Question

i think it over night,but couldn’t find a fast way to do it

first came to place is replace()

for x,y in {'aa':'a','ab':'b'}.items():
    s = s.replace(x,y)

but it won’t work for this case:’aabcc’

correct is ‘abcc’
but output would be ‘bcc’

i can write some similar code like this (i have already write those code in various language like C#,php,actionscript,etc..)

   output = []
   def foo1(x):
      if(x!='a'):
          output.append(x)
      else:
          functionPointer = readafterA
   def readafterA(x):
      if(x=='a'):
          output.append('a')
      elif(x=='b'):
          output.append('b')
      else:
          output.append('a'+x)
      functionPointer = foo1
   functionPointer = normalExecute
   for singlechar in s:
      functionPointer(singlechar)
   print(''.join(output))

i don’t want write those code any more ,becuz:

1st,those code would messed up when the dictionary go big.

2nd,i believe there has to be some more elegant ways to do this job in Python

edit: i have fixed it in first paragraph,but that was not the point.

Answer 1

You can just not use a dict. Specify the tuples in the order you want to substitute them.

>>> s = 'aabcc'
>>> for x,y in (('ab','b'), ('aa','a')):
...     s = s.replace(x,y)
... 
>>> s
'abcc'

You could also use regular expressions

>>> s = 'aabcc'
>>> import re
>>> re.sub("a([ab])","\\1", s)
'abcc'