Reputation: 2158
I have gotten a very strange data. I have dictionary with keys and values where I want to use this dictionary to search if these keywords are ONLY starting and/or end of the text not middle of the sentence. I tried to create simple data frame below to show the problem case and python codes that I have tried so far. How do I get it go search for only starting or ending of the sentence? This one searches whole text sub-strings.
Code:
d = {'apple corp':'Company','app':'Application'} #dictionary
l1 = [1, 2, 3,4]
l2 = [
"The word Apple is commonly confused with Apple Corp which is a business",
"Apple Corp is a business they make computers",
"Apple Corp also writes App",
"The Apple Corp also writes App"
]
df = pd.DataFrame({'id':l1,'text':l2})
df['text'] = df['text'].str.lower()
df
Original Dataframe:
id text
1 The word Apple is commonly confused with Apple Corp which is a business
2 Apple Corp is a business they make computers
3 Apple Corp also writes App
4 The Apple Corp also writes App
Code Tried out:
def matcher(k):
x = (i for i in d if i in k)
# i.startswith(k) getting error
return ';'.join(map(d.get, x))
df['text_value'] = df['text'].map(matcher)
df
Error:
TypeError: 'in <string>' requires string as left operand, not bool
when I use this x = (i for i in d if i.startswith(k) in k)
Empty values if i tried this x = (i for i in d if i.startswith(k) == True in k)
TypeError: sequence item 0: expected str instance, NoneType found
when i use this x = (i.startswith(k) for i in d if i in k)
Results from Code above ... Create new field 'text_value':
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business Company;Application
2 Apple Corp is a business they make computers Company;Application
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Company;Application
Trying to get an FINAL output like this:
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business NaN
2 Apple Corp is a business they make computers Company
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Application
Upvotes: 0
Views: 193
Reputation: 79288
joined = "|".join(d.keys())
pat = '(?i)^(?:the\\s*)?(' + joined + ')\\b.*?|.*\\b(' + joined + ')$'+'|.*'
get = lambda x: d.get(x.group(1),"") + (';' +d.get(x.group(2),"") if x.group(2) else '')
df.text.str.replace(pat,get)
0
1 Company
2 Company;Application
3 Company;Application
Name: text, dtype: object
Upvotes: 0
Reputation: 8631
You need a matcher
function which can accept flag
and then call that twice to get the results for startswith
and endswith
.
def matcher(s, flag="start"):
if flag=="start":
for i in d:
if s.startswith(i):
return d[i]
else:
for i in d:
if s.endswith(i):
return d[i]
return None
df['st'] = df['text'].apply(matcher)
df['ed'] = df['text'].apply(matcher, flag="end")
df['text_value'] = df[['st', 'ed']].apply(lambda x: ';'.join(x.dropna()),1)
df = df[['id','text', 'text_value']]
The text_value
column looks like:
0
1 Company
2 Company;Application
3 Application
Name: text_value, dtype: object
Upvotes: 1