but*_*ife 2 python python-itertools
我看过在线课程,他们有如下示例:
from itertools import count
# creates a count iterator object
iterator =(count(start = 0, step = 2))
# prints an even list of integers
print("Even list:",
list(next(iterator) for _ in range(5)))
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...您可以使用range或 来编写np.arange。这是另一个例子:
# list containing some strings
my_list =["x", "y", "z"]
# count spits out integers for
# each value in my list
for i in zip(count(start = 1, step = 1), my_list):
print(i)
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...这基本上只是enumerate。所以我的问题是:您能否举一个使用 无法完成(或必须更加笨拙地完成)的itertools.count示例?itertools.islicerange
在这种情况下,count实例是偶尔使用的,而不是在单个循环中立即使用。
class Foo:
_x = count() # Infinite supply of unique integer values
def __init__(self):
self._id = f'Foo #{next(self._x)}'
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islice这是一个用于防止 O(n) 内存使用的情况:
def is_sorted(some_list):
return all(i <= j for i, j in zip(some_list, islice(some_list, 1, None)))
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如果你把它写成
def is_sorted(some_list):
return all(i <= j for i, j in zip(some_list, some_list[1:]))
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在测试第一对之前,您将不得不制作几乎完整的副本some_list,这对于像[2, 1] + [3] * 10000.
两者都不是必要的,因为每一个都是可以简单定义的:
def count(start=0, step=1):
while True:
yield start
start += step
# A more accurate translation would be more complicated than necessary for our purposes here.
# The real version would have to be able to handle stop=None
# and choose 1 and -1 as default values for step, depending
# on whether stop is less than or greater than start.
def islice(itr, start, stop, step):
for _ in range(start):
next(itr)
while start < stop:
yield next(itr)
start += step
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