我有一个项目,我正在尝试做一个复杂的加密方法。使用嵌套循环很常见吗?还是我错过了什么?
我打算创建一个尝试所有字符串来查找密码的方法。例如,当我输入['A','B']
时,它应该逐个创建这些:
['A', 'B', 'AA', 'AB', 'BA', 'BB', 'AAA', 'AAB', 'ABA', 'ABB', 'BAA', 'BAB', 'BBA', 'BBB', (triple permutations), (quadruple permutations), and it goes on ...]
我的代码:
def rulefinder():
global externalrul1
global externalrul2
rul2 = uniquestring1[0]
rul1 = uniquestring1[0]
for n in range (0,3):
for m in range (0, len(uniquestring1)):
for z in range(0, n+1):
for k in range (0,3):
for i in range(0, len(uniquestring1)):
for o in range(0, k+1):
for y in range (0, len(uniquestring1)):
rul2 = rul2[:-1] + uniquestring1[y]
for x in range (0, len(uniquestring1)):
rul1= rul1[:-1] + uniquestring1[x]
code=""
for cha in Genx1:
if cha==uniquestring1[0]:
code +=codereplacer(rul1)
elif cha==uniquestring1[1]:
code +=codereplacer(rul2)
print(code)
print(uniquestring1[0],rul1)
print(uniquestring1[1],rul2)
print(LastString)
if code == LastString:
axiom1=uniquestring1[0]
axiom2=uniquestring1[1]
externalrul1=rul1
externalrul2=rul2
print('rules are found')
print("First RULE:", uniquestring1[0], rul1)
print("Second RULE:", uniquestring1[1], rul2)
findsubgeneration(code, axiom1, rul1, axiom2, rul2)
return
rul1 = rul1[:o] + uniquestring1[i] + rul1[(o + 1):]
rul1 += codereplacer(uniquestring1[i])
rul2 = rul2[:z] + uniquestring1[m] + rul2[(z + 1):]
rul1 =""
rul2 += codereplacer(uniquestring1[m])
你正在以非常MATLAB的方式做事(很多循环,只迭代索引,而不是迭代的元素)。 Pythonic方式效率更高(在发动机罩下使用发电机),并且更加清洁:
import itertools
l = ['A','B']
n = 5 # The size of your "expanding permutations"
res = []
for i in range(1,n):
res.extend(map(''.join, list(itertools.product(l, repeat=i))))
print res
结果:
['A', 'B', 'AA', 'AB', 'BA', 'BB', 'AAA', 'AAB', 'ABA', 'ABB', 'BAA', 'BAB', 'BBA', 'BBB',
'AAAA', 'AAAB', 'AABA', 'AABB', 'ABAA', 'ABAB', 'ABBA', 'ABBB', 'BAAA', 'BAAB', 'BABA', 'BABB', 'BBAA', 'BBAB', 'BBBA', 'BBBB']
警告:只要你正在处理一个小清单,你应该没事,但随着l
的增长,结果会成倍增长,并可能会耗尽你的记忆。因此,不是在extend
列表中执行res
,而是可以在循环中将结果写入磁盘中。