re.findall(r'([a-z])1', 'abbbbcppq')]
['bb', 'bb', 'pp']
or
>>> [i[0] for i in re.findall(r'(([a-z])2)', 'abbbbcppq')]
['bb', 'bb', 'pp']
Note that , re.findall here should return the list of tuples with the characters which are matched by the first group as first element and the second group as second element. For our case chars within first group would be enough so I mentioned i[0].
Maybe you can use the generator to achieve this
def adj(s):
last_c = None
for c in s:
if c == last_c:
yield c * 2
last_c = c
s = 'ugknbfddgicrmopn'
v = [x for x in adj(s)]
print(v)
# output: ['dd']
A1 = "abcdededdssffffccfxx"
print A1[1]
for i in range(len(A1)-1):
if A1[i+1] == A1[i]:
if not A1[i+1] == A1[i-1]:
print A1[i] *2
“or maybe there’s some better ways”
Since regex is often misunderstood by the next developer to encounter your code (may even be you),
And since simpler != shorter,
How about the following pseudo-code:
function findMultipleLetters(inputString) {
foreach (letter in inputString) {
dictionaryOfLettersOccurrance[letter]++;
if (dictionaryOfLettersOccurrance[letter] == 2) {
multipleLetters.add(letter);
}
}
return multipleLetters;
}
multipleLetters = findMultipleLetters("ugknbfddgicrmopn");
It is pretty easy without regular expressions:
In [4]: [k for k, v in collections.Counter("abracadabra").items() if v==2]
Out[4]: ['b', 'r']
I’m trying to catch if one letter that appears twice in a string using RegEx (or maybe there’s some better ways?), for example my string is:
ugknbfddgicrmopn
The output would be:
dd
However, I’ve tried something like:
re.findall('[a-z]{2}', 'ugknbfddgicrmopn')
but in this case, it returns:
['ug', 'kn', 'bf', 'dd', 'gi', 'cr', 'mo', 'pn'] # the except output is `['dd']`
I also have a way to get the expect output:
>>> l = []
>>> tmp = None
>>> for i in 'ugknbfddgicrmopn':
... if tmp != i:
... tmp = i
... continue
... l.append(i*2)
...
...
>>> l
['dd']
>>>
But that’s too complex…
If it’s 'abbbcppq', then only catch:
abbbcppq
^^ ^^
So the output is:
['bb', 'pp']
Then, if it’s 'abbbbcppq', catch bb twice:
abbbbcppq
^^^^ ^^
So the output is:
['bb', 'bb', 'pp']
As a Pythonic way You can use zip function within a list comprehension:
>>> s = 'abbbcppq'
>>>
>>> [i+j for i,j in zip(s,s[1:]) if i==j]
['bb', 'bb', 'pp']
If you are dealing with large string you can use iter() function to convert the string to an iterator and use itertols.tee() to create two independent iterator, then by calling the next function on second iterator consume the first item and use call the zip class (in Python 2.X use itertools.izip() which returns an iterator) with this iterators.
>>> from itertools import tee
>>> first = iter(s)
>>> second, first = tee(first)
>>> next(second)
'a'
>>> [i+j for i,j in zip(first,second) if i==j]
['bb', 'bb', 'pp']
Benchmark with RegEx recipe:
# ZIP
~ $ python -m timeit --setup "s='abbbcppq'" "[i+j for i,j in zip(s,s[1:]) if i==j]"
1000000 loops, best of 3: 1.56 usec per loop
# REGEX
~ $ python -m timeit --setup "s='abbbcppq';import re" "[i[0] for i in re.findall(r'(([a-z])2)', 'abbbbcppq')]"
100000 loops, best of 3: 3.21 usec per loop
After your last edit as mentioned in comment if you want to only match one pair of b in strings like "abbbcppq" you can use finditer() which returns an iterator of matched objects, and extract the result with group() method:
>>> import re
>>>
>>> s = "abbbcppq"
>>> [item.group(0) for item in re.finditer(r'([a-z])1',s,re.I)]
['bb', 'pp']
Note that re.I is the IGNORECASE flag which makes the RegEx match the uppercase letters too.
Using back reference, it is very easy:
import re
p = re.compile(ur'([a-z])1{1,}')
re.findall(p, u"ugknbfddgicrmopn")
#output: [u'd']
re.findall(p,"abbbcppq")
#output: ['b', 'p']
For more details, you can refer to a similar question in perl: Regular expression to match any character being repeated more than 10 times
>>> l = ['ug', 'kn', 'bf', 'dd', 'gi', 'cr', 'mo', 'pn']
>>> import re
>>> newList = [item for item in l if re.search(r"([a-z]{1})1", item)]
>>> newList
['dd']
You need use capturing group based regex and define your regex as raw string.
>>> re.search(r'([a-z])1', 'ugknbfddgicrmopn').group()
'dd'
>>> [i+i for i in re.findall(r'([a-z])1', 'abbbbcppq')]
['bb', 'bb', 'pp']
or
>>> [i[0] for i in re.findall(r'(([a-z])2)', 'abbbbcppq')]
['bb', 'bb', 'pp']
Note that , re.findall here should return the list of tuples with the characters which are matched by the first group as first element and the second group as second element. For our case chars within first group would be enough so I mentioned i[0].
Maybe you can use the generator to achieve this
def adj(s):
last_c = None
for c in s:
if c == last_c:
yield c * 2
last_c = c
s = 'ugknbfddgicrmopn'
v = [x for x in adj(s)]
print(v)
# output: ['dd']
A1 = "abcdededdssffffccfxx"
print A1[1]
for i in range(len(A1)-1):
if A1[i+1] == A1[i]:
if not A1[i+1] == A1[i-1]:
print A1[i] *2
“or maybe there’s some better ways”
Since regex is often misunderstood by the next developer to encounter your code (may even be you),
And since simpler != shorter,
How about the following pseudo-code:
function findMultipleLetters(inputString) {
foreach (letter in inputString) {
dictionaryOfLettersOccurrance[letter]++;
if (dictionaryOfLettersOccurrance[letter] == 2) {
multipleLetters.add(letter);
}
}
return multipleLetters;
}
multipleLetters = findMultipleLetters("ugknbfddgicrmopn");
It is pretty easy without regular expressions:
In [4]: [k for k, v in collections.Counter("abracadabra").items() if v==2]
Out[4]: ['b', 'r']