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Match strings with alternating characters

I want to match strings in which every second character is same.
for example 'abababababab'
I have tried this : '''(([a-z])[^/2])*'''
The output should return the complete string as it is like 'abababababab'

This is actually impossible to do in a real regular expression with an amount of states polynomial to the alphabet size, because the expression is not a Chomsky level-0 grammar.
However, Python's regexes are not actually regular expressions, and can handle much more complex grammars than that. In particular, you could put your grammar as the following.
(..)\1*
(..) is a sequence of 2 characters. \1* matches the exact pair of characters an arbitrary (possibly null) number of times.
I interpreted your question as wanting every other character to be equal (ababab works, but abcbdb fails). If you needed only the 2nd, 4th, ... characters to be equal you can use a similar one.
.(.)(.\1)*

You could match the first [a-z] followed by capturing ([a-z]) in a group. Then repeat 0+ times matching again a-z and a backreference to group 1 to keep every second character the same.
^[a-z]([a-z])(?:[a-z]\1)*$
Explanation
^ Start of the string
[a-z]([a-z]) Match a-z and capture in group 1 matching a-z
)(?:[a-z]\1)* Repeat 0+ times matching a-z followed by a backreference to group 1
$ End of string
Regex demo

Though not a regex answer, you could do something like this:
def all_same(string):
return all(c == string[1] for c in string[1::2])
string = 'abababababab'
print('All the same {}'.format(all_same(string)))
string = 'ababacababab'
print('All the same {}'.format(all_same(string)))
the string[1::2] says start at the 2nd character (1) and then pull out every second character (the 2 part).
This returns:
All the same True
All the same False

This is a bit complicated expression, maybe we would start with:
^(?=^[a-z]([a-z]))([a-z]\1)+$
if I understand the problem right.
Demo

Regular expression misses match at beginning of string

I have strings of as and bs. I want to extract all overlapping subsequences, where a subsequence is a single a surrounding by any number of bs. This is the regex I wrote:
import re
pattern = """(?= # inside lookahead for overlapping results
(?:a|^) # match at beginning of str or after a
(b* (?:a) b*) # one a between any number of bs
(?:a|$)) # at end of str or before next a
"""
a_between_bs = re.compile(pattern, re.VERBOSE)
It seems to work as expected, except when the very first character in the string is an a, in which case this subsequence is missed:
a_between_bs.findall("bbabbba")
# ['bbabbb', 'bbba']
a_between_bs.findall("abbabb")
# ['bbabb']
I don't understand what is happening. If I change the order of how a potential match could start, the results also change:
pattern = """(?=
(?:^|a) # a and ^ swapped
(b* (?:a) b*)
(?:a|$))
"""
a_between_bs = re.compile(pattern, re.VERBOSE)
a_between_bs.findall("abbabb")
# ['abb']
I would have expected this to be symmetric, so that strings ending in an a might also be missed, but this doesn't appear to be the case. What is going on?
Edit:
I assumed that solutions to the toy example above would translate to my full problem, but that doesn't seem to be the case, so I'm elaborating now (sorry about that). I am trying to extract "syllables" from transcribed words. A "syllable" is a vowel or a diphtongue, preceded and followed by any number of consonants. This is my regular expression to extract them:
vowels = 'æɑəɛiɪɔuʊʌ'
diphtongues = "|".join(('aj', 'aw', 'ej', 'oj', 'ow'))
consonants = 'θwlmvhpɡŋszbkʃɹdnʒjtðf'
pattern = f"""(?=
(?:[{vowels}]|^|{diphtongues})
([{consonants}]* (?:[{vowels}]|{diphtongues}) [{consonants}]*)
(?:[{vowels}]|$|{diphtongues})
)
"""
syllables = re.compile(pattern, re.VERBOSE)
The tricky bit is that the diphtongues end in consonants (j or w), which I don't want to be included in the next syllable. So replacing the first non-capturing group by a double negative (?<![{consonants}]) doesn't work. I tried to instead replace that group by a positive lookahead (?<=[{vowels}]|^|{diphtongues}), but regex won't accept different lengths (even removing the diphtongues doesn't work, apparently ^ is of a different length).
So this is the problematic case with the pattern above:
syllables.findall('æbə')
# ['bə']
# should be: ['æb', 'bə']
Edit 2:
I've switched to using regex, which allows variable-width lookbehinds, which solves the problem. To my surprise, it even appears to be faster than the re module in the standard library. I'd still like to know how to get this working with the re module, though. (:

