有没有办法键入可以以二进制方式搜索元组的提示?

Question

你能告诉Python字符串列表是排序的，可以用二进制方式搜索吗？

假设你有一个字符串元组，如下所示：

from typing import Tuple
words: Tuple[str] = ("alphabet", "bike", "car", "house", "word")

首先，我输入是否正确提示了这一点？ 其次，有没有办法告诉python这个元组可以用二进制方式搜索(因为它是排序的(还是没有必要的？

Answer 1

类型注释应Tuple[t1, t2, ...]其中t1, t2...是相应元组值的类型。您键入的方式暗示了元组，这意味着它是一个包含单个字符串项的元组。

创建类型提示系统是为了将类似静态的声明引入 python，所以需要注意两件事

• 如果你有2-3个项目的元组(例如，Tuple[number, number]表示2D点(，注释元组是有意义的。您拥有的是一个集合，最好注释为Iterable[str].
无法表示集合已排序，因为类型批注
• 仅批注类型，这与集合是否排序无关。

Answer 2

你不能说元组是可以使用类型提示进行二进制搜索的。但是，您可以编写一个装饰器来包装函数，以强制对所有输入元组进行排序。如果您有许多函数需要排序元组作为输入，这可能是一个好主意。它不会修改元组(它不能 - 元组不可变(

def is_sorted(t):
""" Returns true if the tuple, t, is sorted """
return sorted(list(t)) == list(t)
def enforce_sort(func):
def inner(*args, **kwargs):
new_args = [None]*len(args) # Since args is a tuple, we can't modify it
for n, arg in enumerate(args):
new_args[n] = arg 
if isinstance(arg, tuple):
## Do you want to sort it?
new_args[n] = tuple(sorted(arg))
## Or do you wait to raise an exception if its not sorted?
# if not is_sorted(arg):
#     raise ValueError("Input tuple must be sorted")
for k, v in kwargs.items():
if isinstance(v, tuple):
## Do you want to sort it?
kwargs[k] = tuple(sorted(v))
## Or do you want to raise an exception if its not sorted?
# if not is_sorted(v):
#    raise ValueError("Input tuple must be sorted")
return func(*new_args, **kwargs)
return inner
@enforce_sort
def show_twoples(t1, t2):
""" prints two tuples, t1 and t2 """
print(f"t1: {t1}")
print(f"t2: {t2}")
a = (1,2,3,4,5) # Sorted
b = (9,8,6,4,2) # Not sorted
c = [1,2,6,2,3] # Not sorted, but not a tuple so it won't be affected
print(f"nOriginal: t1 = {a}, t2 = {b}")
show_twoples(a, b)
print(f"nOriginal: t1 = {b}, t2 = {a}")
show_twoples(b, a)
print(f"nOriginal: t1 = {a} t2 = {c}")
show_twoples(a, c)
print(f"nOriginal: t1 = {c}, t2 = {b}")
show_twoples(t1 = c, t2 = b)

输出：

Original: t1 = (1, 2, 3, 4, 5), t2 = (9, 8, 6, 4, 2)
t1: (1, 2, 3, 4, 5)
t2: (2, 4, 6, 8, 9)
Original: t1 = (9, 8, 6, 4, 2), t2 = (1, 2, 3, 4, 5)
t1: (2, 4, 6, 8, 9)
t2: (1, 2, 3, 4, 5)
Original: t1 = (1, 2, 3, 4, 5) t2 = [1, 2, 6, 2, 3]
t1: (1, 2, 3, 4, 5)
t2: [1, 2, 6, 2, 3]
Original: t1 = [1, 2, 6, 2, 3], t2 = (9, 8, 6, 4, 2)
t1: [1, 2, 6, 2, 3]
t2: (2, 4, 6, 8, 9)

根据您想要的内容，您可以在装饰器中自动对元组进行排序(这就是它现在正在做的事情(，或者您可以在元组未排序时引发异常(通过切换注释掉的部分(。