I am a beginner, been learning python for a few months as my very first programming language. I am looking to find a pattern from a text file. My first attempt has been using regex, which does work but has a limitation:
import re
noun_list = ['bacon', 'cheese', 'eggs', 'milk', 'list', 'dog']
CC_list = ['and', 'or']
noun_list_pattern1 = r'\b\w+\b,\s\b\w+\b,\sand\s\b\w+\b|\b\w+\b,\s\b\w+\b,\sor\s\b\w+\b|\b\w+\b,\s\b\w+\b\sand\s\b\w+\b|\b\w+\b,\s\b\w+\b,\saor\s\b\w+\b'
with open('test_sentence.txt', 'r') as input_f:
read_input = input_f.read()
word = re.findall(noun_list_pattern1, read_input)
for w in word:
print w
else:
pass
So at this point you may be asking why are the lists in this code since they are not being used. Well, I have been racking my brains out, trying all sort of for loops and if statements in functions to try and find a why to replicate the regex pattern, but using the lists.
The limitation with regex is that the \b\w+\w\ code which is found a number of times in `noun_list_pattern' actually only finds words - any words - but not specific nouns. This could raise false positives. I want to narrow things down more by using the elements in the list above instead of the regex.
Since I actually have 4 different regex in the regex pattern (it contains 4 |), I will just go with 1 of them here. So I would need to find a pattern such as:
'noun in noun_list' + ', ' + 'noun in noun_list' + ', ' + 'C in CC_list' + ' ' + 'noun in noun_list
Obviously, the above code quoted line is not real python code, but is an experession of my thoughts about the match needed. Where I say noun in noun_list I mean an iteration through the noun_list; C in CC_list is an iteration through the CC_list; , is a literal string match for a comma and whitespace.
Hopefully I have made myself clear!
Here is the content of the test_sentence.txt file that I am using:
I need to buy are bacon, cheese and eggs.
I also need to buy milk, cheese, and bacon.
What's your favorite: milk, cheese or eggs.
What's my favorite: milk, bacon, or eggs.
Break your problem down a little. First, you need a pattern that will match the words from your list, but no other. You can accomplish that with the alternation operator | and the literal words. red|green|blue, for example, will match "red", "green", or "blue", but not "purple". Join the noun list with that character, and add the word boundary metacharacters along with parentheses to group the alternations:
noun_patt = r'\b(' + '|'.join(nouns) + r')\b'
Do the same for your list of conjunctions:
conj_patt = r'\b(' + '|'.join(conjunctions) + r')\b'
The overall match you want to make is "one or more noun_patt match, each optionally followed by a comma, followed by a match for the conj_patt and then one more noun_patt match". Easy enough for a regex:
patt = r'({0},? )+{1} {0}'.format(noun_patt, conj_patt)
You don't really want to use re.findall(), but re.search(), since you're only expecting one match per line:
for line in lines:
... print re.search(patt, line).group(0)
...
bacon, cheese and eggs
milk, cheese, and bacon
milk, cheese or eggs
milk, bacon, or eggs
As a note, you're close to, if not rubbing up against, the limits of regular expressions as far as parsing English. Any more complex than this, and you will want to look into actual parsing, perhaps with NLTK.
In actuality, you don't necessarily need regular expressions, as there are a number of ways to do this using just your original lists.
noun_list = ['bacon', 'cheese', 'eggs', 'milk', 'list', 'dog']
conjunctions = ['and', 'or']
#This assumes that file has been read into a list of newline delimited lines called `rawlines`
for line in rawlines:
matches = [noun for noun in noun_list if noun in line] + [conj for conj in conjunctions if conj in line]
if len(matches) == 4:
for match in matches:
print match
The reason the match number is 4, is that 4 is the correct number of matches. (Note, that this could also be the case for repeated nouns or conjunctions).
EDIT:
This version prints the lines that are matched and the words matched. Also fixed the possible multiple word match problem:
words_matched = []
matching_lines = []
for l in lst:
matches = [noun for noun in noun_list if noun in l] + [conj for conj in conjunctions if conj in l]
invalid = True
valid_count = 0
for match in matches:
if matches.count(match) == 1:
valid_count += 1
if valid_count == len(matches):
invalid = False
if not invalid:
words_matched.append(matches)
matching_lines.append(l)
for line, matches in zip(matching_lines, words_matched):
print line, matches
However, if this doesn't suit you, you can always build the regex as follows (using the itertools module):
#The number of permutations choices is 3 (as revealed from your examples)
for nouns, conj in itertools.product(itertools.permutations(noun_list, 3), conjunctions):
matches = [noun for noun in nouns]
matches.append(conj)
#matches[:2] is the sublist containing the first 2 items, -1 is the last element, and matches[2:-1] is the element before the last element (if the number of nouns were more than 3, this would be the elements between the 2nd and last).
regex_string = '\s,\s'.join(matches[:2]) + '\s' + matches[-1] + '\s' + '\s,\s'.join(matches[2:-1])
print regex_string
#... do regex related matching here
The caveat of this method is that it is pure brute-force as it generates all the possible combinations (read permutations) of both lists which can then be tested to see if each line matches. Hence, it is horrendously slow, but in this example that matches the ones given (the non-comma before the conjunction), this will generate exact matches perfectly.
