Python-替换后如何还原原始字符串

Question

假设，原始字符串是：

a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

现在，在上面字符串中，所有" D6"的实例都通过：

替换为" 9f"

b = a.replace('D6','9F')
b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

现在，假设我们只有B的价值，而不是A。我们需要在B上执行一些操作以检索A。

的原始值

另外，我们知道" A"的值应满足某种数学条件。

数学条件是：

数字z应该由A(z％a == 0(

对

z = 55057004365075793824891923502198296150348187500859129529014955509148421282041969078213265169463529503768779794209446773790749529176461595867792548236095966024387560672845152234957439383409540755826755640123124159246487058454615922008741879614211920551517049373314503998980825185719370304183623398662036133862488876163410866971729000216470924616148028986990798495248878127793311548452974671645100371499570058070179424193067736979204502413302335974105838586819414807952974885796840178274113497125765593996690493177955553456655538977929256055738007112424150644005452979891672942537126552535517394691741201589304958975238

我们需要利用" a"的数学属性来从b。

恢复它

我知道我们不能在b上使用repent((函数来恢复" A"，因为" 9f"的某些原始实例可能被替换。

我认为我们需要在不同位置执行替换操作，直到我们满足这种情况为止。

对于给定的字符串，我们有：

b.count('9F')
6

因此，我们必须以不同的位置组合替换" 9F"，直到我们回来'一个"满足数学条件"，" C"。

我做了类似的事情：

count = b.count('9F')
for i in range(1,count+1):
    print "trying: %d" %(i)
    tmp = b.replace('9F','D6',i)
    num = int(tmp, 16)
    if z % num == True:
        print num

这似乎是与排列和组合有关的问题。

Answer 1

您已经在原始字符串中具有4 CC_1实例。那些引起了Wierd行为

a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
print(a.count('9F'))
#4

否则string.replace工作完美

a = 'hello'
b = a.replace('l','a')
print(b)
#heaao
c = b.replace('a','l')
print(c)
#hello
print( a == c)
#True

给您True(将l转换为a，然后将a转换为l(

可能的解决方案，将现有的9F替换为字符串中不存在的东西，例如XY，然后将其用于替换

a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
#Replace 9F to XY, then D6 to 9F
b = a.replace('9F','XY').replace('D6','9F')
#Replace 9F to D6, then XY to 9F
c = b.replace('9F', 'D6').replace('XY', '9F')
print(a == c)
#True

Answer 2

您不会在这两个操作之后检索原始字符串，因为您在原始字符串中也可能有" 9F"，也不会：

test = "D69F"
changed = test.replace("D6", "9F")
print(changed)
# 9F9F
undo_change = changed.replace("9F", "D6")
print(undo_change)
# D6D6

如果它不会干扰您的中间步骤，则可以掩盖您的替代品，例如使用test.replace("D6", "§9F")。