I wish to let the user ask a simple question, so I can extract a few standard elements from the string entered.
Examples of strings to be entered:
Who is the director of The Dark Knight?
What is the capital of China?
Who is the president of USA?
As you can see sometimes it is "Who", sometimes it is "What". I'm most likely looking for the "|" operator. I'll need to extract two things from these strings. The word after "the" and before "of", as well as the word after "of".
For example:
1st sentence: I wish to extract "director" and place it in a variable called Relation, and extract "The Dark Knight" and place it in a variable called Concept.
Desired output:
RelationVar = "director"
ConceptVar = "The Dark Knight"
2nd sentence: I wish to extract "capital", assign it to variable "Relation".....and extract "China" and place it in variable "Concept".
RelationVar = "capital"
ConceptVar = "China"
Any ideas on how to use the re.match function? or any other method?
You're correct that you want to use | for who/what. The rest of the regex is very simple, the group names are there for clarity but you could use r"(?:Who|What) is the (.+) of (.+)[?]" instead.
>>> r = r"(?:Who|What) is the (?P<RelationVar>.+) of (?P<ConceptVar>.+)[?]"
>>> l = ['Who is the director of The Dark Knight?', 'What is the capital of China?', 'Who is the president of USA?']
>>> [re.match(r, i).groupdict() for i in l]
[{'RelationVar': 'director', 'ConceptVar': 'The Dark Knight'}, {'RelationVar': 'capital', 'ConceptVar': 'China'}, {'RelationVar': 'president', 'ConceptVar': 'USA'}]
Change (?:Who|What) to (Who|What) if you also want to capture whether the question uses who or what.
Actually extracting the data and assigning it to variables is very simple:
>>> m = re.match(r, "What is the capital of China?")
>>> d = m.groupdict()
>>> relation_var = d["RelationVar"]
>>> concept_var = d["ConceptVar"]
>>> relation_var
'capital'
>>> concept_var
'China'
Here is the script, you can simply use | to optional match one inside the brackets.
This worked fine for me
import re
list = ['Who is the director of The Dark Knight?','What is the capital of China?','Who is the president of USA?']
for string in list:
a = re.compile(r'(What|Who) is the (.+) of (.+)')
nodes = a.findall(string);
Relation = nodes[0][0]
Concept = nodes[0][1]
print Relation
print Concept
print '----'
Best Regards:)
Related
I am writing a code using python to extract the name of a road,street, highway, for example a sentence like "There is an accident along Uhuru Highway", I want my code to be able to extract the name of the highway mentioned, I have written the code below.
sentence="there is an accident along uhuru highway"
listw=[word for word in sentence.lower().split()]
for i in range(len(listw)):
if listw[i] == "highway":
print listw[i-1] + " "+ listw[i]
I can achieve this but my code is not optimized, i am thinking of using regular expressions, any help please
'uhuru highway' can be found as follows
import re
m = re.search(r'\S+ highway', sentence) # non-white-space followed by ' highway'
print(m.group())
# 'uhuru highway'
If the location you want to extract will always have highway after it, you can use:
>>> sentence = "there is an accident along uhuru highway"
>>> a = re.search(r'.* ([\w\s\d\-\_]+) highway', sentence)
>>> print(a.group(1))
>>> uhuru
You can do the following without using regexes:
sentence.split("highway")[0].strip().split(' ')[-1]
First split according to "highway". You'll get:
['there is an accident along uhuru', '']
And now you can easily extract the last word from the first part.
I am trying to replace any i's in a string with capital I's. I have the following code:
str.replace('i ','I ')
However, it does not replace anything in the string. I am looking to include a space after the I to differentiate between any I's in words and out of words.
Thanks if you can provide help!
The exact code is:
new = old.replace('i ','I ')
new = old.replace('-i-','-I-')
new = old.replace('i ','I ')
new = old.replace('-i-','-I-')
You throw away the first new when you assign the result of the second operation over it.
Either do
new = old.replace('i ','I ')
new = new.replace('-i-','-I-')
or
new = old.replace('i ','I ').replace('-i-','-I-')
or use regex.
I think you need something like this.
>>> import re
>>> s = "i am what i am, indeed."
>>> re.sub(r'\bi\b', 'I', s)
'I am what I am, indeed.'
This only replaces bare 'i''s with I, but the 'i''s that are part of other words are left untouched.
