Python中利用all()来优化减少判断的实例分析

def checker(nums):
    for x in range(len(nums)):
        if x+1 < len(nums):
            if nums[x] > nums[x+1]:
                return False
    return True

def checker(nums):
    return all(i <= j for i, j in zip(nums, nums[1:]))
#如果不想为zip的第二个参数创建个list，可以这样写
from itertools import islice

def checker(nums):
    return all(i <= j for i, j in zip(nums, islice(nums, 1, None)))

>>>a = [1,2,3] #此处可迭代对象为列表
>>> b = [4,5,6]
>>> c = [4,5,6,7,8]
>>> zipped = zip(a,b)     # 打包为元组的列表
>>> zipped
[(1, 4), (2, 5), (3, 6)]
>>> zip(a,c)              # 元素个数与最短的列表一致
[(1, 4), (2, 5), (3, 6)]
>>> zip(*zipped)          # 与 zip 相反，可理解为解压
[(1, 2, 3), (4, 5, 6)]

>>> a = [1,2,3] #此处可迭代对象为列表
>>> b = [4,5,6]
>>> c = [4,5,6,7,8]
>>> zipped = zip(a,b)
>>> zipped  #返回的是一个对象
>>> list(zipped)
[(1, 4), (2, 5), (3, 6)] #使用list()函数转换为列表
>>> list(zip(a,c))
[(1, 4), (2, 5), (3, 6)]
>>> zipped = zip(a,b)
>>> list(zip(*zipped)) #解压也使用list进行转换
[(1, 2, 3), (4, 5, 6)]