How to combine list elements in a tuple?-CodePudding

I am struggling to combine specific list elements from a tuple. Would love any feedback or help! I am new to Python, so apologies if this is not a good question.

If I have a tuple list like this:

tuple_1 = [('A', 'B', 'C', 'D'), ('A', 'H'), ('B', 'C', 'D', 'A')]

I want to combine elements 'B', 'C', and 'D' from every tuple in the list:

tuple_1_new = [('A', 'BCD'), ('A', 'H'), ('BCD', 'A')]

My code looks like this:

next_insert = [(iter(x)) for x in tuple_1]
tuple_1_new = [i   next(next_insert)   next(next_insert) if i == "B" else i for i in next_insert]

but when I print(tuple_1_new), it is giving me an output of:

[<tuple_iterator object at ...>, <tuple_iterator object at ...>, <tuple_iterator object at ...>]

I feel like my code is correct, but I'm confused with this output. Again, sorry if this is a dumb question. Would appreciate any help - thanks!

CodePudding user response：

def foo(arr):
    w = "".join(arr)    
    ind = w.find("BCD")
    if ind >= 0:
        ans = list(arr)
        return tuple(ans[:ind]   ["BCD"]   ans[ind   3:])
    return arr

[foo(x) for x in tuple_1]
# [('A', 'BCD'), ('A', 'H'), ('BCD', 'A')]

CodePudding user response：

Another solution, using generator:

tuple_1 = [("A", "B", "C", "D"), ("A", "H"), ("B", "C", "D", "A")]

def fn(x):
    while x:
        if x[:3] == ("B", "C", "D"):
            yield "".join(x[:3])
            x = x[3:]
        else:
            yield x[0]
            x = x[1:]


out = [tuple(fn(t)) for t in tuple_1]
print(out)

Prints:

[('A', 'BCD'), ('A', 'H'), ('BCD', 'A')]

CodePudding user response：

A list comprehension answer:

[tuple([t for t in tup if t not in ['B', 'C', 'D']]   [''.join([t for t in tup if t in ['B', 'C', 'D']])]) for tup in tuple_1]

Although not quite getting the desired output, prints:

[('A', 'BCD'), ('A', 'H', ''), ('A', 'BCD')]

Note: In a 'simple' list comprehension, the 'for x in iterable_name' creates a processing loop using 'x' (or a collection of names if expanding a tuple or performing a zip extraction) as variable(s) in each loop. When performing list comprehension within a list comprehension (for loop inside for loop), each loop will contribute one or more variables, which obviously must not incur a name collision.

CodePudding user response：

Assuming the strings are single letters like in your example:

tuple_1_new = [tuple(' '.join(t).replace('B C D', 'BCD').split())
               for t in tuple_1]

Or with Regen:

tuple_1_new = [tuple(re.findall('BCD|.', ''.join(t)))
               for t in tuple_1]