Con*_*nor 2 python walrus-operator
通过使用海象运算符,我实现了合并排序:
def mergesort(array):
if len(array) == 1:
output = array
else:
pivot = len(array) // 2
left = mergesort(array[pivot:])
right = mergesort(array[:pivot])
output = []
while (l := len(left)) or (r := len(right)):
if l and r and left[0] < right[0]:
output.append(left.pop(0))
elif r:
output.append(right.pop(0))
else:
output.append(left.pop(0))
return output
mergesort([66, 93, 85, 46, 56, 88, 56, 75, 55, 99, 87])
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但这会返回错误UnboundLocalError: local variable 'r' referenced before assignment:
---------------------------------------------------------------------------
UnboundLocalError Traceback (most recent call last)
/tmp/ipykernel_134/1678992391.py in <module>
----> 1 mergesort(array)
/tmp/ipykernel_134/4030760045.py in mergesort(array)
5 else:
6 pivot = len(array) // 2
----> 7 left = mergesort(array[pivot:])
8 right = mergesort(array[:pivot])
9
...
/tmp/ipykernel_134/4030760045.py in mergesort(array)
10 output = []
11 while (l := len(left)) or (r := len(right)):
---> 12 if l and r and left[0] < right[0]:
13 output.append(left.pop(0))
14 elif r:
UnboundLocalError: local variable 'r' referenced before assignment
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r为什么我的 for 循环中没有包含 while l?
布尔 OR or是惰性的,所以当l结果为 true 时,(r := len(right))甚至不会被执行。
在这种情况下,您可以使用非惰性按位或 |,尽管这有点滥用。
或者只使用列表的真值而不是它们的长度:
while left or right:
if left and right and left[0] < right[0]:
output.append(left.pop(0))
elif right:
output.append(right.pop(0))
else:
output.append(left.pop(0))
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顺便说一句,最好使用<=而不是<,这样它就成为归并排序应有的稳定排序。
附录:享受懒惰的乐趣:
while left or right:
which = (left or right)[0] <= (right or left)[0] and left or right
output.append(which.pop(0))
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另,请注意,我切换到while ... and ...循环后并附加剩余的非空循环:
while left and right:
which = left if left[0] <= right[0] else right
output.append(which.pop(0))
output += left or right
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或者回到你的风格:
while left and right:
if left[0] <= right[0]:
output.append(left.pop(0))
else:
output.append(right.pop(0))
output.extend(left or right)
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