将“元组列表”转换为平面列表或矩阵

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>>> 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.

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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]

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Or you can flatten the list like this:

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

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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]

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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.

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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]

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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

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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

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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

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




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