Question

而不是完整的shuffle，我在python中寻找部分 shuffle函数。

示例：“string”必须产生“stnrig”，但不能产生“nrsgit”

如果我可以定义必须重新排列的特定“百分比”字符，那会更好。

目的是测试字符串比较算法。我想确定“shuffle的百分比”，超过这个百分比，（my）算法会将两个（混洗）字符串标记为完全不同。

更新：

这是我的代码。欢迎改进！

import random

percent_to_shuffle = int(raw_input("Give the percent value to shuffle : "))
to_shuffle = list(raw_input("Give the string to be shuffled : "))

num_of_chars_to_shuffle = int((len(to_shuffle)*percent_to_shuffle)/100)

for i in range(0,num_of_chars_to_shuffle):
    x=random.randint(0,(len(to_shuffle)-1))
    y=random.randint(0,(len(to_shuffle)-1))
    z=to_shuffle[x]
    to_shuffle[x]=to_shuffle[y]
    to_shuffle[y]=z

print ''.join(to_shuffle)

Answer 1

这是一个比它看起来更简单的问题。并且语言有正确的工具，不要像往常一样留在你和想法之间：

import random

def pashuffle(string, perc=10):
    data = list(string)
    for index, letter in enumerate(data):
        if random.randrange(0, 100) < perc/2:
            new_index = random.randrange(0, len(data))
            data[index], data[new_index] = data[new_index], data[index]
    return "".join(data)

Answer 2

你的问题很棘手，因为有一些优势需要考虑：

带有重复字符的字符串（即你如何改变“aaaab”？）
如何衡量链接字符交换或重新排列块？

在任何情况下，定义为将字符串混合到一定百分比的度量标准可能与您在算法中使用的度量标准相同，以查看它们的接近程度。

我的代码用于随机播放n字符：

import random
def shuffle_n(s, n):
  idx = range(len(s))
  random.shuffle(idx)
  idx = idx[:n]
  mapping = dict((idx[i], idx[i-1]) for i in range(n))
  return ''.join(s[mapping.get(x,x)] for x in range(len(s)))

基本上选择n位置随机交换，然后将它们与列表中的下一个交换...这样就可以确保不生成逆交换，并且确切地n个字符是交换（如果有重复的字符，运气不好）。

用'string'解释运行，3作为输入：

idx is [0, 1, 2, 3, 4, 5]
we shuffle it, now it is [5, 3, 1, 4, 0, 2]
we take just the first 3 elements, now it is [5, 3, 1]
those are the characters that we are going to swap
s t r i n g
  ^   ^   ^
t (1) will be i (3)
i (3) will be g (5)
g (5) will be t (1)
the rest will remain unchanged
so we get 'sirgnt'

这种方法的坏处是它不会产生所有可能的变化，例如，它无法从'string'制作'gnrits'。这可以通过调整索引的分区来修复，如下所示：

import random

def randparts(l):
    n = len(l)
    s = random.randint(0, n-1) + 1
    if s >= 2 and n - s >= 2: # the split makes two valid parts
        yield l[:s]
        for p in randparts(l[s:]):
            yield p
    else: # the split would make a single cycle
        yield l

def shuffle_n(s, n):
    idx = range(len(s))
    random.shuffle(idx)
    mapping = dict((x[i], x[i-1])
        for i in range(len(x))
        for x in randparts(idx[:n]))
    return ''.join(s[mapping.get(x,x)] for x in range(len(s)))

Answer 3

也许是这样：

>>> s = 'string'
>>> shufflethis = list(s[2:])
>>> random.shuffle(shufflethis)
>>> s[:2]+''.join(shufflethis)
'stingr'

根据fortran的想法，我将其添加到收藏中。这很快：

def partial_shuffle(st, p=20):
    p = int(round(p/100.0*len(st)))

    idx = range(len(s))
    sample = random.sample(idx, p)

    res=str()
    samptrav = 1

    for i in range(len(st)):
        if i in sample:
            res += st[sample[-samptrav]]
            samptrav += 1
            continue
        res += st[i]

    return res

Answer 4

import random

def partial_shuffle(a, part=0.5):
    # which characters are to be shuffled:
    idx_todo = random.sample(xrange(len(a)), int(len(a) * part))

    # what are the new positions of these to-be-shuffled characters:
    idx_target = idx_todo[:]
    random.shuffle(idx_target)

    # map all "normal" character positions {0:0, 1:1, 2:2, ...}
    mapper = dict((i, i) for i in xrange(len(a)))

    # update with all shuffles in the string: {old_pos:new_pos, old_pos:new_pos, ...}
    mapper.update(zip(idx_todo, idx_target))

    # use mapper to modify the string:
    return ''.join(a[mapper[i]] for i in xrange(len(a)))

for i in xrange(5):
    print partial_shuffle('abcdefghijklmnopqrstuvwxyz', 0.2)

打印

abcdefghljkvmnopqrstuxwiyz
ajcdefghitklmnopqrsbuvwxyz
abcdefhwijklmnopqrsguvtxyz
aecdubghijklmnopqrstwvfxyz
abjdefgcitklmnopqrshuvwxyz

Answer 5

邪恶并使用已弃用的API：

import random
# adjust constant to taste
# 0 -> no effect, 0.5 -> completely shuffled, 1.0 -> reversed
# Of course this assumes your input is already sorted ;)
''.join(sorted(
    'abcdefghijklmnopqrstuvwxyz',
    cmp = lambda a, b: cmp(a, b) * (-1 if random.random() < 0.2 else 1)
))

如何在Python中进行随机但部分的随机播放？

5 个答案: