I write a code that has str data
def characters(self, content):
self.contentText = content.split()
# self.contentText is List here
I am sending self.contentText
list to another module as:
self.contentText = Formatter.formatter(self.contentText)
In this method, I am writing below code:
remArticles = remArticles = {' a ':'', ' the ':'', ' and ':'', ' an ':'', '& nbsp;':''}
contentText = [i for i in contentText if i not in remArticles.keys()]
But it is not replacing. Is it that remArticles should be list and not dict
But I tried replacing it with list too. It wouldn't simply replace.
ofcourse with list, below will be the code:
contentText = [i for i in contentText if i not in remArticles]
This is continuation from Accessing Python List Type
Initially I was trying:
for i in remArticles:
print type(contentText)
print "1"
contentText = contentText.replace(i, remArticles[i])
print type(contentText)
But that threw errors:
contentText = contentText.replace(i, remArticles[i])
AttributeError: 'list' object has no attribute 'replace'
Your question is not clear but if your goal is to convert a string to a list, remove unwanted words, and then turn the list back into a string, then you can do it like this:
def clean_string(s):
words_to_remove = ['a', 'the', 'and', 'an', ' ']
list_of_words = s.split()
cleaned_list = [word for word in list_of_words if word not in words_to_remove]
new_string = ' '.join(cleaned_list)
return new_string
This is how you could do the same without converting to a list:
def clean_string(s):
words_to_remove = ['a', 'the', 'and', 'an', ' ']
for word in words_to_remove:
s = s.replace(word, '')
return s
And if you wanted more flexibility in removing some words but replacing others, you could do the following with a dictionary:
def clean_string(s):
words_to_replace = {'a': '', 'the': '', 'and': '&', 'an': '', ' ': ' '}
for old, new in words_to_replace.items():
s = s.replace(old, new)
return s
Your problem is that your map contains spaces within the keys. Following code solves your problem:
[i for i in contentText if i not in map(lambda x: x.strip(), remArticles.keys())]
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.