获取字符串列表，其中字符已在排列中删除 [英] Getting a list of strings with character removed in permutation

问题描述

我想从排列中的字符串中删除一个字符....

I want to remove a character from a string in permutation....

让我们说我有一个功能

def (string,char):
    # remove char from string

假设我将aAabbAA作为字符串，将A作为char，然后我希望将作为输出的字符串[aabb,aAabb,aabbA,aabbA, aabbAA,aAabbA ,aAabbA ]删除3次，2次，1次.

Say I have aAabbAA as string and A as char then I want the strings [aabb,aAabb,aabbA,aabbA, aabbAA,aAabbA ,aAabbA ] as output that is A gets removed 3 times , 2 times , 1 times.

我能做到的最好方法是什么?

What is the best way in which I can do that ??

非常感谢....

推荐答案

这是使用递归的一个疯狂主意:

Here is one crazy idea using recursion:

def f(s, c, start):
    i = s.find(c, start)
    if i < 0:
        return [s]
    else:
        return f(s, c, i+1) + f(s[:i]+s[i+1:], c, i)

s = 'aAabbAA'
print f(s, 'A', 0)
# ['aAabbAA', 'aAabbA', 'aAabbA', 'aAabb', 'aabbAA', 'aabbA', 'aabbA', 'aabb']

使用set:

def f(s, c, start):
    i = s.find(c, start)
    if i < 0:
        return set([s])
    else:
        return set.union(f(s, c, i+1), f(s[:i]+s[i+1:], c, i))

s = 'aAabbAA'
print f(s, 'A', 0)
# set(['aAabbA', 'aabbAA', 'aAabbAA', 'aabb', 'aAabb', 'aabbA'])

使用三元运算符:

def f(s, c, start):
    i = s.find(c, start)
    return [s] if i < 0 else f(s, c, i+1) + f(s[:i]+s[i+1:], c, i)

s = 'aAabbAA'
print f(s, 'A', 0)
# ['aAabbAA', 'aAabbA', 'aAabbA', 'aAabb', 'aabbAA', 'aabbA', 'aabbA', 'aabb']

timeit:

In [32]: timeit.timeit('x = f("aAabbAA", "A", 0)', 
                       'from test3 import f', number=10000) 
Out[32]: 0.11674594879150391

In [33]: timeit.timeit('x = deperm("aAabbAA", "A")', 
                       'from test4 import deperm', number=10000) 
Out[33]: 0.35839986801147461

In [34]: timeit.timeit('x = f("aAabbAA"*6, "A", 0)', 
                       'from test3 import f', number=1) 
Out[34]: 0.45998811721801758

In [35]: timeit.timeit('x = deperm("aAabbAA"*6, "A")', 
                       'from test4 import deperm', number=1) 
Out[35]: 7.8437530994415283

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