Можно ли передать словарь при замене строк в Python?
В дополнение к тому, что было сказано ранее, здесь есть функциональный способ с Рамда :
import { assocPath } from 'ramda';
const o1 = { a: { b: { c: 1 }, bb: { cc: 22 } } };
const o2 = assocPath(['a', 'b', 'c'])(42)(o1);
console.log(o1 !== o2, o1.a !== o2.a); // new copies of "changed" objects
console.log(o1.a.bb === o2.a.bb); // deep unchanged properties are copied by reference
2 ответа
closest is probably:
somere.sub(lambda m: replacements[m.group()], text)
for example:
>>> za = re.compile('z\w')
>>> za.sub(lambda m: dict(za='BLU', zo='BLA')[m.group()], 'fa za zo bu')
'fa BLU BLA bu'
with a .get instead of []-indexing if you want to supply a default for matches that are missing in replacements.
Edit: what rick really wants is to have a dict with keys to be taken as regular expression patterns, such as '\d+S', and (hopefully) constant string values (hopefully w/o backreferences). The cookbook recipe can be adapted for this purpose:
def dict_sub(d, text):
""" Replace in 'text' non-overlapping occurences of REs whose patterns are keys
in dictionary 'd' by corresponding values (which must be constant strings: may
have named backreferences but not numeric ones). The keys must not contain
anonymous matching-groups.
Returns the new string."""
# Create a regular expression from the dictionary keys
regex = re.compile("|".join("(%s)" % k for k in d))
# Facilitate lookup from group number to value
lookup = dict((i+1, v) for i, v in enumerate(d.itervalues()))
# For each match, find which group matched and expand its value
return regex.sub(lambda mo: mo.expand(lookup[mo.lastindex]), text)
Example use:
d={'\d+S': 'wot', '\d+T': 'zap'}
t='And 23S, and 45T, and 66T but always 029S!'
print dict_sub(d, t)
emits:
And wot, and zap, and zap but always wot!
You could avoid building lookup and just use mo.expand(d.values()[mo.lastindex-1]), but that might be a tad slow if d is very large and there are many matches (sorry, haven't precisely measured/benchmarked both approaches, so this is just a guess;-).
Сделать это достаточно просто:
replacements = dict(hello='goodbye', good='bad')
s = "hello, good morning";
for old, new in replacements.items():
s = s.replace(old, new)
Вы найдете много мест, где функции PHP принимают массив значений, и нет прямого эквивалента Python, но с массивами гораздо проще работать (списки) в Python, поэтому это не проблема.