I need a shorter way to do this.
input: an ordered list of string pairs
L
; and a strings
task: for each pair
(a,b)
, replace each occurrencea
ins
byb
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
L
is (("xyz", "P"), ("xy", Q)) ands
is "wxyzabc" then the result should be "wPabc". If the order ofL
were reversed, the result would be "wQzabc".bonus: it's a great thing if it can also handle regex patterns as replacement pairs
My idea:
- 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.