I suggest fixing this with a double negation:
(?= # inside lookahead for overlapping results
(?<![^a]) # match at beginning of str or after a
(b*ab*) # one a between any number of bs
(?![^a]) # at end of str or before next a
)
See the regex demo
Note I replaced the grouping constructs with lookarounds: (?:a|^) with (?<![^a]) and (?:a|$) with (?![^a]). The latter is not really important, but the first is very important here.
The (?:a|^) at the beginning of the outer lookahead pattern matches a or start of the string, whatever comes first. If a is at the start, it is matched and when the input is abbabb, you get bbabb since it matches the capturing group pattern and there is an end of string position right after. The next iteration starts after the first a, and cannot find any match since the only a left in the string has no a after bs.
Note that order of alternative matters. If you change to (?:^|a), the match starts at the start of the string, b* matches empty string, ab* grabs the first abb in abbabb, and since there is a right after, you get abb as a match. There is no way to match anything after the first a.

Remember that python "short-circuits", so, if it matches "^", its not going to continue looking to see if it matches "a" too. This will "consume" the matching character, so in cases where it matches "a", "a" is consumed and not available for the next group to match, and because using the (?:) syntax is non-capturing, that "a" is "lost", and not available to be captured by the next grouping (b*(?:a)b*), whereas when "^" is consumed by the first grouping, that first "a" would match in the second grouping.

The Behavior of Alternative Match "|" with .* in a Regex

I seldom use | together with .* before. But today when I use both of them together, I find some results really confusing. The expression I use is as follows (in python):
>>> s = "abcdefg"
>>> re.findall(r"((a.*?c)|(.*g))",s)
[('abc',''),('','defg')]
The result of the first caputure is all right, but the second capture is beyond my expectation, for I have expected the second capture would be "abcdefg" (the whole string).
Then I reverse the two alternatives:
>>> re.findall(r"(.*?g)|(a.*?c)",s)
[('abcdefg', '')]
It seems that the regex engine only reads the string once - when the whole string is read in the first alternative, the regex engine will stop and no longer check the second alternative. However, in the first case, after dealing with the first alternative, the regex engine only reads from "a" to "c", and there are still "d" to "g" left in the string, which matches ".*?g" in the second alternative. Have I got it right? What's more, as for an expression with alternatives, the regex engine will check the first alternative first, and if it matches the string, it will never check the second alternative. Is it correct?
Besides, if I want to get both "abc" and "abcdefg" or "abc" and "bcde" (the two results overlap) like in the first case, what expression should I use?
Thank you so much!

You cannot have two matches starting from the same location in the regex (the only regex flavor that does it is Perl6).
In re.findall(r"((a.*?c)|(.*g))",s), re.findall will grab all non-overlapping matches in the string, and since the first one starts at the beginning, ends with c, the next one can only be found after c, within defg.
The (.*?g)|(a.*?c) regex matches abcdefg because the regex engine parses the string from left to right, and .*? will get any 0+ chars as few as possible but up to the first g. And since g is the last char, it will match and capture the whole string into Group 1.
To get abc and abcdefg, you may use, say
(a.*?c)?.*g
See the regex demo
Python demo:
import re
rx = r"(a.*?c)?.*g"
s = "abcdefg"
m = re.search(rx, s)
if m:
print(m.group(0)) # => abcdefg
print(m.group(1)) # => abc
It might not be what you exactly want, but it should give you a hint: you match the bigger part, and capture a subpart of the string.

Re-read the docs for the re.findall method.
findall "return[s] all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found."
Specifically, non-overlapping matches, and left-to-right. So if you have a string abcdefg and one pattern will match abc, then any other patterns must (1) not overlap; and (2) be further to the right.
It's perfectly valid to match abc and defg per the description. It would be a bug to match abc and abcdefg or even abc and cdefg because they would overlap.

How to check that selected regex from char set in two sections are equal?

Test String:gcaaaattatacccacatttctttttaaaatttcagcaaaattttaaactatacg
What I want to detect: any combination of two characters including "a" in them and the "a" cannot be the first character.
Example: gcaaaattata cccaca tttc tttttaaaattt cagcaaaattttaaac tata cg
My Regex: [{g,t,c}]{2,}a[{a,g,t,c}]
Problem: When it matches a character from the first set {g,t,c} it will match any character from the second list.
My Question: How to match from the second list what is already selected from the first set for an output like the example above.
Update
Further explanation:
- The combination is of two characters only including "a"
- "a" must be in between and cannot be the start. So , "ttttaaa" starting with t but nothing intercept the "a"s, if it was the same character "t" so match the pattern, If any character not "a" or "t" stop matching.
So these are working: "tttaaat","tattttatatat"
These are not working: "taaaaaaa","attttta"
I'm writing in python if that could help.

