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.
Related
I am trying to remove some whole words (but case insensitive) in a pyspark dataframe column.
import re
s = "I like the book. i'v seen it. Iv've" # add a new phrase
exclude_words = ["I", "I\'v", "I\'ve"]
exclude_words_re = re.compile(r"\b(" + r"|".join(exclude_words) +r")\b|\s", re.I|re.M)
exclude_words_re.sub("" , s)
I added
"Iv've"
but, got:
'like the book. seen it.'
"Iv've" should not be removed because it does not match any whole words in exclude_words.
2 changes to implement to your code:
Use proper regex flags to ignore case
Add \b to only include whole words.
import re
s = "I like the book. i'v seen it. Iv've I've"
exclude_words = ["I", "I\'v", "I\'ve"]
exclude_words_re = re.compile(r"(^|\b)((" + r"|".join(exclude_words) +r"))(\s|$)", re.I|re.M)
exclude_words_re.sub("" , s)
"like the book. seen it. Iv've "
I am trying to write some Python code that will replace some unwanted string using RegEx. The code I have written has been taken from another question on this site.
I have a text:
text_1=u'I\u2019m \u2018winning\u2019, I\u2019ve enjoyed none of it. That\u2019s why I\u2019m withdrawing from the market,\u201d wrote Arment.'
I want to remove all the \u2019m, \u2019s, \u2019ve and etc..
The code that I've written is given below:
rep={"\n":" ","\n\n":" ","\n\n\n":" ","\n\n\n\n":" ",u"\u201c":"", u"\u201d":"", u"\u2019[a-z]":"", u"\u2013":"", u"\u2018":""}
rep = dict((re.escape(k), v) for k, v in rep.iteritems())
pattern = re.compile("|".join(rep.keys()))
text = pattern.sub(lambda m: rep[re.escape(m.group(0))], text_1)
The code works perfectly for:
"u"\u201c":"", u"\u201d":"", u"\u2013":"" and u"\u2018":""
However, It doesn't work that great for:
u"\u2019[a-z] : The presence of [a-z] turns rep into \\[a\\-z\\] which doesnt match.
The output I am looking for is:
text_1=u'I winning, I enjoyed none of it. That why I withdrawing from the market,wrote Arment.'
How do I achieve this?
The information about the newlines completely changes the answer. For this, I think building the expression using a loop is actually less legible than just using better formatting in the pattern itself.
replacements = {'newlines': ' ',
'deletions': ''}
pattern = re.compile(u'(?P<newlines>\n+)|'
u'(?P<deletions>\u201c|\u201d|\u2019[a-z]?|\u2013|\u2018)')
def lookup(match):
return replacements[match.lastgroup]
text = pattern.sub(lookup, text_1)
The problem here is actually the escaping, this code does what you want more directly:
remove = (u"\u201c", u"\u201d", u"\u2019[a-z]?", u"\u2013", u"\u2018")
pattern = re.compile("|".join(remove))
text = pattern.sub("", text_1)
I've added the ? to the u2019 match, as I suppose that's what you want as well given your test string.
For completeness, I think I should also link to the Unidecode package which may actually be more closely what you're trying to achieve by removing these characters.
The simplest way is this regex:
X = re.compile(r'((\\)(.*?) ')
text = re.sub(X, ' ', text_1)
Short question:
I have a string:
title="Announcing Elasticsearch.js For Node.js And The Browser"
I want to find all pairs of words where each word is properly capitalized.
So, expected output should be:
['Announcing Elasticsearch.js', 'Elasticsearch.js For', 'For Node.js', 'Node.js And', 'And The', 'The Browser']
What I have right now is this:
'[A-Z][a-z]+[\s-][A-Z][a-z.]*'
This gives me the output:
['Announcing Elasticsearch.js', 'For Node.js', 'And The']
How can I change my regex to give desired output?
