随机洗牌 [英] random shuffles
问题描述
x.sort（cmp = lambda x，y：cmp（random.random（），0.5））
选择随机抽取的x均匀分布？
直观地说，假设list.sort（）对
进行最小数量的比较实现排序，我我说答案是肯定的。但我感觉不太舒服
凭直觉......谁能想到更坚实的论证呢？
- BB
-
" On na ?? t tous les m？¨tresdum？ame monde"
does
x.sort(cmp = lambda x,y : cmp(random.random(),0.5))
pick a random shuffle of x with uniform distribution ?
Intuitively, assuming list.sort() does a minimal number of comparisons to
achieve the sort, I''d say the answer is yes. But I don''t feel quite confortable
with the intuition... can anyone think of a more solid argumentation ?
- BB
--
"On na??t tous les m?¨tres du m?ame monde"
推荐答案
Boris Borcic写道：
Boris Borcic wrote:
确实
x.sort（cmp = lambda x，y：cmp（ random.random（），0.5））
随机抽取x均匀分布？
直观地，假设list.sort（ ）对b / b
进行最小数量的比较，我会说答案是肯定的。但是，我觉得自己并不觉得这种直觉很舒服......有人能想到一个更坚实的讨论吗？
does
x.sort(cmp = lambda x,y : cmp(random.random(),0.5))
pick a random shuffle of x with uniform distribution ?
Intuitively, assuming list.sort() does a minimal number of comparisons to
achieve the sort, I''d say the answer is yes. But I don''t feel quite
confortable with the intuition... can anyone think of a more solid
argumentation ?
轶事证据表明答案是否定的：
Anecdotal evidence suggests the answer is no:
>> histo = {}
我在xrange（1000）：
>>histo = {}
for i in xrange(1000):
.... t =元组（排序（范围（3），lambda x，y：cmp（random.random（），0.5）））
.... histo [t] = histo .get（t，0）+ 1
....
.... t = tuple(sorted(range(3), lambda x, y: cmp(random.random(), 0.5)))
.... histo[t] = histo.get(t, 0) + 1
....
>>已排序（histo.values（））
>>sorted(histo.values())
[60,62,64,122,334， 358]
对：
[60, 62, 64, 122, 334, 358]
versus:
> > histo = {}
for x in xrange（1000）：
>>histo = {}
for i in xrange(1000):
.... t = tuple（ sorted（range（3），key = lambda x：random.random（）））
... .histo [t] = histo.get（t，0）+ 1
....
.... t = tuple(sorted(range(3), key=lambda x: random.random()))
.... histo[t] = histo.get(t, 0) + 1
....
>> sorted（histo.values（））
>>sorted(histo.values())
[147,158] ，160,164,183,188]
彼得
[147, 158, 160, 164, 183, 188]
Peter
Boris Borcic写道：
Boris Borcic wrote:
确实
x.sort（cmp = lambda x，y：cmp（random.random（），0.5））
随机抽取x均匀分布？
直观地说，假设list.sort（）与
达到排序，我会说答案是肯定的。但我感觉不太舒服
凭直觉......谁能想到更坚实的论证？
does
x.sort(cmp = lambda x,y : cmp(random.random(),0.5))
pick a random shuffle of x with uniform distribution ?
Intuitively, assuming list.sort() does a minimal number of comparisons to
achieve the sort, I''d say the answer is yes. But I don''t feel quite confortable
with the intuition... can anyone think of a more solid argumentation ?
为什么不使用提供的shuffle方法？
random.shuffle（x）
Why not use the supplied shuffle method?
random.shuffle(x)
Dustan写道：
Dustan wrote:
Boris Borcic写道：
Boris Borcic wrote:
确实
x.sort（cmp = lambda x，y：cmp（random.random（），0.5））
随机选择x随机均匀分布？
直观地说，假设list.sort（）与
进行最小数量的比较实现这种，我会说答案是肯定的。但我感觉不太舒服
凭直觉......谁能想到更坚实的论证？
does
x.sort(cmp = lambda x,y : cmp(random.random(),0.5))
pick a random shuffle of x with uniform distribution ?
Intuitively, assuming list.sort() does a minimal number of comparisons to
achieve the sort, I''d say the answer is yes. But I don''t feel quite confortable
with the intuition... can anyone think of a more solid argumentation ?
为什么不使用提供的shuffle方法？
random.shuffle（x）
Why not use the supplied shuffle method?
random.shuffle(x)
或查看此主题：
http://groups.google.com/group/comp....vc=2&q=shuffle
Iain
or check out this thread:
http://groups.google.com/group/comp....vc=2&q=shuffle
Iain
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