生成可变序列长度的列表排列

Question

Suppose I have a list that looks like this:

beast = ['E', 'O', 'R', 'E', 'O', 'S', 'P', 'E', 'P', 'P', 'O', 'R', 'S', 'P', 'O', 'E']

Now, here are the basic patterns that I want to extract:

basic_oros = ['O', 'R', 'O']

pi_poro = ['P', 'O', 'R', 'O']

pi_orpo = ['O', 'R', 'P', 'O']

pi_porpo = ['P', 'O', 'R', 'P', 'O']

si_orso = ['O', 'R', 'S', 'O']

spi_orspo = ['O', 'R', 'S', 'P', 'O']

spi_porso = ['P', 'O', 'R', 'S', 'O']

spi_porspo = ['P', 'O', 'R', 'S', 'P', 'O']

The problem is, if I extract only ['P', 'O', 'R', 'S', 'P', 'O'] out of beast , I miss out on the previous P before that pattern occurs.

(Better example - There can also be ['O', 'O', 'R', 'O', 'O', 'O'] )

Therefore, every letter in those basic patterns can be expanded in a sense of:

Os can occur together thrice.

Ps can occur together twice.

Rs can occur together twice.

Ss can occur only once.

I want to generate a list of all possible permutations but have no idea where to start.

I wrote a naive python code that replaces every occurrence of O with three Os and etc. But that results in -

['O', 'O', 'O', 'R' 'O', 'O', 'O']

Which won't work when I need to find: ['O', 'R', 'O', 'O']

Any help is appreciated. Thank you.

Answer 1

This sounds like a job for regexes. If you join your beast list into a string with ''.join(beast) , you can build regexes to search it for substrings matching specific patterns. For example, using the {m,n} quantifier to search for a certain number of repetitions of a subpattern, you can use

re.search(r'O{1,3}R{1,2}O{1,3}', ''.join(beast))

to search for a pattern consisting of 1 to 3 Os, 1 to 2 Rs, and 1 to 3 more Os.

See the re module documentation for more information on Python's regex syntax and functions, and see any number of online tutorials for help getting used to writing regexes.