在 Python 中生成和压缩两个列表的最干净有效的方法
Most clean and efficient way to generate and zip two lists in Python
给出这两个列表
zeros = [2,3,1,2]
ones = [3,4,5]
（条件总是len(zeros) == len(ones) + 1）
我想创建一个列表，其中 0 和 1 交替出现在列表中提到的数量级。我可以通过以下方式实现：
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，这是最有效/最干净的方法吗？我得到了 2.66 µs ± 78.8 ns 的性能，但我仍然认为这可以在单行中完成，而且效率可能更高
您可以使用 itertools.chain、zip_longest 和 itertools.repeat 来创建一个不太混乱的单行。
>>> 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]
在我的机器上，这花费了 3.34µs。更重要的是，它是 list 的包装器，花费了这段时间。迭代器本身会按需生成元素，如果您实际上并不一次需要它们的话。
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) 创建一系列 repeat 对象，代表您的 0 的运行；同样创建 1 组。zip_longest 将它们压缩成一对序列，添加一个什么都不做的空列表来平衡 zeros chain.from_iterable 使该序列变平（从 (a, b), (c, d) 到 (a, b, c, d)。- 外部
chain.from_iterable 然后将 repeat 对象展平为一个序列，list 变成一个列表。
您还可以使用 itertools 文档中的 roundrobin 方法简化单线，它处理合并零组和一个组，以及第一轮展平。
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))))
带列表理解的 Zip 应该可以解决问题。
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]
注意 ones+[0] 是为了确保您不会在 zip 操作中删除零列表中的最后一个值。
两个明显更快的解决方案，使用“技巧”特殊处理 第一个 零而不是最后一个，用它来初始化 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 指出的那样，元组使速度提高了大约 30%。）
基准测试结果：
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
基准代码：
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()
给出这两个列表
zeros = [2,3,1,2]
ones = [3,4,5]
（条件总是len(zeros) == len(ones) + 1）
我想创建一个列表，其中 0 和 1 交替出现在列表中提到的数量级。我可以通过以下方式实现：
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，这是最有效/最干净的方法吗？我得到了 2.66 µs ± 78.8 ns 的性能，但我仍然认为这可以在单行中完成，而且效率可能更高
您可以使用 itertools.chain、zip_longest 和 itertools.repeat 来创建一个不太混乱的单行。
>>> 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]
在我的机器上，这花费了 3.34µs。更重要的是，它是 list 的包装器，花费了这段时间。迭代器本身会按需生成元素，如果您实际上并不一次需要它们的话。
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)创建一系列repeat对象，代表您的 0 的运行；同样创建 1 组。zip_longest将它们压缩成一对序列，添加一个什么都不做的空列表来平衡zeros 中的额外值
chain.from_iterable使该序列变平（从(a, b), (c, d)到(a, b, c, d)。- 外部
chain.from_iterable然后将repeat对象展平为一个序列，list变成一个列表。
您还可以使用 itertools 文档中的 roundrobin 方法简化单线，它处理合并零组和一个组，以及第一轮展平。
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))))
带列表理解的 Zip 应该可以解决问题。
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]
注意 ones+[0] 是为了确保您不会在 zip 操作中删除零列表中的最后一个值。
两个明显更快的解决方案，使用“技巧”特殊处理 第一个 零而不是最后一个，用它来初始化 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 指出的那样，元组使速度提高了大约 30%。）
基准测试结果：
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
基准代码：
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()