2015-05-16 149 views
1

我正在创建一个循环遍历字母表的脚本,之后再添加一个循环的新字母。 这里是我到目前为止的代码:Python:通过字母循环

import string 
for a in (string.ascii_letters + string.digits): 
    print (a) 
for a in (string.ascii_letters + string.digits): 
    for b in (string.ascii_letters + string.digits): 
     print(a + b) 
for a in (string.ascii_letters + string.digits): 
    for b in (string.ascii_letters + string.digits): 
     print(a + b) 
     for c in (string.ascii_letters + string.digits): 
      print(a + b + c) 

什么是这样做的更有效的方法?

回答

2

itertools模块将让你定义这样的字符串的无限流。

import string 
import itertools 

def alphanumeric_strings(): 
    for n in itertools.count(): 
     for t in itertools.product(string.ascii_letters + string.digits, repeat=n): 
      yield ''.join(t) 

for x in alphanumeric_strings(): 
    print x 

count简单地生成一个整数从1计数的无限流,这样在每次调用product时候,你会得到一个元组的流与一个比去年更位数。对join的调用将像('a', 'B', '1')这样的元组转换为字符串aB1

1

您可以使用itertools.product,这就是“等同于嵌套的for循环在发电机表达”:

import itertools 
import string 
ALPHABET = string.ascii_letters + string.digits 
for a in itertools.product(ALPHABET, repeat=1): 
    print(a) 
for a,b in itertools.product(ALPHABET, repeat=2): 
    print(a + b) 
# etc. 
1

使用itertools:

import string 
from itertools import count, product 

def words(chars=string.ascii_letters + string.digits): 
    for n in count(1): 
     yield from map(''.join, product(chars, repeat=n)) 

不使用itertools:

import string 

def words(chars=string.ascii_letters + string.digits): 
    yield from chars 
    for word in words(chars): 
     for char in chars: 
      yield word + char 

两个产生从给定的人物所做的所有的话:

for word in words('aB3'): 
    print(word) 

打印:

a 
B 
3 
aa 
aB 
a3 
Ba 
BB 
B3 
3a 
3B 
33 
aaa 
aaB 
... 
0

的另一种方法:

abc = {0:"", 1:"", 2:""} 

end = "" 

for i in range(len(abc)): 

    for letter in …: 

     abc[i] = letter 

     end = abc[0] + abc[1] + abc[2] 

     print(end) 

而不itertools和字母可变量的一种简单方法。只需添加到abc字典。

0

在Python 2下面的作品和3

# For Python 2/3 compatibility 
from __future__ import print_function 

import itertools 
import string 


def all_words(alphabet, max_length=None): 
    if max_length: 
     count_range = range(1, max_length+1) 
    else: 
     count_range = itertools.count(1) 

    for length in count_range: 
     for chars in itertools.product(alphabet, repeat=length): 
      yield ''.join(chars) 

# Demonstration 

def main(): 
    demo_alphabet = 'abc' 
    print(list(all_words(demo_alphabet, 3))) 

    print() 

    alphabet = string.ascii_letters + string.digits 
    print(list(itertools.islice(all_words(alphabet, 2), 124))) 


if __name__ == '__main__': 
    main() 

的输出是:

['a', 'b', 'c', 'aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc', 
'aaa', 'aab', 'aac', 'aba', 'abb', 'abc', 'aca', 'acb', 'acc', 'baa', 
'bab', 'bac', 'bba', 'bbb', 'bbc', 'bca', 'bcb', 'bcc', 'caa', 'cab', 
'cac', 'cba', 'cbb', 'cbc', 'cca', 'ccb', 'ccc'] 

['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', '1', '2', '3', 
'4', '5', '6', '7', '8', '9', 'aa', 'ab', 'ac', 'ad', 'ae', 'af', 'ag', 
'ah', 'ai', 'aj', 'ak', 'al', 'am', 'an', 'ao', 'ap', 'aq', 'ar', 'as', 
'at', 'au', 'av', 'aw', 'ax', 'ay', 'az', 'aA', 'aB', 'aC', 'aD', 'aE', 
'aF', 'aG', 'aH', 'aI', 'aJ', 'aK', 'aL', 'aM', 'aN', 'aO', 'aP', 'aQ', 
'aR', 'aS', 'aT', 'aU', 'aV', 'aW', 'aX', 'aY', 'aZ', 'a0', 'a1', 'a2', 
'a3', 'a4', 'a5', 'a6', 'a7', 'a8', 'a9']