I need a shorter way to do this.
input: an ordered list of string pairs
L; and a string
task: for each pair
(a,b), replace each occurrence
note 1: the replacements have to be independent. The result of one replacement can not be taken as a trigger for a subsequent replacement
Example: L=(("a","b"),("b","c")); s="ab"; desired output: "bc", and not "cc".
note 2: L is ordered; the beginning of the list has higher priority, so if
Lis (("xyz", "P"), ("xy", Q)) and
sis "wxyzabc" then the result should be "wPabc". If the order of
Lwere reversed, the result would be "wQzabc".
bonus: it's a great thing if it can also handle regex patterns as replacement pairs
- iterate over L, and replace each source to a sequence of joker characters of equal length (the joker does not appear anywhere in the string), while noting the position where we found it.
- then split the string to a list of strings of length 1, replace the first joker of each sequence with the desired target replacement, join the list back together and remove jokers.
Here is my implementation in Python:
def multireplace(text, changes): # find a possible joker character for num in range(256): joker = chr(num) if text.find(joker) != -1: continue jokerUsable = True for old,new in changes: if old.find(joker) != -1 or new.find(joker) != -1: jokerUsable = False break if jokerUsable: break # list of lists, each list is about one replacement pair, # and its elements will be the indices where they have to be done index_lists= for old, new in changes: indices =  pattern = re.escape(old) for m in re.finditer(pattern, text): indices.append(m.start()) text = text.replace(old, joker*len(old)) index_lists.append(indices) character_list = list(text) for i in range(len(index_lists)): old, new = changes[i] for place in index_lists[i]: character_list[place] = new # remove jokers i=0 while i < len(character_list): if character_list[i] == joker: del character_list[i] else: i+=1 return "".join(character_list)
This is very cumbersome, and seems to be a standard problem, so I was wondering if there is a standard way of doing this.