在Python2.7.6中列出迭代器和在迭代器上迭代的区别
我最近一直在玩一些涉及迭代器的代码:在Python2.7.6中列出迭代器和在迭代器上迭代的区别,python,python-2.7,Python,Python 2.7,我最近一直在玩一些涉及迭代器的代码: """IntegerPartitions.py Generate and manipulate partitions of integers into sums of integers. D. Eppstein, August 2005. https://www.ics.uci.edu/~eppstein/PADS/IntegerPartitions.py """ def mckay(n): """ Integer partition
"""IntegerPartitions.py
Generate and manipulate partitions of integers into sums of integers.
D. Eppstein, August 2005.
https://www.ics.uci.edu/~eppstein/PADS/IntegerPartitions.py
"""
def mckay(n):
"""
Integer partitions of n, in reverse lexicographic order.
The output and asymptotic runtime are the same as mckay(n),
but the algorithm is different: it involves no division,
and is simpler than mckay, but uses O(n) extra space for
a recursive call stack.
"""
if n == 0:
yield []
if n <= 0:
return
for p in mckay(n-1):
if len(p) == 1 or (len(p) > 1 and p[-1] < p[-2]):
p[-1] += 1
yield p
p[-1] -= 1
p.append(1)
yield p
p.pop()
分区似乎不会显示,除非我逐个打印它们。这是语言中的错误(我的版本是UbuntuTrusty上的Python2.7.6),还是我遗漏了什么?我在谷歌上四处搜索,似乎找不到任何与此相关的信息
我认为这可能与递归调用有关,但我用下面的代码进行了尝试,并发现了类似的结果
def mckay(n):
"""
Integer partitions of n, in reverse lexicographic order.
Note that the generated output consists of the same list object,
repeated the correct number of times; the caller must leave this
list unchanged, and must make a copy of any partition that is
intended to last longer than the next call into the generator.
The algorithm follows Knuth v4 fasc3 p38 in rough outline.
"""
if n == 0:
yield []
if n <= 0:
return
partition = [n]
last_nonunit = (n > 1) - 1
while True:
yield partition
if last_nonunit < 0:
return
if partition[last_nonunit] == 2:
partition[last_nonunit] = 1
partition.append(1)
last_nonunit -= 1
continue
replacement = partition[last_nonunit] - 1
total_replaced = replacement + len(partition) - last_nonunit
reps,rest = divmod(total_replaced,replacement)
partition[last_nonunit:] = reps*[replacement]
if rest:
partition.append(rest)
last_nonunit = len(partition) - (partition[-1]==1) - 1
问题是函数
mckay
正在修改同一个列表对象,因此当您对其调用list()
时,实际上会得到一个包含4个项目的列表,这些项目实际上指向同一个对象。因此,正如列表对象最后是空的一样,您得到的只是带有空列表的列表
>>> p = mckay(4)
>>> [id(x) for x in p]
[139854369904832, 139854369904832, 139854369904832, 139854369904832, 139854369904832]
>>> for x in mckay(4):
print x, '-->', id(x)
[4] --> 140446845125552
[3, 1] --> 140446845125552
[2, 2] --> 140446845125552
[2, 1, 1] --> 140446845125552
[1, 1, 1, 1] --> 140446845125552
>>> x # The actual list object is empty at the end of the iteration
[]
>>> id(x)
140446845125552
但是,当您在其上循环时,您只是立即打印返回的对象,因此输出不同,这里的修复方法是生成一个浅拷贝:
yield p[:]
>>> p = mckay(4)
>>> [id(x) for x in p]
[139854369904832, 139854369904832, 139854369904832, 139854369904832, 139854369904832]
>>> for x in mckay(4):
print x, '-->', id(x)
[4] --> 140446845125552
[3, 1] --> 140446845125552
[2, 2] --> 140446845125552
[2, 1, 1] --> 140446845125552
[1, 1, 1, 1] --> 140446845125552
>>> x # The actual list object is empty at the end of the iteration
[]
>>> id(x)
140446845125552
yield p[:]