Adapt as required.
Related
I have the following string:
"The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"
List of words to be extracted:
["town","teddy","chicken","boy went"]
NB: town and teddy are wrongly spelt in the given sentence.
I have tried the following but I get other words that are not part of the answer:
import difflib
sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"
list1 = ["town","teddy","chicken","boy went"]
[difflib.get_close_matches(x.lower().strip(), sent.split()) for x in list1 ]
I am getting the following result:
[['twn', 'to'], ['tddy'], ['chicken.', 'picked'], ['went']]
instead of:
'twn', 'tddy', 'chicken','boy went'
Notice in the documentation for difflib.get_closest_matches():
difflib.get_close_matches(word, possibilities, n=3, cutoff=0.6)
Return a list of the best "good enough" matches. word is a sequence for which close matches are desired (typically a string), and
possibilities is a list of sequences against which to match word
(typically a list of strings).
Optional argument n (default 3) is the maximum number of close matches to return; n must be greater than 0.
Optional argument cutoff (default 0.6) is a float in the range [0, 1]. Possibilities that don’t score at least that similar to word are
ignored.
At the moment, you are using the default n and cutoff arguments.
You can specify either (or both), to narrow down the returned matches.
For example, you could use a cutoff score of 0.75:
result = [difflib.get_close_matches(x.lower().strip(), sent.split(), cutoff=0.75) for x in list1]
Or, you could specify that only at most 1 match should be returned:
result = [difflib.get_close_matches(x.lower().strip(), sent.split(), n=1) for x in list1]
In either case, you could use a list comprehension to flatten the lists of lists (since difflib.get_close_matches() always returns a list):
matches = [r[0] for r in result]
Since you also want to check for close matches of bigrams, you can do so by extracting pairings of adjacent "words", and pass them to difflib.get_close_matches() as part of the possibilities argument.
Here is a full working example of this in action:
import difflib
import re
sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"
list1 = ["town", "teddy", "chicken", "boy went"]
# this extracts overlapping pairings of "words"
# i.e. ['The boy', 'boy went', 'went to', 'to twn', ...
pairs = re.findall(r'(?=(\b[^ ]+ [^ ]+\b))', sent)
# we pass the sent.split() list as before
# and concatenate the new pairs list to the end of it also
result = [difflib.get_close_matches(x.lower().strip(), sent.split() + pairs, n=1) for x in list1]
matches = [r[0] for r in result]
print(matches)
# ['twn', 'tddy', 'chicken.', 'boy went']
If you read Python documentation fordifflib.get_close_matches()
https://docs.python.org/3/library/difflib.html
It returns all possible best matches.
Method signature:
difflib.get_close_matches(word, possibilities, n=3, cutoff=0.6)
Here n is the maximum number of close matches to return. So I think you can pass this as 1.
>>> [difflib.get_close_matches(x.lower().strip(), sent.split(),1)[0] for x in list1]
['twn', 'tddy', 'chicken.', 'went']
I'm attempting to capitalize all words in a section of text that only appear once. I have the bit that finds which words only appear once down, but when I go to replace the original word with the .upper version, a bunch of other stuff gets capitalized too. It's a small program, so here's the code.
from collections import Counter
from string import punctuation
path = input("Path to file: ")
with open(path) as f:
word_counts = Counter(word.strip(punctuation) for line in f for word in line.replace(")", " ").replace("(", " ")
.replace(":", " ").replace("", " ").split())
wordlist = open(path).read().replace("\n", " ").replace(")", " ").replace("(", " ").replace("", " ")
unique = [word for word, count in word_counts.items() if count == 1]
for word in unique:
print(word)
wordlist = wordlist.replace(word, str(word.upper()))
print(wordlist)
The output should be 'Genesis 37:1 Jacob lived in the land of his father's SOJOURNINGS, in the land of Canaan., as sojournings is the first word that only appears once. Instead, it outputs GenesIs 37:1 Jacob lIved In the land of hIs FATher's SOJOURNINGS, In the land of Canaan. Because some of the other letters appear in keywords, it tries to capitalize them as well.