You can use this:
#!/usr/bin/python
import re
title="Announcing Elasticsearch.js For Node.js And The Browser TEst"
pattern = r'(?=((?<![A-Za-z.])[A-Z][a-z.]*[\s-][A-Z][a-z.]*))'
print re.findall(pattern, title)
A "normal" pattern can't match overlapping substrings, all characters are founded once for all. However, a lookahead (?=..) (i.e. "followed by") is only a check and match nothing. It can parse the string several times. Thus if you put a capturing group inside the lookahead, you can obtain overlapping substrings.
There's probably a more efficient way to do this, but you could use a regex like this:
(\b[A-Z][a-z.-]+\b)
Then iterate through the capture groups like so testing with this regex: (^[A-Z][a-z.-]+$) to ensure the matched group(current) matches the matched group(next).
Working example:
import re
title = "Announcing Elasticsearch.js For Node.js And The Browser"
matchlist = []
m = re.findall(r"(\b[A-Z][a-z.-]+\b)", title)
i = 1
if m:
for i in range(len(m)):
if re.match(r"(^[A-Z][a-z.-]+$)", m[i - 1]) and re.match(r"(^[A-Z][a-z.-]+$)", m[i]):
matchlist.append([m[i - 1], m[i]])
print matchlist
Output:
[
['Browser', 'Announcing'],
['Announcing', 'Elasticsearch.js'],
['Elasticsearch.js', 'For'],
['For', 'Node.js'],
['Node.js', 'And'],
['And', 'The'],
['The', 'Browser']
]
If your Python code at the moment is this
title="Announcing Elasticsearch.js For Node.js And The Browser"
results = re.findall("[A-Z][a-z]+[\s-][A-Z][a-z.]*", title)
then your program is skipping odd numbered pairs. An easy solution would be to research the pattern after skipping the first word like this:
m = re.match("[A-Z][a-z]+[\s-]", title)
title_without_first_word = title[m.end():]
results2 = re.findall("[A-Z][a-z]+[\s-][A-Z][a-z.]*", title_without_first_word)
Now just combine results and result2 together.
I have a list like this:
Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa]
I want to strip the unwanted characters using python so the list would look like:
Tomato
Populus trichocarpa
I can do the following for the first one:
name = ">Tomato4439"
name = name.strip(">1234567890")
print name
Tomato
However, I am not sure what to do with the second one. Any suggestion would be appreciated.
given:
s='Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa]'
this:
s = s.split()
[s[0].strip('0123456789,'), s[-2].replace('[',''), s[-1].replace(']','')]
will give you
['Tomato', 'Populus', 'trichocarpa']
It might be worth investigating regular expressions if you are going to do this frequently and the "rules" might not be that static as regular expressions are much more flexible dealing with the data in that case. For the sample problem you present though, this will work.
import re
a = "Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa]"
re.sub(r"^([A-Za-z]+).+\[([^]]+)\]$", r"\1 \2", a)
This gives
'Tomato Populus trichocarpa'
If the strings you're trying to parse are consistent semantically, then your best option might be classifying the different "types" of strings you have, and then creating regular expressions to parse them using python's re module.
>>> import re
>>> line = "Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa]"
>>> match = re.match("^([a-zA-Z]+).*\[([a-zA-Z ]+)\].*",line)
>>> match.groups()
('Tomato', 'Populus trichocarpa')
edited to not include the [] on the 2nd part... this should work for any thing that matches the pattern of your query (eg starts with name, ends with something in []) it would also match
"Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa apples]" for example
Previous answers were simpler than mine, but:
Here is one way to print the stuff that you don't want.
tag = "Tomato4439, >gi|224089052|ref|XP_002308615.1| predicted protein [Populus trichocarpa]"
import re, os
find = re.search('>(.+?) \[', tag).group(1)
print find
Gives you
gi|224089052|ref|XP_002308615.1| predicted protein
Then you can use the replace function to remove that from the original string. And the translate function to remove the extra unwanted characters.
tag2 = tag.replace(find, "")
tag3 = str.translate(tag2, None, ">[],")
print tag3
Gives you
Tomato4439 Populus trichocarpa
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.