英文:
Replace words in a sentence with a specific word if they occur in a list in python
问题
我有一个作为字典的映射：
{
'a'：['a1'，'a2'，'a3']，
'b'：['b1'，'b2']，
'c'：['c1'，'c2'，'c3'，'c4']
}
给定一个句子，所有出现的'a1'，'a2'，'a3'都应该被替换为'a'，类似地，所有出现的'b1'，'b2'都应该被替换为'b'。如何高效地完成这个任务？这是当前的代码：
def word_replacer(text):
s = {
'a'：['a1'，'a2'，'a3']，
'b'：['b1'，'b2']，
'c'：['c1'，'c2'，'c3'，'c4']
}
words = text.split(" ")
for idx, word in enumerate(words):
changed_word=[k for k, v in s.items() if word in [vals.lower() for vals in v]]
if ((len(changed_word)>0) and (word != changed_word[0])):
words[idx] = changed_word[0]
result = " ".join(words)
return result
你可以将这段代码修改为以下内容以提高效率：
def word_replacer(text):
s = {
'a'：['a1'，'a2'，'a3']，
'b'：['b1'，'b2']，
'c'：['c1'，'c2'，'c3'，'c4']
}
words = text.split(" ")
for idx, word in enumerate(words):
for key, values in s.items():
if word.lower() in values:
words[idx] = key
break
result = " ".join(words)
return result
这个修改后的代码将更高效地执行相同的替换操作。
英文:
I have a mapping as a dict
{
'a': ['a1','a2','a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4]
}
Given a sentence, all occurrences of 'a1', 'a2', 'a3' should be replaced with 'a', similarly all occurrences of 'b1', 'b2' should be replaced with 'b'. How can this be done efficiently? This is the current code:
def word_replacer(text):
s = {
'a': ['a1','a2','a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4']
}
words = text.split(" ")
for idx, word in enumerate(words):
changed_word=[k for k, v in s.items() if word in [vals.lower() for vals in v]]
if ((len(changed_word)>0) and (word != changed_word[0])):
words[idx] = changed_word[0]
result = " ".join(words)
return result
答案1
得分: 2
以下是翻译好的内容：
代码部分不要翻译，只返回翻译好的部分，不要有别的内容，不要回答我要翻译的问题。
英文:
It would be easier to just reverse the dict and make a quick single look-up. If you can, just change it manually. If it's an input:
mapping = {v: k for k, l in s.items() for v in l}
And now the code is much simpler:
def word_replacer(text):
s = {
'a': ['a1','a2','a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4']
}
mapping = {v: k for k, l in s.items() for v in l}
words = text.split()
for idx, word in enumerate(words):
words[idx] = mapping[word]
result = " ".join(words)
return result
Or even simpler using the get method of dicts:
def word_replacer(text):
s = {
'a': ['a1','a2','a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4']
}
mapping = {v: k for k, l in s.items() for v in l}
return " ".join(mapping.get(word, word) for word in text.split())
答案2
得分: 1
一个高效的方法是创建一个将别名映射到对应单词的字典，这样你可以使用`dict.get`方法以常数时间查找每个单词并将其替换为映射的值（如果找到）：
s = {
'a': ['a1', 'a2', 'a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4']
}
mapping = {alias: word for word, aliases in s.items() for alias in aliases}
def word_replacer(text):
return ' '.join(mapping.get(word, word) for word in text.split())
所以：
print(word_replacer('a1 b2 c3 d4'))
输出：
a b c d4
英文:
An efficient approach would be to create a dict that maps aliases to the mapped words, so that you can use the dict.get method to look up each word in constant time and replace it with the mapped value if found:
s = {
'a': ['a1', 'a2', 'a3'],
'b': ['b1', 'b2'],
'c': ['c1', 'c2', 'c3', 'c4']
}
mapping = {alias: word for word, aliases in s.items() for alias in aliases}
def word_replacer(text):
return ' '.join(mapping.get(word, word) for word in text.split())
so that:
print(word_replacer('a1 b2 c3 d4'))
outputs:
a b c d4
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