生成帶面具

Question

所有可能的數字比方說，我有這樣的字符串：

a = "123**7*9" 

，我需要生成的所有可能的組合：

12300709...12399799 

如何做到這一點與Python？

Answer 1

您可以使用itertools.product和字符串格式化：

>>> from itertools import product 
>>> strs = "123**7*9" 
>>> c = strs.count("*")    #count the number of "*"'s 
>>> strs = strs.replace("*","{}") #replace '*'s with '{}' for formatting 
>>> for x in product("",repeat=c): 
...  print strs.format(*x)    #use `int()` to get an integer 

12300709 
12300719 
12300729 
12300739 
12300749 
12300759 
12300769 
12300779 
12300789 
12300799 
.... 

Answer 2

你也可以像這樣做只使用標準庫：

a = "123**7*9" 
a = a.replace("*", "%d") 
for x in range(10): 
    for y in range(10): 
     for z in range(10): 
      print a % (x,y,z) 

編輯，BOOM：

a = "123**7*9" 
c = a.count("*") 
a = a.replace("*", "%s") 
for x in range(10**c): 
    print a % tuple(list("%03d" % x)) 

Answer 3

遞歸變體：

def combinate(pattern, order=0): 
    if pattern: 
     for val in combinate(pattern[:-1], order+1): 
      last_value = pattern[-1] 
      if last_value == '*': 
       for gen in xrange(10): 
        value = gen * (10**order) + val 
        yield value 
      else: 
       value = int(last_value)*(10**order)+val 
       yield value 
    else: 
     yield 0 


for i in combinate('1*1**2'): 
    print i