Any ideas?
I rewrote the code pretty significantly since some of the chained replace calls might prove to be unreliable.
import string
# The sentence.
sentence = "Genesis 37:1 Jacob lived in the land of his father's SOJOURNINGS, in the land of Canaan."
rm_punc = sentence.translate(None, string.punctuation) # remove punctuation
words = rm_punc.split(' ') # split spaces to get a list of words
# Find all unique word occurrences.
single_occurrences = []
for word in words:
# if word only occurs 1 time, append it to the list
if words.count(word) == 1:
single_occurrences.append(word)
# For each unique word, find it's index and capitalize the letter at that index
# in the initial string (the letter at that index is also the first letter of
# the word). Note that strings are immutable, so we are actually creating a new
# string on each iteration. Also, sometimes small words occur inside of other
# words, e.g. 'an' inside of 'land'. In order to make sure that our call to
# `index()` doesn't find these small words, we keep track of `start` which
# makes sure we only ever search from the end of the previously found word.
start = 0
for word in single_occurrences:
try:
word_idx = start + sentence[start:].index(word)
except ValueError:
# Could not find word in sentence. Skip it.
pass
else:
# Update counter.
start = word_idx + len(word)
# Rebuild sentence with capitalization.
first_letter = sentence[word_idx].upper()
sentence = sentence[:word_idx] + first_letter + sentence[word_idx+1:]
print(sentence)
Text replacement by patters calls for regex.
Your text is a bit tricky, you have to
remove digits
remove punktuations
split into words
care about capitalisation: 'It's' vs 'it's'
only replace full matches 'remote' vs 'mote' when replacing mote
etc.
This should do this - see comments inside for explanations:
bible.txt is from your link
from collections import Counter
from string import punctuation , digits
import re
from collections import defaultdict
with open(r"SO\AllThingsPython\P4\bible.txt") as f:
s = f.read()
# get a set of unwanted characters and clean the text
ps = set(punctuation + digits)
s2 = ''.join( c for c in s if c not in ps)
# split into words
s3 = s2.split()
# create a set of all capitalizations of each word
repl = defaultdict(set)
for word in s3:
repl[word.upper()].add(word) # f.e. {..., 'IN': {'In', 'in'}, 'THE': {'The', 'the'}, ...}
# count all words _upper case_ and use those that only occure once
single_occurence_upper_words = [w for w,n in Counter( (w.upper() for w in s3) ).most_common() if n == 1]
text = s
# now the replace part - for all upper single words
for upp in single_occurence_upper_words:
# for all occuring capitalizations in the text
for orig in repl[upp]:
# use regex replace to find the original word from our repl dict with
# space/punktuation before/after it and replace it with the uppercase word
text = re.sub(f"(?<=[{punctuation} ])({orig})(?=[{punctuation} ])",upp, text)
print(text)
Output (shortened):
Genesis 37:1 Jacob lived in the land of his father's SOJOURNINGS, in the land of Canaan.
2 These are the GENERATIONS of Jacob.
Joseph, being seventeen years old, was pasturing the flock with his brothers. He was a boy with the sons of Bilhah and Zilpah, his father's wives. And Joseph brought a BAD report of them to their father. 3 Now Israel loved Joseph more than any other of his sons, because he was the son of his old age. And he made him a robe of many colors. [a] 4 But when his brothers saw that their father loved him more than all his brothers, they hated him
and could not speak PEACEFULLY to him.
<snipp>
The regex uses lookahead '(?=...)' and lookbehind '(?<=...)'syntax to make sure we replace only full words, see regex syntax.
This is my code.
SENTENCE = "He sad might have lung cancer. It’s just a rumor."
sent=(sent_tokenize(SENTENCE))
The output is
['He sad might have lung cancer.', 'It’s just a rumor.']
I want to get this array as
['He sad might have lung cancer.']
['It’s just a rumor.']
Is their any way of doing this and if so how?
Since you want to split according to a sentence, you can simply do this:
sentence_list = SENTENCE.split('.')
for sentence in sentence_list:
single_sentence = [sentence + '.']
If you actually want all lists containing a single sentence in the same data structure, you'd have to use a list of lists or a dictionary:
my_sentences = []
sentence_list = SENTENCE.split('.')
for sentence in sentence_list:
my_sentences.append([sentence + '.'])
To shorten this out using a list comprehension:
my_sentences = [[sentence + '.'] for sentence in SENTENCE.split('.')]
with the only culprit being that the SENTENCE splitting part will happen more often so it'll be slower working with a massive amount of sentences.
