Transform “list of tuples” into a flat list or a matrix

Question

With Sqlite, a \"select..from\" command returns the results \"output\", which prints (in python):

>>print output
[(12.2817, 12.2817), (0, 0), (8.52, 8.         


        

Answer 1

By far the fastest (and shortest) solution posted:

list(sum(output, ()))

About 50% faster than the itertools solution, and about 70% faster than the map solution.

Answer 2

Or you can flatten the list like this:

reduce(lambda x,y:x+y, map(list, output))

Answer 3

In case of arbitrary nested lists(just in case):

def flatten(lst):
    result = []
    for element in lst: 
        if hasattr(element, '__iter__'):
            result.extend(flatten(element))
        else:
            result.append(element)
    return result

>>> flatten(output)
[12.2817, 12.2817, 0, 0, 8.52, 8.52]

Answer 4

List comprehension approach that works with Iterable types and is faster than other methods shown here.

flattened = [item for sublist in l for item in sublist]

l is the list to flatten (called output in the OP's case)

---

timeit tests:

l = list(zip(range(99), range(99)))  # list of tuples to flatten

List comprehension

[item for sublist in l for item in sublist]

timeit result = 7.67 µs ± 129 ns per loop

List extend() method

flattened = []
list(flattened.extend(item) for item in l)

timeit result = 11 µs ± 433 ns per loop

sum()

list(sum(l, ()))

timeit result = 24.2 µs ± 269 ns per loop

Answer 5

This is what numpy was made for, both from a data structures, as well as speed perspective.

import numpy as np

output = [(12.2817, 12.2817), (0, 0), (8.52, 8.52)]
output_ary = np.array(output)   # this is your matrix 
output_vec = output_ary.ravel() # this is your 1d-array

Answer 6

In Python 3 you can use the * syntax to flatten a list of iterables:

>>> t = [ (1,2), (3,4), (5,6) ]
>>> t
[(1, 2), (3, 4), (5, 6)]
>>> import itertools
>>> list(itertools.chain(*t))
[1, 2, 3, 4, 5, 6]
>>>