一般来说，有没有一种有效的方法可以知道Python中的迭代器中有多少个元素，而不用遍历每个元素并计数?
当前回答
我决定在现代版本的Python上重新运行基准测试，并发现几乎完全颠倒了基准测试
我运行了以下命令:
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return len(tuple(x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return len(list(x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return sum(map(lambda i: 1, x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return sum(1 for _ in x)" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " d = deque(enumerate(x, 1), maxlen=1)" -s " return d[0][0] if d else 0" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " counter = count()" -s " deque(zip(x, counter), maxlen=0)" -s " return next(counter)" -- "itlen(it)"
它们等价于为以下每个itlen*(it)函数计时:
it = iter(range(1000000))
from collections import deque
from itertools import count
def itlen1(x):
return len(tuple(x))
def itlen2(x):
return len(list(x))
def itlen3(x):
return sum(map(lambda i: 1, x))
def itlen4(x):
return sum(1 for _ in x)
def itlen5(x):
d = deque(enumerate(x, 1), maxlen=1)
return d[0][0] if d else 0
def itlen6(x):
counter = count()
deque(zip(x, counter), maxlen=0)
return next(counter)
在装有AMD Ryzen 7 5800H和16 GB RAM的Windows 11、Python 3.11机器上，我得到了以下输出:
10000000 loops, best of 5: 103 nsec per loop
10000000 loops, best of 5: 107 nsec per loop
10000000 loops, best of 5: 138 nsec per loop
10000000 loops, best of 5: 164 nsec per loop
10000000 loops, best of 5: 338 nsec per loop
10000000 loops, best of 5: 425 nsec per loop
这表明len(list(x))和len(tuple(x))是绑定的;后面跟着sum(map(lambda i: 1, x));然后紧靠sum(1 for _ in x);那么其他答案中提到的其他更复杂的方法和/或在基数中使用的方法至少要慢两倍。
其他回答
不，任何方法都需要解析每个结果。你可以这样做
iter_length = len(list(iterable))
但是在无限迭代器上运行它当然不会返回。它还将消耗迭代器，如果你想使用内容，它将需要重置。
告诉我们你想要解决的真正问题可能会帮助我们找到更好的方法来实现你的实际目标。
编辑:使用list()会将整个可迭代对象一次性读入内存，这可能是不可取的。另一种方法是做
sum(1 for _ in iterable)
正如另一个人发布的那样。这样可以避免把它保存在记忆中。
这段代码应该工作:
>>> iter = (i for i in range(50))
>>> sum(1 for _ in iter)
50
尽管它确实遍历每一项并计算它们，但这是最快的方法。
它也适用于迭代器中没有项的情况:
>>> sum(1 for _ in range(0))
0
当然，对于一个无限的输入，它会一直运行，所以请记住迭代器可以是无限的:
>>> sum(1 for _ in itertools.count())
[nothing happens, forever]
此外，请注意，这样做将耗尽迭代器，并且进一步尝试使用它将看不到任何元素。这是Python迭代器设计的一个不可避免的结果。如果你想保留元素，你就必须把它们存储在一个列表或其他东西中。
一个简单的方法是使用内置函数set()或list():
答:set()在迭代器中没有重复项的情况下(最快的方式)
iter = zip([1,2,3],['a','b','c'])
print(len(set(iter)) # set(iter) = {(1, 'a'), (2, 'b'), (3, 'c')}
Out[45]: 3
or
iter = range(1,10)
print(len(set(iter)) # set(iter) = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Out[47]: 9
B: list()以防迭代器中有重复的项
iter = (1,2,1,2,1,2,1,2)
print(len(list(iter)) # list(iter) = [1, 2, 1, 2, 1, 2, 1, 2]
Out[49]: 8
# compare with set function
print(len(set(iter)) # set(iter) = {1, 2}
Out[51]: 2
我决定在现代版本的Python上重新运行基准测试，并发现几乎完全颠倒了基准测试
我运行了以下命令:
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return len(tuple(x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return len(list(x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return sum(map(lambda i: 1, x))" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " return sum(1 for _ in x)" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " d = deque(enumerate(x, 1), maxlen=1)" -s " return d[0][0] if d else 0" -- "itlen(it)"
py -m timeit -n 10000000 -s "it = iter(range(1000000))" -s "from collections import deque" -s "from itertools import count" -s "def itlen(x):" -s " counter = count()" -s " deque(zip(x, counter), maxlen=0)" -s " return next(counter)" -- "itlen(it)"
它们等价于为以下每个itlen*(it)函数计时:
it = iter(range(1000000))
from collections import deque
from itertools import count
def itlen1(x):
return len(tuple(x))
def itlen2(x):
return len(list(x))
def itlen3(x):
return sum(map(lambda i: 1, x))
def itlen4(x):
return sum(1 for _ in x)
def itlen5(x):
d = deque(enumerate(x, 1), maxlen=1)
return d[0][0] if d else 0
def itlen6(x):
counter = count()
deque(zip(x, counter), maxlen=0)
return next(counter)
在装有AMD Ryzen 7 5800H和16 GB RAM的Windows 11、Python 3.11机器上，我得到了以下输出:
10000000 loops, best of 5: 103 nsec per loop
10000000 loops, best of 5: 107 nsec per loop
10000000 loops, best of 5: 138 nsec per loop
10000000 loops, best of 5: 164 nsec per loop
10000000 loops, best of 5: 338 nsec per loop
10000000 loops, best of 5: 425 nsec per loop
这表明len(list(x))和len(tuple(x))是绑定的;后面跟着sum(map(lambda i: 1, x));然后紧靠sum(1 for _ in x);那么其他答案中提到的其他更复杂的方法和/或在基数中使用的方法至少要慢两倍。
def count_iter(iter):
sum = 0
for _ in iter: sum += 1
return sum