The solution using re.split() function:
import re
s = "He sad might have lung cancer. It’s just a rumor."
parts = [l if l[-1] == '.' else l + '.' for l in re.split(r'\.\s?(?!$)', s)]
print(parts)
The output:
['He sad might have lung cancer.', 'It’s just a rumor.']
r'\.\s?(?!$)' pattern, defines separator as . except that which is at the end of the text (?!$)
l if l[-1] == '.' else l + '.' - recovering . at the end of each line(as the dilimiter was not captured while splitting)
I am trying to get all names that start with a capital letter and ends with a full-stop on the same line where the number of characters are between 3 and 5
My text is as follows:
King. Great happinesse
Rosse. That now Sweno, the Norwayes King,
Craues composition:
Nor would we deigne him buriall of his men,
Till he disbursed, at Saint Colmes ynch,
Ten thousand Dollars, to our generall vse
King. No more that Thane of Cawdor shall deceiue
Our Bosome interest: Goe pronounce his present death,
And with his former Title greet Macbeth
Rosse. Ile see it done
King. What he hath lost, Noble Macbeth hath wonne.
I am testing it out on this link. I am trying to get all words between 3 and 5 but haven't succeeded.
Does this produce your desired output?
import re
re.findall(r'[A-Z].{2,4}\.', text)
When text contains the text in your question it will produce this output:
['King.', 'Rosse.', 'King.', 'Rosse.', 'King.']
The regex pattern matches any sequence of characters following an initial capital letter. You can tighten that up if required, e.g. using [a-z] in the pattern [A-Z][a-z]{2,4}\. would match an upper case character followed by between 2 to 4 lowercase characters followed by a literal dot/period.
If you don't want duplicates you can use a set to get rid of them:
>>> set(re.findall(r'[A-Z].{2,4}\.', text))
set(['Rosse.', 'King.'])
You may have your own reasons for wanting to use regexs here, but Python provides a rich set of string methods and (IMO) it's easier to understand the code using these:
matched_words = []
for line in open('text.txt'):
words = line.split()
for word in words:
if word[0].isupper() and word[-1] == '.' and 3 <= len(word)-1 <=5:
matched_words.append(word)
print matched_words
I'm using Python to search some words (also multi-token) in a description (string).
To do that I'm using a regex like this
result = re.search(word, description, re.IGNORECASE)
if(result):
print ("Trovato: "+result.group())
But what I need is to obtain the first 2 word before and after the match. For example if I have something like this:
Parking here is horrible, this shop sucks.
"here is" is the word that I looking for. So after I matched it with my regex I need the 2 words (if exists) before and after the match.
In the example:
Parking here is horrible, this
"Parking" and horrible, this are the words that I need.
ATTTENTION
The description cab be very long and the pattern "here is" can appear multiple times?
How about string operations?
line = 'Parking here is horrible, this shop sucks.'
before, term, after = line.partition('here is')
before = before.rsplit(maxsplit=2)[-2:]
after = after.split(maxsplit=2)[:2]
Result:
>>> before
['Parking']
>>> after
['horrible,', 'this']
Try this regex: ((?:[a-z,]+\s+){0,2})here is\s+((?:[a-z,]+\s*){0,2})
with re.findall and re.IGNORECASE set
Demo
I would do it like this (edit: added anchors to cover most cases):
(\S+\s+|^)(\S+\s+|)here is(\s+\S+|)(\s+\S+|$)
Like this you will always have 4 groups (might have to be trimmed) with the following behavior:
If group 1 is empty, there was no word before (group 2 is empty too)
If group 2 is empty, there was only one word before (group 1)
If group 1 and 2 are not empty, they are the words before in order
If group 3 is empty, there was no word after
If group 4 is empty, there was only one word after
If group 3 and 4 are not empty, they are the words after in order
Corrected demo link
Based on your clarification, this becomes a bit more complicated. The solution below deals with scenarios where the searched pattern may in fact also be in the two preceding or two subsequent words.
line = "Parking here is horrible, here is great here is mediocre here is here is "
print line
pattern = "here is"
r = re.search(pattern, line, re.IGNORECASE)
output = []
if r:
while line:
before, match, line = line.partition(pattern)
if match:
if not output:
before = before.split()[-2:]
else:
before = ' '.join([pattern, before]).split()[-2:]
after = line.split()[:2]
output.append((before, after))
print output
Output from my example would be:
[(['Parking'], ['horrible,', 'here']), (['is', 'horrible,'], ['great', 'here']), (['is', 'great'], ['mediocre', 'here']), (['is', 'mediocre'], ['here', 'is']), (['here', 'is'], [])]