找出与lis中下一个元素不同的元素的位置

Question

我想要元素在列表中与下一个元素不同的位置

nums = np.array(["cat","cat","dog","dog","dog","goldfish","bee","bee","cat","cat"])

所以我在列表中使用zip和enumerate

print [c for a,b,c in zip(nums, nums[1:],enumerate(nums)) if a != b]
[(1, 'cat'), (4, 'dog'), (5, 'goldfish'), (7, 'bee')]

我只想要这个职位

>>[1,4,5,7]

如何删除这里的元组？或者有没有更好的（更快的）方法来获得列表中不同元素的位置？你知道吗

Answer 1

只需比较1个偏移的切片，就可以得到不匹配的索引-

np.flatnonzero(nums[1:] != nums[:-1])

Answer 2

如果您正在寻找非numpy解决方案：

In [19]: a = ["cat","cat","dog","dog","dog","goldfish","bee","bee","cat","cat"]

In [20]: [x for x in range(len(a)-1) if a[x] != a[x+1]]
Out[20]: [1, 4, 5, 7]