flatten list in python

Question

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.

I guess then just list(chain.from_iterable(chain.from_iterable(lis))) is enough, no need of that list() call. ` — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Oct 15, 2012 at 9:57
is it always just a list containing single element (nested) sublists? — moooeeeep
– moooeeeep, Commented Oct 15, 2012 at 9:57
This is very close: Flatten (an irregular) list of lists in Python — Paolo Moretti
– Paolo Moretti, Commented Oct 15, 2012 at 10:01

waitingkuo · Accepted Answer · 2012-10-15 09:51:03Z

2

How about

result = [x[0][0] for x in exampleArray]

answered Oct 15, 2012 at 9:51

waitingkuo

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1 Comment

Tim McNamara Over a year ago

Very good specific example. Not the best general one though, obviously.

Ashwini Chaudhary · Accepted Answer · 2012-10-15 09:52:03Z

1

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]

answered Oct 15, 2012 at 9:52

Ashwini Chaudhary

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Comments

Andy Hayden · Accepted Answer · 2012-10-15 09:59:41Z

1

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)

answered Oct 15, 2012 at 9:59

Andy Hayden

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Meitham · Accepted Answer · 2016-06-29 09:58:09Z

1

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)

edited Jun 29, 2016 at 9:58

answered Oct 15, 2012 at 9:52

Meitham

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1 Comment

user1741339 Over a year ago

thanks just wanted to see if there was will remove the extra list I added to the code

mechmind · Accepted Answer · 2012-10-15 11:56:04Z

0

For fixed level of depth (as in example) you can just sum enough times:

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

answered Oct 15, 2012 at 11:56

mechmind

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