Python:比较三个列表的值以创建新列表

Question

我正在尝试比较Python中三个列表的值，并根据这些比较创建一个新列表。为了使其更令人困惑，当语句= true。

时，第三列表计数（ 1）

例如：

list1 = [5,10,15,20,25]
list2 = [3,6,9,12,15]
list3 = [0,1,11,21,28]
list4 = []
if list1 >= list3 and list2 <= list3:
    list4 = 0
else:
    list4 = list4 + 1

所以List4应该看起来像：List4 [1,2,0,1,2]

任何帮助将不胜感激！

谢谢

Answer 1

您可以使用Generator函数和zip：

def solve(list1, list2, list3):
    last = 0
    for a, b, c in zip(list1, list2, list3):
        if a >= c and b <= c:
            yield 0
            last = 0
        else:
            last += 1     
            yield last
...             
>>> list(solve(list1, list2, list3))
[1, 2, 0, 1, 2]

Answer 2

list1 = [5,10,15,20,25]
list2 = [3,6,9,12,15]
list3 = [0,1,11,21,28]
answer = []
app = 0
for a,b,c in zip(list1, list2, list3):
    if a>=c>=b:
        app = 0
    else:
        app += 1
    answer.append(app)

证明：

>>> list1 = [5,10,15,20,25]
>>> list2 = [3,6,9,12,15]
>>> list3 = [0,1,11,21,28]
>>> answer = []
>>> 
>>> app = 0
>>> for a,b,c in zip(list1, list2, list3):
...     if a>=c>=b:
...         app = 0
...     else:
...         app += 1
...     answer.append(app)
... 
>>> answer
[1, 2, 0, 1, 2]