减少不必要的 for 循环

注：参考于码农高天

1. 使用comprehension、generator

from timeit import timeit


def with_for():
    lst = []
    for i in range(100):
        lst.append(i)
    return lst


def without_for():
    return [i for i in range(100)]


def without_for_faster():
    return (i for i in range(100))

# 使用append时python没法直接确定使用的是list的append，需要先从lst中找到append
print(f"with   :{timeit(with_for, number=10000)}")
# 使用list comprehension,直接调用list.append
print(f"without:{timeit(without_for, number=10000)}")
# 使用 generator 更快
print(f"without:{timeit(without_for_faster, number=10000)}")
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2. 使用builtin函数

from timeit import timeit

lst = [i for i in range(100)]


def with_for():
    max_num = 0
    for num in lst:
        if num > max_num:
            max_num = num
    return max_num


def without_for():
    return max(lst)

# 保存、循环、判断在C中进行会更快
print(f"with   :{timeit(with_for, number=10000)}")
# 使用builtin函数会更快，例如max、min
# 运行python bytecode的时候会比在cpython中耗费时间
print(f"without:{timeit(without_for, number=10000)}")
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3. 使用any、all

from timeit import timeit

lst = [i for i in range(100)]


def with_for():
    for num in lst:
        if num > 50:
            return True
    return False


def without_for():
    return any(num > 50 for num in lst)

def without_for_list_comprehension():
    return any([num > 50 for num in lst])


# with   :0.008582599999999996
# without:0.022688
# without:0.03122409999999999

print(f"with   :{timeit(with_for, number=10000)}")
# generator 运行在python层面  generator本身的建立和调用需要花费时间
print(f"without:{timeit(without_for, number=10000)}")
# 使用any、all 尽量不使用comprehension
print(f"without:{timeit(without_for_list_comprehension, number=10000)}")
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from timeit import timeit

lst = [False] * 100


def with_for():
    for b in lst:
        if b:
            return True
    return False


def without_for():
    return any(lst)


print(f"with   :{timeit(with_for, number=10000)}")
print(f"without:{timeit(without_for, number=10000)}")
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4. filter、map、zip

from timeit import timeit

lst = [i for i in range(100)]


def good(num):
    return num >= 60


def with_for():
    ret = []
    for num in lst:
        if good(num):
            ret.append(num)
    return ret


def without_for():
    return [num for num in lst if good(num)]


def without_for_builtin():
    return filter(good, lst)


print(f"with   :{timeit(with_for, number=10000)}")
print(f"without:{timeit(without_for, number=10000)}")
print(f"without:{timeit(without_for_builtin, number=10000)}")
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from timeit import timeit

lst = [i for i in range(100)]


def change(num):
    return num * 2


def with_for():
    ret = []
    for num in lst:
        ret.append(change(num))
    return ret


def without_for():
	# return (change(num) for num in lst)
    return map(change, lst)


print(f"with   :{timeit(with_for, number=10000)}")
print(f"without:{timeit(without_for, number=10000)}")
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from timeit import timeit

lst = [i for i in range(100)]
lst2 = [i for i in range(100)]


def change(num, num2):
    return num + num2


def with_for():
    ret = []
    for idx in range(len(lst)):
        ret.append(change(lst[idx], lst2[idx]))
    return ret


def without_for():
    # return (change(lst[i], lst2[i])
    #         for i in range(len(lst)))
    return map(change, lst, lst2)


print(f"with   :{timeit(with_for, number=10000)}")
print(f"without:{timeit(without_for, number=10000)}")
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from timeit import timeit

lst = [i for i in range(100)]
lst2 = [i for i in range(100)]


def with_for():
    ret = []
    for i in range(len(lst)):
        ret.append((lst[i], lst2[i]))
    return ret


def without_for():
    return zip(lst, lst2)


print(f"with   :{timeit(with_for, number=10000)}")
print(f"without:{timeit(without_for, number=10000)}")
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5. built-in函数

built-in 函数 汇总