For your example from comments, you may need something like this:
>>> s = 'i am sam\nsam I am\nThat Sam-i-am! indeed'
>>> re.sub(r'\b(-?)i(-?)\b', r'\1I\2', s)
'I am sam\nsam I am\nThat Sam-I-am! indeed'
I need help in regex or Python to extract a substring from a set of string. The string consists of alphanumeric. I just want the substring that starts after the first space and ends before the last space like the example given below.
Example 1:
A:01 What is the date of the election ?
BK:02 How long is the river Nile ?
Results:
What is the date of the election
How long is the river Nile
While I am at it, is there an easy way to extract strings before or after a certain character? For example, I want to extract the date or day like from a string like the ones given in Example 2.
Example 2:
Date:30/4/2013
Day:Tuesday
Results:
30/4/2013
Tuesday
I have actually read about regex but it's very alien to me. Thanks.
I recommend using split
>>> s="A:01 What is the date of the election ?"
>>> " ".join(s.split()[1:-1])
'What is the date of the election'
>>> s="BK:02 How long is the river Nile ?"
>>> " ".join(s.split()[1:-1])
'How long is the river Nile'
>>> s="Date:30/4/2013"
>>> s.split(":")[1:][0]
'30/4/2013'
>>> s="Day:Tuesday"
>>> s.split(":")[1:][0]
'Tuesday'
>>> s="A:01 What is the date of the election ?"
>>> s.split(" ", 1)[1].rsplit(" ", 1)[0]
'What is the date of the election'
>>>
There's no need to dig into regex if this is all you need; you can use str.partition
s = "A:01 What is the date of the election ?"
before,sep,after = s.partition(' ') # could be, eg, a ':' instead
If all you want is the last part, you can use _ as a placeholder for 'don't care':
_,_,theReallyAwesomeDay = s.partition(':')
I have a list of phrases (input by user) I'd like to locate them in a text file, for examples:
titles = ['Blue Team', 'Final Match', 'Best Player',]
text = 'In today Final match, The Best player is Joe from the Blue Team and the second best player is Jack from the Red team.'
1./ I can find all the occurrences of these phrases like so
titre = re.compile(r'(?P<title>%s)' % '|'.join(titles), re.M)
list = [ t for t in titre.split(text) if titre.search(t) ]
(For simplicity, I am assuming a perfect spacing.)
2./ I can also find variants of these phrases e.g. 'Blue team', final Match', 'best player' ... using re.I, if they ever appear in the text.
But I want to restrict to finding only variants of the input phrases with their first letter upper-cased e.g. 'Blue team' in the text, regardless how they were entered as input, e.g. 'bluE tEAm'.
Is it possible to write something to "block" the re.I flag for a portion of a phrase? In pseudo code I imagine generate something like '[B]lue Team|[F]inal Match'.
Note: My primary goal is not, for example, calculating frequency of the input phrases in the text but extracting and analyzing the text fragments between or around them.
I would use re.I and modify the list-comp to:
l = [ t for t in titre.split(text) if titre.search(t) and t[0].isupper() ]
I think regular expressions won't let you specify just a region where the ignore case flag is applicable. However, you can generate a new version of the text in which all the characters have been lower cased, but the first one for every word:
new_text = ' '.join([word[0] + word[1:].lower() for word in text.split()])
This way, a regular expression without the ignore flag will match taking into account the casing only for the first character of each word.
How about modifying the input so that it is in the correct case before you use it in the regular expression?
I'm trying to parse the title tag in an RSS 2.0 feed into three different variables for each entry in that feed. Using ElementTree I've already parsed the RSS so that I can print each title [minus the trailing )] with the code below:
feed = getfeed("http://www.tourfilter.com/dallas/rss/by_concert_date")
for item in feed:
print repr(item.title[0:-1])
I include that because, as you can see, the item.title is a repr() data type, which I don't know much about.
A particular repr(item.title[0:-1]) printed in the interactive window looks like this:
'randy travis (Billy Bobs 3/21'
'Michael Schenker Group (House of Blues Dallas 3/26'
The user selects a band and I hope to, after parsing each item.title into 3 variables (one each for band, venue, and date... or possibly an array or I don't know...) select only those related to the band selected. Then they are sent to Google for geocoding, but that's another story.
