0

I want to flatten the list:

exampleArray = [[[151.68694121866872]], 
                [[101.59534468349297]], 
                [[72.16055999176308]]]

to:

[151.68694121866872, 101.59534468349297, 72.16055999176308]

Right now I am doing this:

resultArray= list(chain.from_iterable(list(chain.from_iterable(exampleArray))))

Even though it works I wanted to know if there's a better way.

moooeeeep
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user1741339
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5 Answers5

2

How about

result = [x[0][0] for x in exampleArray]
waitingkuo
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In [6]: from itertools import chain

In [7]: lis=[[[151.68694121866872]], [[101.59534468349297]], [[72.16055999176308]]]

In [8]: list(chain(*(chain(*lis))))

Out[8]: [151.68694121866872, 101.59534468349297, 72.16055999176308]
Ashwini Chaudhary
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There isn't a better way, this is the most recommended way. See official recipes

def flatten(listOfLists):
    "Flatten one level of nesting"
    return chain.from_iterable(listOfLists)
Meitham
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You don't need to convert the itertools.chain object (an iterable) into a list:

resultArray= list(chain.from_iterable(list(chain.from_iterable(exampleArray))))
# could be rewritten as
resultArray= list(chain.from_iterable(chain.from_iterable(exampleArray)))

.

You could write a deepness function using recursion:

def deep_chain_from_iterable(it, n):
    if n == 0:
        return list(it)
    else:
        return deep_chain_from_iterable(itertools.chain.from_iterable(it),n-1)

deep_chain_from_iterable(exampleArray, 2)
Andy Hayden
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For fixed level of depth (as in example) you can just sum enough times:

sum(sum(listoflistsoflists, []), [])
mechmind
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