比较相邻的值，像对删除一样删除并比较新列表

Question

a = [1n，1s，1s，2e，2W，2W，1n，2W]说我有这样的清单。有没有办法以以下方式进行比较。

Pseudo code: Iterate over list [1N, 1S, 1S, 2E, 2W, 1N, 2W], 1==1, delete those values. Iterate over
new list [1S, 2E, 2W, 1N, 2W], 1!=2, move on, 2==2 delete those values. Iterate
over new list [1S, 1N, 2W], 1==1, delete those values. Answer = 2W

到目前为止我拥有的东西。

def dirReduc(arr):
    templist = []
    for i in range(1, len(arr)):
        a = arr[i - 1]
        b = arr[i]
        if a == b:
            templist = (arr[b:])
    (templist)
a = [1, 1, 1, 2, 2, 1, 2]
print(dirReduc(a)

测试案例会产生正确的值，但是我需要运行循环，直到只能得到两个。那就是我卡住

Answer 1

如果您可以理解问题，则只需要一段时间就可以根据需要进行迭代。

a = [1, 1, 1, 2, 2, 1, 2]
finished = False
while not finished:    # Iterate until finished = True
    finished = True    # That only happens when no repeated elements are found
    for i in range(len(a)-1):
        if a[i] == a[i+1]:
            a.pop(i)   # When removing the element i from a,
            a.pop(i)   # now the i + 1 is in the place of i
            print(a)
            finished = False
            break

它将产生：

[1, 2, 2, 1, 2]
[1, 1, 2]
[2]