import itertools
import string
def crackdict(max, min=1, chars=None):
assert max >= min >= 1
if chars is None:
import string
chars = string.printable[:-5]
p = []
for i in xrange(min, max + 1):
p.append(itertools.product(string.printable[:-5], repeat=i))
return itertools.chain(*p)
#Just for test:
max, min = 4, 1
d = crackdict(max, min)
i want to print the result like this 0 1 2 3 4 5 6
but now the print like [('0',), ('1',), ('2',), ('3',), ('4',), ('5',), ('6',), ('7',), ('8',), ('9',), ('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',), ('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',), ('!',), ('"',), ('#',), ('$',), ('%',), ('&',), ("'",), ('(',), (')',), ('*',), ('+',), (',',), ('-',), ('.',), ('/',), (':',), (';',), ('<',), ('=',), ('>',), ('?',), ('@',), ('[',), ('\',), (']',), ('^',), ('_',), ('`',), ('{',), ('|',), ('}',), ('~',), (' ',), ('0', '0'), ('0', '1'), ('0', '2'), ('0', '3'), ('0', '4'), ('0', '5'), ('0', '6'), ('0', '7'), ('0', '8'), ('0', '9'), ('0', 'a'), ('0', 'b'), ('0', 'c'), ('0', 'd'), ('0', 'e'), ('0', 'f'), ('0', 'g'), ('0', 'h'), ('0', 'i'), ('0', 'j'), ('0', 'k'), ('0', 'l'), ('0', 'm'), ('0', 'n'), ('0', 'o'), ('0', 'p'), ('0', 'q'), ('0', 'r'), ('0', 's'), ('0', 't'), ('0', 'u'), ('0', 'v'), ('0', 'w'), ('0', 'x'), ('0', 'y'), ('0', 'z'), ('0', 'A'), ('0', 'B'), ('0', 'C'), ('0', 'D'), ('0', 'E'), ('0', 'F'), ('0', 'G'), ('0', 'H'), ('0', 'I'), ('0', 'J'), ('0',
how can i do it ? thanks!
('1', '2', '3')similar in any way to "aaaaaa"? Do you think you could make your example consistent with itself?