Q: Transform "list of tuples" into a flat list or a matrix With Sqlite, a select .. from command returns the results output, which prints:
>>print output
[(12.2817, 12.2817), (0, 0), (8.52, 8.52)]

It seems to be a list of tuples. I would like to either convert output to a simple list:
[12.2817, 12.2817, 0, 0, 8.52, 8.52]

or a 2x3 matrix:
12.2817 12.2817
0          0 
8.52     8.52

to be read via output[i][j]
The flatten command does not do the job for the 1st option, and I have no idea for the second one...
A fast solution would be appreciated, as the real data is much bigger.

A: >>> flat_list = []
>>> nested_list = [(1, 2, 4), (0, 9)]
>>> for a_tuple in nested_list:
...     flat_list.extend(list(a_tuple))
... 
>>> flat_list
[1, 2, 4, 0, 9]
>>> 

you could easily move from list of tuple to single list as shown above.

A: use itertools chain:
>>> import itertools
>>> list(itertools.chain.from_iterable([(12.2817, 12.2817), (0, 0), (8.52, 8.52)]))
[12.2817, 12.2817, 0, 0, 8.52, 8.52]


A: Or you can flatten the list like this:
reduce(lambda x,y:x+y, map(list, output))


A: 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

A: 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


A: In Python 2.7, and all versions of Python3, you can use itertools.chain to flatten a list of iterables. Either with the * syntax or the class method.
>>> 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]
>>> list(itertools.chain.from_iterable(t))
[1, 2, 3, 4, 5, 6]


A: 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]


A: def flatten_tuple_list(tuples):
    return list(sum(tuples, ()))


tuples = [(5, 6), (6, 7, 8, 9), (3,)]
print(flatten_tuple_list(tuples))


A: Update: Flattening using extend but without comprehension and without using list as iterator (fastest)
After checking the next answer to this that provided a faster solution via a list comprehension with dual for I did a little tweak and now it performs better, first the execution of list(...) was dragging a big percentage of time, then changing a list comprehension for a simple loop shaved a bit more as well.
The new solution is:
l = []
for row in output: l.extend(row)

The old one replacing list with [] (a bit slower but not much):
[l.extend(row) for row in output]


Older (slower):
Flattening with list comprehension
l = []
list(l.extend(row) for row in output)


some timeits for new extend and the improvement gotten by just removing list(...) for [...]:
import timeit
t = timeit.timeit
o = "output=list(zip(range(1000000000), range(10000000))); l=[]"
steps_ext = "for row in output: l.extend(row)"
steps_ext_old = "list(l.extend(row) for row in output)"
steps_ext_remove_list = "[l.extend(row) for row in output]"
steps_com = "[item for sublist in output for item in sublist]"

print(f"{steps_ext}\n>>>{t(steps_ext, setup=o, number=10)}")
print(f"{steps_ext_remove_list}\n>>>{t(steps_ext_remove_list, setup=o, number=10)}")
print(f"{steps_com}\n>>>{t(steps_com, setup=o, number=10)}")
print(f"{steps_ext_old}\n>>>{t(steps_ext_old, setup=o, number=10)}")

Time it results:
for row in output: l.extend(row)                  
>>> 7.022608777000187

[l.extend(row) for row in output]
>>> 9.155910597999991

[item for sublist in output for item in sublist]
>>> 9.920002304000036

list(l.extend(row) for row in output)
>>> 10.703829122000116


A: 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.
