使用 itertools.product 与…结合 itertools.chain 将各种长度组合在一起:

from itertools import chain, product
def bruteforce(charset, maxlength):
    return (''.join(candidate)
        for candidate in chain.from_iterable(product(charset, repeat=i)
        for i in range(1, maxlength + 1)))

演示:

>>> list(bruteforce('abcde', 2))
['a', 'b', 'c', 'd', 'e', 'aa', 'ab', 'ac', 'ad', 'ae', 'ba', 'bb', 'bc', 'bd', 'be', 'ca', 'cb', 'cc', 'cd', 'ce', 'da', 'db', 'dc', 'dd', 'de', 'ea', 'eb', 'ec', 'ed', 'ee']

这将有效地使用输入集生成逐渐变大的单词,最大长度可达maxlength。

做 不 尝试在存储器中产生长度为10的26个字符的列表;相反,对生成的结果进行迭代:

for attempt in bruteforce(string.ascii_lowercase, 10):
    # match it against your password, or whatever
    if matched:
        break

如果你真的想强行使用它,试试这个,但它会花费你荒谬的时间:

your_list = 'abcdefghijklmnopqrstuvwxyz'
complete_list = []
for current in xrange(10):
    a = [i for i in your_list]
    for y in xrange(current):
        a = [x+i for i in your_list for x in a]
    complete_list = complete_list+a

在一个较小的例子中,list=“ab”,我们最多只能到达5,这将打印以下内容:

['a', 'b', 'aa', 'ba', 'ab', 'bb', 'aaa', 'baa', 'aba', 'bba', 'aab', 'bab', 'abb', 'bbb', 'aaaa', 'baaa', 'abaa', 'bbaa', 'aaba', 'baba', 'abba', 'bbba', 'aaab', 'baab', 'abab', 'bbab', 'aabb', 'babb', 'abbb', 'bbbb', 'aaaaa', 'baaaa', 'abaaa', 'bbaaa', 'aabaa', 'babaa', 'abbaa', 'bbbaa', 'aaaba','baaba', 'ababa', 'bbaba', 'aabba', 'babba', 'abbba', 'bbbba', 'aaaab', 'baaab', 'abaab', 'bbaab', 'aabab', 'babab', 'abbab', 'bbbab', 'aaabb', 'baabb', 'ababb', 'bbabb', 'aabbb', 'babbb', 'abbbb', 'bbbbb']

我发现了另一种使用itertools创建字典的非常简单的方法。

generator=itertools.combinations_with_replacement('abcd', 4 )

这将遍历“a”、“b”、“c”和“d”的所有组合,并创建总长度为1到4的组合。即a、b、c、d、aa、ab。。。。。。。。。,dddc,dddd。generator是一个itertool对象,您可以像这样正常地循环,

for password in generator:
        ''.join(password)

每个密码都是元组类型的infact,您可以像往常一样使用它们。

如果你真的想要一个brueforce算法,不要在你的计算机内存中保存任何大列表,除非你想要一个会因MemoryError而崩溃的慢速算法。

你可以试着这样使用itertools.product:

from string import ascii_lowercase
from itertools import product

charset = ascii_lowercase  # abcdefghijklmnopqrstuvwxyz
maxrange = 10


def solve_password(password, maxrange):
    for i in range(maxrange+1):
        for attempt in product(charset, repeat=i):
            if ''.join(attempt) == password:
                return ''.join(attempt)


solved = solve_password('solve', maxrange)  # This worked for me in 2.51 sec

itertools.product(*iterables) 返回您输入的可迭代项的笛卡尔乘积。

[i for i in product('bar', (42,))] 退货,例如。 [('b', 42), ('a', 42), ('r', 42)]

这个 repeat 参数允许您准确地做出您所要求的:

[i for i in product('abc', repeat=2)]

退货

[('a', 'a'),
 ('a', 'b'),
 ('a', 'c'),
 ('b', 'a'),
 ('b', 'b'),
 ('b', 'c'),
 ('c', 'a'),
 ('c', 'b'),
 ('c', 'c')]

笔记 以下为:

你想要一个暴力算法,所以我给了你。现在,当密码开始变大时,这是一个非常长的方法,因为它呈指数级增长(花了62秒才找到“已解决”一词)。

itertools 非常适合这样做:

itertools.chain.from_iterable((''.join(l)
                               for l in itertools.product(charset, repeat=i))
                              for i in range(1, maxlen + 1))

使用递归的解决方案:

def brute(string, length, charset):
    if len(string) == length:
        return
    for char in charset:
        temp = string + char
        print(temp)
        brute(temp, length, charset)

用法:

brute("", 4, "rce")

import string, itertools

    #password = input("Enter password: ")

    password = "abc"

    characters = string.printable

    def iter_all_strings():
        length = 1
        while True:
            for s in itertools.product(characters, repeat=length):
                yield "".join(s)
            length +=1

    for s in iter_all_strings():
        print(s)
        if s == password:
            print('Password is {}'.format(s))
            break

使用itertools和字符串模块的简单解决方案

# modules to easily set characters and iterate over them
import itertools, string 

# character limit so you don't run out of ram
maxChar = int(input('Character limit for password: '))  

# file to save output to, so you can look over the output without using so much ram
output_file = open('insert filepath here', 'a+') 

# this is the part that actually iterates over the valid characters, and stops at the 
# character limit.
x = list(map(''.join, itertools.permutations(string.ascii_lowercase, maxChar))) 

# writes the output of the above line to a file 
output_file.write(str(x)) 

# saves the output to the file and closes it to preserve ram
output_file.close() 

我用管道将输出输出到一个文件中以保存ram,并使用输入函数将字符限制设置为“hiiworld”之类的值。下面是相同的脚本,但使用了字母、数字、符号和空格的更流畅的字符集。

import itertools, string

maxChar = int(input('Character limit for password: '))
output_file = open('insert filepath here', 'a+')

x = list(map(''.join, itertools.permutations(string.printable, maxChar)))
x.write(str(x))
x.close()

from random import choice

sl = 4  #start length
ml = 8 #max length 
ls = '9876543210qwertyuiopasdfghjklzxcvbnm' # list
g = 0
tries = 0

file = open("file.txt",'w') #your file

for j in range(0,len(ls)**4):
    while sl <= ml:
        i = 0
        while i < sl:
            file.write(choice(ls))
            i += 1
        sl += 1
        file.write('\n')
        g += 1
    sl -= g
    g = 0
    print(tries)
    tries += 1


file.close()

试试这个:

import os
import sys

Zeichen=["a","b","c","d","e","f","g","h"­,"i","j","k","l","m","n","o","p","q­","r","s","­;t","u","v","w","x","y","z"]
def start(): input("Enter to start")
def Gen(stellen): if stellen==1: for i in Zeichen: print(i) elif stellen==2: for i in Zeichen:    for r in Zeichen: print(i+r) elif stellen==3: for i in Zeichen: for r in Zeichen: for t in Zeichen:     print(i+r+t) elif stellen==4: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in Zeichen:    print(i+r+t+u) elif stellen==5: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in    Zeichen: for o in Zeichen: print(i+r+t+u+o) else: print("done")

#*********************
start()
Gen(1)
Gen(2)
Gen(3)
Gen(4)
Gen(5)