我有3名名單:產品與descreasing值
a = [10, 9, 8, 7, 6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
我需要得到具有itertools.product()
獲得的所有排列,但只有當值下降:
[10, 8, 6] # is good
[6, 8, 4] # is not good, since 8 > 6
有一個簡單的這樣做的方式還是應該與列表理解和條件?
我有3名名單:產品與descreasing值
a = [10, 9, 8, 7, 6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
我需要得到具有itertools.product()
獲得的所有排列,但只有當值下降:
[10, 8, 6] # is good
[6, 8, 4] # is not good, since 8 > 6
有一個簡單的這樣做的方式還是應該與列表理解和條件?
你可以通過遍歷的itertools.product
迭代器只提取了在反向排序那些返回的項目與列表理解這樣做:
[item for item in product(a,b,c) if sorted(item, reverse = True) == list(item)]
例子:
from itertools import product
a = [10,9,8,7,6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
[item for item in product(a,b,c) if sorted(item, reverse = True) == list(item)]
# [(10, 8, 6), (10, 8, 5), (10, 8, 4), (10, 8, 3), (10, 8, 2) ...continues
這是一個簡單的一線解決方案
>>> mylist = [10, 9, 8, 7, 6]
>>> all(earlier >= later for earlier, later in zip(mylist, mylist[1:]))
True
>>> mylist = [10, 9, 7, 8, 6]
>>> all(earlier >= later for earlier, later in zip(mylist, mylist[1:]))
False
我在這裏找到這裏:
如果你不想使用列表理解爲某種原因:
def decreasing(l):
return all(a >= b for a, b in zip(l[:-1], l[1:]))
filter(decreasing, product(a, b, c))
你可以參考下面的代碼不包含列表解析:
from itertools import product
a = [10, 9, 8, 7, 6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
for result in product(a,b,c):
if sorted(result, reverse = True) == list(result):
print result