[英]Most clean and efficient way to generate and zip two lists in Python
鑒於這兩個列表
zeros = [2,3,1,2]
ones = [3,4,5]
(條件總是len(zeros) == len(ones) + 1
)
我想創建一個列表,其中交替包含列表中提到的大小的 0 和 1。 我可以通過以下方式實現:
zeros_list = [[0]*n for n in zeros]
ones_list = [[1]*n for n in ones]
output = [z for x in zip(zeros_list, ones_list) for y in x for z in y]
output += [0]*zeros[-1]
print(output)
> [0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0]
然而,這是最有效/最干凈的方式嗎? 我得到了 2.66 µs ± 78.8 ns 的性能,但我仍然認為這可以在單行中完成,而且效率可能更高
兩個明顯更快的解決方案,使用“技巧”來特殊處理第一個零而不是最后一個零,使用它來初始化output
。
def superb_rain(zeros, ones):
zeros = iter(zeros)
output = [0] * next(zeros)
for o in ones:
output += (1,) * o
output += (0,) * next(zeros)
return output
def superb_rain2(zeros, ones):
z = iter(zeros).__next__
output = [0] * z()
for o in ones:
output += (1,) * o
output += (0,) * z()
return output
(正如@schwobaseggl 指出的那樣,元組使它的速度提高了大約 30%。)
基准測試結果:
0.14 us 0.13 us 0.13 us baseline
3.04 us 3.02 us 2.98 us original
3.27 us 3.19 us 3.29 us chepner_1
5.03 us 5.12 us 5.25 us chepner_2
4.66 us 4.74 us 4.68 us chepner_2__superb_rain
2.52 us 2.53 us 2.47 us Alain_T
3.35 us 3.27 us 3.42 us python_user
1.02 us 0.99 us 1.04 us superb_rain
1.07 us 1.11 us 1.09 us superb_rain2
基准代碼:
import timeit
from itertools import zip_longest, cycle, islice, repeat, chain
def baseline(zeros, ones):
pass
def original(zeros, ones):
zeros_list = [[0]*n for n in zeros]
ones_list = [[1]*n for n in ones]
output = [z for x in zip(zeros_list, ones_list) for y in x for z in y]
output += [0]*zeros[-1]
return output
def chepner_1(zeros, ones):
return list(chain.from_iterable(chain.from_iterable(zip_longest((repeat(0, x) for x in zeros), (repeat(1, x) for x in ones), fillvalue=[]))))
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next in nexts:
yield next()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
def chepner_2(zeros, ones):
zero_groups = (repeat(0, x) for x in zeros)
one_groups = (repeat(1, x) for x in ones)
return list(chain.from_iterable(roundrobin(zero_groups, one_groups)))
def chepner_2__superb_rain(zeros, ones):
return list(chain.from_iterable(map(repeat, cycle([0, 1]), roundrobin(zeros, ones))))
def Alain_T(zeros, ones):
return [B for N,P in zip(zeros+[0],ones+[0]) for B in ([0]*N+[1]*P)]
def python_user(zeros, ones):
res = [None] * (len(ones) + len(zeros))
res[::2] = ([0]*n for n in zeros)
res[1::2] = ([1]*n for n in ones)
res = [y for x in res for y in x]
return res
def superb_rain(zeros, ones):
zeros = iter(zeros)
output = [0] * next(zeros)
for o in ones:
output += (1,) * o
output += (0,) * next(zeros)
return output
def superb_rain2(zeros, ones):
z = iter(zeros).__next__
output = [0] * z()
for o in ones:
output += (1,) * o
output += (0,) * z()
return output
funcs = [
baseline,
original,
chepner_1,
chepner_2,
chepner_2__superb_rain,
Alain_T,
python_user,
superb_rain,
superb_rain2,
]
zeros = [2,3,1,2]
ones = [3,4,5]
number = 10**5
expect = original(zeros, ones)
for func in funcs:
print(func(zeros, ones) == expect, func.__name__)
print()
tss = [[] for _ in funcs]
for _ in range(4):
for func, ts in zip(funcs, tss):
t = min(timeit.repeat(lambda: func(zeros, ones), number=number)) / number
ts.append(t)
print(*('%.2f us ' % (1e6 * t) for t in ts[1:]), func.__name__)
print()
具有列表理解的 Zip 應該可以解決問題。
zeros = [2,3,1,2]
ones = [3,4,5]
output = [B for N,P in zip(zeros,ones+[0]) for B in [0]*N+[1]*P]
print(output)
[0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0]
請注意, ones+[0]
是為了確保您不會在 zip 操作中刪除零列表中的最后一個值。
您可以使用itertools.chain
、 zip_longest
和itertools.repeat
創建一個不太容易混淆的單行代碼。
>>> list(chain.from_iterable(chain.from_iterable(zip_longest((repeat(0, x) for x in zeros), (repeat(1, x) for x in ones), fillvalue=[]))))
[0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0]
在我的機器上,這花費了 3.34µs。 更重要的是,這次是要list
的包裝器。 迭代器本身會按需生成元素,如果您實際上並不一次需要它們的話。
list(chain.from_iterable(
chain.from_iterable(zip_longest((repeat(0, x) for x in zeros),
(repeat(1, x) for x in ones),
fill_value=[]))))
(repeat(0, x) for x in zeros)
創建一系列repeat
對象,代表您運行的 0; 同樣創建 1 組。zip_longest
將它們壓縮成一對序列,添加一個什么都不做的空列表來平衡zeros
中的額外值chain.from_iterable
將該序列展平 (from (a, b), (c, d)
to (a, b, c, d)
。chain.from_iterable
然后將repeat
的對象壓平成一個序列,這個list
變成一個列表。 您還可以使用itertools
文檔中的roundrobin
方法簡化單行代碼,它處理合並零組和一個組,以及第一輪展平。
from itertools import cycle, islice, repeat, chain
zeros = [2,3,1,2]
ones = [3,4,5]
# From the itertools documentation
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next in nexts:
yield next()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
zero_groups = (repeat(0, x) for x in zeros)
one_groups = (repeat(1, x) for x in ones)
print(list(chain.from_iterable(roundrobin(zero_groups, one_groups))))
鑒於這兩個列表
zeros = [2,3,1,2]
ones = [3,4,5]
(條件總是len(zeros) == len(ones) + 1
)
我想創建一個列表,其中包含列表中提到的數量級的交替 0 和 1。 我可以通過以下方式實現:
zeros_list = [[0]*n for n in zeros]
ones_list = [[1]*n for n in ones]
output = [z for x in zip(zeros_list, ones_list) for y in x for z in y]
output += [0]*zeros[-1]
print(output)
> [0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0]
然而,這是最有效/最干凈的方式嗎? 我得到了 2.66 µs ± 78.8 ns 的性能,但我仍然認為這可以在單線中完成,而且效率可能更高
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