CODEWITHSUNDEEP
pythonnested-lists
dia
dia

Reputation: 329

how to flatten lists while leaving some nesting (in python)

I have a list of lists of lists with strings, something like this (representing chapters, paragraphs and sentences of a text)):

[ [[ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'] ],
   [ ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ]],
  [[ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'] ],
   [ ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ]] ]

I know how to flatten this list completly (for example by [x for y in z for x in y]), but what I would like to do is to flatten it partially, to finally look like this:

[ [ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'],
    ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ],
  [ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'],
    ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ] ]

I managed to solve this by some for loops:

semiflattend_list=list()
for chapter in chapters:
    senlist=list()
    for paragraph in chapter:
        for sentences in paragraph:
            senlist.append(sentences)
    semiflattend_list.append(senlist)

But I wonder if there is a better, shorter solution? (I don't think, zip is a way to go, because my lists are different in size.)

Upvotes: 1

Views: 89

Answers (1)

cwallenpoole
cwallenpoole

Reputation: 82028

The easiest change I can see is using the itertools.chain method:

q = [
     [[ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'] ],
       [ ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ]],
     [[ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'] ],
       [ ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ]]
    ]

r = [list(itertools.chain(*g)) for g in q]
print(r)

[[['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'], ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3']],
 [['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'], ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3']]]

So, what does [list(itertools.chain(*g)) for g in q] mean:

# If I only had this
[g for g in q]
# I would get the same I started with.
# What I really want is to expand the nested lists

# * before an iterable (basically) converts the iterable into its parts.
func foo(bar, baz):
   print( bar + " " + baz )

lst = ["cat", "dog"]
foo(*lst) # prints "cat dog"

# itertools.chain accepts an arbitrary number of lists, and then outputs 
# a generator of the results:
c = itertools.chain([1],[2])
# c is now <itertools.chain object at 0x10e1fce10>
# You don't want an generator though, you want a list. Calling `list` converts that:
o = list( c )
# o is now [1,2]
# Now, together:
myList = [[2],[3]]
flattened = list(itertools.chain(*myList))
# flattened is now [2,3]

Upvotes: 1

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