我想从排列中的字符串中删除字符.

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,然后我希望将字符串[aabb,aAabb,aabbA,aabbA,aabbAA,aAabbA,aAabbA]作为输出,将A删除3次,2次,1次.

我能做到的最好方法是什么？

非常感谢….

解决方法:

这是使用递归的一个疯狂想法：

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']

编辑：使用设置：

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'])

编辑2：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']

编辑3：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