这可能不会赢得任何效率奖,但您会获得列表理解的风格点数。
这就是我解决问题的方式。列出大小为 3 的滑动窗口。
>>> nums = [1, 3, 5, 6, 7, 8, 9, 10, 15, 19, 20, 22, 23, 24, 26, 27, 28, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48]
>>> [nums[i:i+3] for i in xrange(len(nums))]
[[1, 3, 5], [3, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8, 9], [8, 9, 10], [9, 10, 15], [10, 15, 19], [15, 19, 20], [19, 20, 22], [20, 22, 23], [22, 23, 24], [23, 24, 26], [24, 26, 27], [26, 27, 28], [27, 28, 32], [28, 32, 33], [32, 33, 35], [33, 35, 37], [35, 37, 38], [37, 38, 39], [38, 39, 40], [39, 40, 41], [40, 41, 42], [41, 42, 43], [42, 43, 44], [43, 44, 47], [44, 47, 48], [47, 48], [48]]
下一步,摆脱连续的项目,现在这很容易。这个谓词将巧妙地过滤掉连续的项目。
>>> [nums[i] for i in xrange(len(nums)) if nums[i:i+3] != range(nums[i],nums[i]+3)]
[1, 3, 9, 10, 15, 19, 20, 23, 24, 27, 28, 32, 33, 35, 43, 44, 47, 48]
编辑:
Eric 提出了一个很好的观点,上面的解决方案并不完全有效。如果你想让这个工作,那么谓词将需要一些加强。首先,我导出了这些方程。他们执行窗口操作。说服自己他们是真的:
a = [1,2,3,4,5]
i = 2
a[i-0:i+3] == range(a[i-0], a[i]+3) # left
a[i-1:i+2] == range(a[i-1], a[i]+2) # center
a[i-2:i+1] == range(a[i-2], a[i]+1) # right
然后你可以把它塞到一边...
[a for i,a in enumerate(nums) if all(nums[i-j:i+k] != range(nums[i-j], nums[i]+k) for j,k in zip(xrange(0,3,1), xrange(3,0,-1)))]
但是,如果您不想被击中,请将谓词拉出到一个函数中:
consec_to_buddies = lambda i, xs: (
xs[i-0:i+3] == range(xs[i-0], xs[i]+3) or
xs[i-1:i+2] == range(xs[i-1], xs[i]+2) or
xs[i-2:i+1] == range(xs[i-2], xs[i]+1)
)
[a for i,a in enumerate(nums) if not consec_to_buddies(i, nums)]
同样,这不是最有效的,因为您将计算每个项目的谓词,即使您已经知道要取出它。你为优雅付出的代价:)