I've seen some examples of regex and I'm reading about them, but it seems very complicated. Is it? I thought maybe someone here would have some insight as to exactly how to do this in an intelligent way. Should I use the re module? Does it matter that the output is currently is repr()s? Is there a better way? I was thinking I'd use a loop like (and this is my pseudoPython, just kind of notes I'm writing):
list = bandRaw,venue,date,latLong
for item in feed:
parse item.title for bandRaw, venue, date
if bandRaw == str(band)
send venue name + ", Dallas, TX" to google for geocoding
return lat,long
list = list + return character + bandRaw + "," + venue + "," + date + "," + lat + "," + long
else
In the end, I need to have the chosen entries in a .csv (comma-delimited) file looking like this:
band,venue,date,lat,long
randy travis,Billy Bobs,3/21,1234.5678,1234.5678
Michael Schenker Group,House of Blues Dallas,3/26,4321.8765,4321.8765
I hope this isn't too much to ask. I'll be looking into it on my own, just thought I should post here to make sure it got answered.
So, the question is, how do I best parse each repr(item.title[0:-1]) in the feed into the 3 separate values that I can then concatenate into a .csv file?
Don't let regex scare you off... it's well worth learning.
Given the examples above, you might try putting the trailing parenthesis back in, and then using this pattern:
import re
pat = re.compile('([\w\s]+)\(([\w\s]+)(\d+/\d+)\)')
info = pat.match(s)
print info.groups()
('Michael Schenker Group ', 'House of Blues Dallas ', '3/26')
To get at each group individual, just call them on the info object:
print info.group(1) # or info.groups()[0]
print '"%s","%s","%s"' % (info.group(1), info.group(2), info.group(3))
"Michael Schenker Group","House of Blues Dallas","3/26"
The hard thing about regex in this case is making sure you know all the known possible characters in the title. If there are non-alpha chars in the 'Michael Schenker Group' part, you'll have to adjust the regex for that part to allow them.
The pattern above breaks down as follows, which is parsed left to right:
([\w\s]+) : Match any word or space characters (the plus symbol indicates that there should be one or more such characters). The parentheses mean that the match will be captured as a group. This is the "Michael Schenker Group " part. If there can be numbers and dashes here, you'll want to modify the pieces between the square brackets, which are the possible characters for the set.
\( : A literal parenthesis. The backslash escapes the parenthesis, since otherwise it counts as a regex command. This is the "(" part of the string.
([\w\s]+) : Same as the one above, but this time matches the "House of Blues Dallas " part. In parentheses so they will be captured as the second group.
(\d+/\d+) : Matches the digits 3 and 26 with a slash in the middle. In parentheses so they will be captured as the third group.
\) : Closing parenthesis for the above.
The python intro to regex is quite good, and you might want to spend an evening going over it http://docs.python.org/library/re.html#module-re. Also, check Dive Into Python, which has a friendly introduction: http://diveintopython3.ep.io/regular-expressions.html.
EDIT: See zacherates below, who has some nice edits. Two heads are better than one!
Regular expressions are a great solution to this problem:
>>> import re
>>> s = 'Michael Schenker Group (House of Blues Dallas 3/26'
>>> re.match(r'(.*) \((.*) (\d+/\d+)', s).groups()
('Michael Schenker Group', 'House of Blues Dallas', '3/26')
As a side note, you might want to look at the Universal Feed Parser for handling the RSS parsing as feeds have a bad habit of being malformed.
Edit
In regards to your comment... The strings occasionally being wrapped in "s rather than 's has to do with the fact that you're using repr. The repr of a string is usually delimited with 's, unless that string contains one or more 's, where instead it uses "s so that the 's don't have to be escaped:
>>> "Hello there"
'Hello there'
>>> "it's not its"
"it's not its"
Notice the different quote styles.
Regarding the repr(item.title[0:-1]) part, not sure where you got that from but I'm pretty sure you can simply use item.title. All you're doing is removing the last char from the string and then calling repr() on it, which does nothing.
Your code should look something like this:
import geocoders # from GeoPy
us = geocoders.GeocoderDotUS()
import feedparser # from www.feedparser.org
feedurl = "http://www.tourfilter.com/dallas/rss/by_concert_date"
feed = feedparser.parse(feedurl)
lines = []
for entry in feed.entries:
m = re.search(r'(.*) \((.*) (\d+/\d+)\)', entry.title)
if m:
bandRaw, venue, date = m.groups()
if band == bandRaw:
place, (lat, lng) = us.geocode(venue + ", Dallas, TX")
lines.append(",".join([band, venue, date, lat, lng]))
result = "\n".join(lines)
EDIT: replaced list with lines as the var name. list is a builtin and should not be used as a variable name. Sorry.