如何解压缩列表元组

Question

I am attempting to unpack the elements of a list that is packed inside a tuple.

myTuple = (['a', 'list', 'of', 'strings'], ['inside', 'a', 'tuple'], ['extra', 'words', 'for', 'filler'])

So for example I want to get this element ('a')

I have tried this:

for (i, words) in list(enumerate(myTuple)):
    print(words)

But this returns the list like this

['a', 'list', 'of', 'strings']
['inside', 'a', 'tuple']
etc...

How can I get the elements inside the list?

Answer 1

You can use the indexing of your tuple and then the lists to access the inner-most elements. For example, to get at the string 'a' , you could call:

myTuple[0][0]

If you wanted to iterate over all the elements in the lists, you could use the chain method form itertools . For example:

from itertools import chain

for i in chain(*myTuple):
    print(i)

Answer 2

You can use reduce , eg. with numpy

from functools import reduce
reduce(append, myTuple)

Out[149]: 
array(['a', 'list', 'of', 'strings', 'inside', 'a', 'tuple', 'extra',
       'words', 'for', 'filler'], dtype='<U7')

Or, base

import operator
from functools import reduce
reduce(operator.add, myType, [])
# or, reduce(lambda x, y: x + y, myTuple)

reduce is a classic list operation function along with map , filter , etc. It takes an initial value, here an empty list, and successively applies a function ( append ) to each element of the sequence ( myTuple ).

Answer 3

Currently you are just iterating through the entire loop and printing out the elements contained within the list.

However, if you want to access a specific element within the list then just refer to it by it's name using .index()

Alternatively you can just print(list(indexposition))