python - 基于替换创建字符串组合

Question

给定一个单词和一个替换字符字典，我需要根据替换形成一个字符组合

输入

word = 'accompanying'  
substitutions={'c':['$'], 'a': ['4'], 'g': ['9']} 

输出

{'a$$ompanyin9', 'ac$ompanyin9','a$companyin9','4ccomp4nying', '4$$omp4nying', 
'4$comp4nying','4c$omp4nying', '4ccomp4nyin9', 'a$$ompanying', 'a$companying', 'ac$ompanying', 
'accompanyin9', 'accompanying', '4$$omp4nyin9', '4$comp4nyin9', '4c$omp4nyin9','etc.,'}

我写了一个代码，但它并没有提供我期望的所有组合

示例代码

from itertools import product
substitutions={'c':['$'], 'a': ['4'], 'g': ['9']}
for key in substitutions.keys():
  if key not in substitutions[key]:
    substitutions[key].append(key)
wordPossibilities = []
word = 'accompanying'
for substitute in [zip(substitutions.keys(),ch) for ch in product(*substitutions.values())]:
  temp=word
  for replacement in substitute:
    temp=temp.replace(*replacement)
  wordPossibilities.append(temp)
print(set(wordPossibilities))

我的输出

{'4$$omp4nyin9', 'a$$ompanyin9', 'a$$ompanying', 'accompanyin9',
'accompanying', '4ccomp4nyin9', '4$$omp4nying', '4ccomp4nying'}

如果找到替换，我的代码将替换提供的字符串中的所有字符。如何根据索引进行替换以找到所有可能的组合？

标签： pythonpython-3.xcombinationsitertools

Answer 1

使用带有递归的生成器是干净和直接的：

word = 'accompanying'  
subs={'c':['$'], 'a': ['4'], 'g': ['9']} 
def get_subs(d, c = []):
  if not d:
     yield ''.join(c)
  else:
     for i in [d[0], *subs.get(d[0], [])]:
        yield from get_subs(d[1:], c+[i])

print(list(get_subs(word)))

输出：

['accompanying', 'accompanyin9', 'accomp4nying', 'accomp4nyin9', 'ac$ompanying', 'ac$ompanyin9', 'ac$omp4nying', 'ac$omp4nyin9', 'a$companying', 'a$companyin9', 'a$comp4nying', 'a$comp4nyin9', 'a$$ompanying', 'a$$ompanyin9', 'a$$omp4nying', 'a$$omp4nyin9', '4ccompanying', '4ccompanyin9', '4ccomp4nying', '4ccomp4nyin9', '4c$ompanying', '4c$ompanyin9', '4c$omp4nying', '4c$omp4nyin9', '4$companying', '4$companyin9', '4$comp4nying', '4$comp4nyin9', '4$$ompanying', '4$$ompanyin9', '4$$omp4nying', '4$$omp4nyin9']

但是，itertools.product可用于更短的解决方案：

from itertools import product as prod
s = ''.join('{}' if i in subs else i for i in word)
result = [s.format(*i) for i in prod(*[[i, *subs[i]] for i in word if i in subs])]

输出：

['accompanying', 'accompanyin9', 'accomp4nying', 'accomp4nyin9', 'ac$ompanying', 'ac$ompanyin9', 'ac$omp4nying', 'ac$omp4nyin9', 'a$companying', 'a$companyin9', 'a$comp4nying', 'a$comp4nyin9', 'a$$ompanying', 'a$$ompanyin9', 'a$$omp4nying', 'a$$omp4nyin9', '4ccompanying', '4ccompanyin9', '4ccomp4nying', '4ccomp4nyin9', '4c$ompanying', '4c$ompanyin9', '4c$omp4nying', '4c$omp4nyin9', '4$companying', '4$companyin9', '4$comp4nying', '4$comp4nyin9', '4$$ompanying', '4$$ompanyin9', '4$$omp4nying', '4$$omp4nyin9']