You could try following:
import re
s = 'gcaaaattatacccacatttctttttaaaatttcagcaaaattttaaactatacg'
for match in re.finditer(r'(g|c|t)\1*a+(\1)(\1|a)*', s):
print(match.group())
Output:
ttata
cccaca
tttttaaaattt
tata
Example on regex101. (g|c|t) matches any of the characters gct and captures it. \1*a+\1 repeats the first character 0 or more times followed by at least one a followed by first character. (\1|a)* at the end then allows any combination of a and first character.

You can use ([gtc])\1*(a+)(\1+\2*)+ which will look for at least one g, t, or c followed by one or more a and then any combination of those two characters
import re
word='gcaaaattatacccacatttctttttaaaatttcagcaaaattttaaactatacg'
matches = re.finditer(r'([gtc])\1*(a+)(\1+\2*)+', word)
for matchNum, match in enumerate(matches):
print(match.group())

one way to accomplish your goal would be to capture the first character and backreference it in the third portion of your expression.
like so:
(?P<first>[gtc])(?P=first)?a(?:a|(?P=first))*
python regex reference

Regular expression for repeating sequence

I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?

Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ from the start till the end of the string
[...] one of the given character
...{3} three time of the phrase before
(...)* 0 till n times of the characters in the brackets

What you're asking it to find with your regex is "at least one triple of letters a, b, c" - that's what "+" gives you. Whatever follows after that doesn't really matter to the regex. You might want to include "$", which means "end of the line", to be sure that the line must all consist of allowed triples. However in the current form your regex would also demand that the last triple ends in a comma, so you should explicitly code that it's not so.
Try this:
re.match('([abc][abc][abc],)*([abc][abc][abc])$'
This finds any number of allowed triples followed by a comma (maybe zero), then a triple without a comma, then the end of the line.
Edit: including the "^" (start of string) symbol is not necessary, because the match method already checks for a match only at the beginning of the string.

The obligatory "you don't need a regex" solution:
all(letter in 'abc,' for letter in data) and all(len(item) == 3 for item in data.split(','))

You need to iterate over sequence of found values.
data_string = "abc,bca,df"
imatch = re.finditer(r'(?P<value>[abc]{3})(,|$)', data_string)
for match in imatch:
print match.group('value')
So the regex to check if the string matches pattern will be
data_string = "abc,bca,df"
match = re.match(r'^([abc]{3}(,|$))+', data_string)
if match:
print "data string is correct"

Your result is not surprising since the regular expression
([abc][abc][abc],)+
tries to match a string containing three characters of [abc] followed by a comma one ore more times anywhere in the string. So the most important part is to make sure that there is nothing more in the string - as scessor suggests with adding ^ (start of string) and $ (end of string) to the regular expression.

An alternative without using regex (albeit a brute force way):
>>> def matcher(x):
total = ["".join(p) for p in itertools.product(('a','b','c'),repeat=3)]
for i in x.split(','):
if i not in total:
return False
return True
>>> matcher("abc,bca,aaa")
True
>>> matcher("abc,bca,xyz")
False
>>> matcher("abc,aaa,bb")
False

If your aim is to validate a string as being composed of triplet of letters a,b,and c:
for ss in ("abc,bbc,abb,baa,bbb",
"acc",
"abc,bbc,abb,bXa,bbb",
"abc,bbc,ab,baa,bbb"):
print ss,' ',bool(re.match('([abc]{3},?)+\Z',ss))
result
abc,bbc,abb,baa,bbb True
acc True
abc,bbc,abb,bXa,bbb False
abc,bbc,ab,baa,bbb False
\Z means: the end of the string. Its presence obliges the match to be until the very end of the string
By the way, I like the form of Sonya too, in a way it is clearer:
bool(re.match('([abc]{3},)*[abc]{3}\Z',ss))

To just repeat a sequence of patterns, you need to use a non-capturing group, a (?:...) like contruct, and apply a quantifier right after the closing parenthesis. The question mark and the colon after the opening parenthesis are the syntax that creates a non-capturing group (SO post).
For example:
(?:abc)+ matches strings like abc, abcabc, abcabcabc, etc.
(?:\d+\.){3} matches strings like 1.12.2., 000.00000.0., etc.
Here, you can use
^[abc]{3}(?:,[abc]{3})*$
^^
Note that using a capturing group is fraught with unwelcome effects in a lot of Python regex methods. See a classical issue described at re.findall behaves weird post, for example, where re.findall and all other regex methods using this function behind the scenes only return captured substrings if there is a capturing group in the pattern.
In Pandas, it is also important to use non-capturing groups when you just need to group a pattern sequence: Series.str.contains will complain that this pattern has match groups. To actually get the groups, use str.extract. and
the Series.str.extract, Series.str.extractall and Series.str.findall will behave as re.findall.