CODEWITHSUNDEEP
pythonregex
user2152572
user2152572

Reputation: 47

list comprehension on multiple lists of lists

I am struck on an awkward lists comprehension problem, which I am not able to solve. So, I have two lists looking like the following:

a=[[....],[....],[....]]
b=[[....],[....],[....]]
len(a)==len(b) including sublists i.e sublists also have the same dimension.

Now I want to do a re.compile which looks something like:

[re.compile(_subelement_in_a).search(_subelement_in_b).group(1)]

and I am wondering how I can achieve the above using list compherension - something like:

[[re.compile(str(x)).search(str(y)).group(1) for x in a] for y in b]

..but obviously the above does not seem to work and I was wondering if anyone could point me in the right direction.

EDIT

I have just realized that the sublists of b have more elements than the sublists of a. So, for example:

a=[[1 items],[1 items],[1 items]]
b=[[10 item], [10 item], [10 item]]

I would still like to do the same as my above question:

[[re.compile(str(x)).search(str(y)).group(1) for x in b] for y in a]

and the output looking like:

c = [[b[0] in a[0] items],[b[1] in a[1] items],[b[2] in a[2] items]]

Example:

a=[["hgjkhukhkh"],["78hkugkgkug"],["ukkhukhylh"]]
b=[[r"""a(.*?)b""",r"""c(.*?)d""",r"""e(.*?)f""",r"""g(.*?)h""",r"""i(.*?)j"""],[r"""k(.*?)l""",r"""m(.*?)n""",r"""o(.*?)p""",r"""q(.*?)r"""],[r"""s(.*?)t""",r"""u(.*?)v""",r"""x(.*?)y""",r"""z(.*?)>"""]]

using one to one mapping. i.e check if:

elements of sublists of b[0] are present in sublist element of a[0]
elements of sublists of b[1] are present in sublist element of a[1]
elements of sublists of b[2] are presnet in sublist element of a[2]

Upvotes: 0

Views: 1204

Answers (2)

hugomg
hugomg

Reputation: 69934

Sounds like you are looking for zip? It takes a pair of lists and turns it into a list of pairs.

[
    [my_operation(x,y) for x,y in zip(xs, ys)]
    for xs, ys in zip(a, b)
]

-- Edit. Requirements changed:

[
    [[regex(p, s) for p in patterns] for s in strings]
    for strings, patterns in zip(a, b)
]

Upvotes: 3

nneonneo
nneonneo

Reputation: 179392

Use zip liberally:

[[re.search(x, y).group(1) for x,y in zip(s,t)] for s,t in zip(a,b)]

The first zip(a,b) produces lists of sublist pairs. The second zip pairs the elements in parallel sublists together.

Upvotes: 2

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