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
Related
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.
I have some sentence like
1:
"RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held
ball is correctly called."
2:
"Nurkic (POR) maintains legal
guarding position and makes incidental contact with Wall (WAS) that
does not affect his driving shot attempt."
I need to use Python regex to find the name "Oubre Jr." ,"Nurkic" and "Nurkic", "Wall".
p = r'\s*(\w+?)\s[(]'
use this pattern,
I can find "['Nurkic', 'Wall']", but in sentence 1, I just can find ['Nurkic'], missed "Oubre Jr."
Who can help me?
You can use the following regex:
(?:[A-Z][a-z][\s\.a-z]*)+(?=\s\()
|-----Main Pattern-----|
Details:
(?:) - Creates a non-capturing group
[A-Z] - Captures 1 uppercase letter
[a-z] - Captures 1 lowercase letter
[\s\.a-z]* - Captures spaces (' '), periods ('.') or lowercase letters 0+ times
(?=\s\() - Captures the main pattern if it is only followed by ' (' string
str = '''RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held ball is correctly called.
Nurkic (POR) maintains legal guarding position and makes incidental contact with Wall (WAS) that does not affect his driving shot attempt.'''
res = re.findall( r'(?:[A-Z][a-z][\s\.a-z]*)+(?=\s\()', str )
print(res)
Demo: https://repl.it/#RahulVerma8/OvalRequiredAdvance?language=python3
Match: https://regex101.com/r/OsLTrY/1
Here is one approach:
line = "RLB shows Oubre Jr (WAS) legally ties up Nurkic (POR), and a held ball is correctly called."
results = re.findall( r'([A-Z][\w+'](?: [JS][r][.]?)?)(?= \([A-Z]+\))', line, re.M|re.I)
print(results)
['Oubre Jr', 'Nurkic']
The above logic will attempt to match one name, beginning with a capital letter, which is possibly followed by either the suffix Jr. or Sr., which in turn is followed by a ([A-Z]+) term.
You need a pattern that you can match - for your sentence you cou try to match things before (XXX) and include a list of possible "suffixes" to include as well - you would need to extract them from your sources
import re
suffs = ["Jr."] # append more to list
rsu = r"(?:"+"|".join(suffs)+")? ?"
# combine with suffixes
regex = r"(\w+ "+rsu+")\(\w{3}\)"
test_str = "RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held ball is correctly called. Nurkic (POR) maintains legal guarding position and makes incidental contact with Wall (WAS) that does not affect his driving shot attempt."
matches = re.finditer(regex, test_str, re.MULTILINE)
names = []
for matchNum, match in enumerate(matches,1):
for groupNum in range(0, len(match.groups())):
names.extend(match.groups(groupNum))
print(names)
Output:
['Oubre Jr.', 'Nurkic ', 'Nurkic ', 'Wall ']
This should work as long as you do not have Names with non-\w in them. If you need to adapt the regex, use https://regex101.com/r/pRr9ZU/1 as starting point.
Explanation:
r"(?:"+"|".join(suffs)+")? ?" --> all items in the list suffs are strung together via | (OR) as non grouping (?:...) and made optional followed by optional space.
r"(\w+ "+rsu+")\(\w{3}\)" --> the regex looks for any word characters followed by optional suffs group we just build, followed by literal ( then three word characters followed by another literal )
I have a string as follows:
theatre = 'Regal Crown Center Stadium 14'
I would like to break this into an acronym based on the first letter in each word but also include both numbers:
desired output = 'RCCS14'
My code attempts below:
acronym = "".join(word[0] for word in theatre.lower().split())
acronym = "".join(word[0].lower() for word in re.findall("(\w+)", theatre))
acronym = "".join(word[0].lower() for word in re.findall("(\w+ | \d{1,2})", theatre))
acronym = re.search(r"\b(\w+ | \d{1,2})", theatre)
In which I wind up with something like: rccs1 but can't seem to capture that last number. There could be instances when the number is in the middle of the name as well: 'Regal Crown Center 14 Stadium' as well. TIA!
See regex in use here
(?:(?<=\s)|^)(?:[a-z]|\d+)
(?:(?<=\s)|^) Ensure what precedes is either a space or the start of the line
(?:[a-z]|\d+) Match either a single letter or one or more digits
The i flag (re.I in python) allows [a-z] to match its uppercase variants.
See code in use here
import re
r = re.compile(r"(?:(?<=\s)|^)(?:[a-z]|\d+)", re.I)
s = 'Regal Crown Center Stadium 14'
print(''.join(r.findall(s)))
The code above finds all instances where the regex matches and joins the list items into a single string.
Result: RCCS14
You can use re.sub() to remove all lowercase letters and spaces.
Regex: [a-z ]+
Details:
[]+ Match a single character present in the list between one and
unlimited times
Python code:
re.sub(r'[a-z ]+', '', theatre)
Output: RCCS14
Code demo
I can't comment since I don't have enough reputation, but S. Jovan answer isn't satisfying since it assumes that each word starts with a capital letter and that each word has one and only one capital letter.
re.sub(r'[a-z ]+', '', "Regal Crown Center Stadium YB FIEUBFB DBUUFG FUEH 14")
will returns 'RCCSYBFIEUBFBDBUUFGFUEH14'
However ctwheels answers will be able to work in this case :
r = re.compile(r"\b(?:[a-z]|\d+)", re.I)
s = 'Regal Crown Center Stadium YB FIEUBFB DBUUFG FUEH 14'
print(''.join(r.findall(s)))
will print
RCCSYFDF14
import re
theatre = 'Regal Crown Center Stadium 14'
r = re.findall("\s(\d+|\S)", ' '+theatre)
print(''.join(r))
Gives me RCCS14
I have a large string and I want to find all the input sequences that are matching in this string.
So for example, I want to find all the possible matches of defensive rebound in:
Player xy had 10 defensive rebounds only in the 3rd quarter of a match that was a defensive battle between 2 teams that have a defensive rebound rate of over 80% and moreover the average number of rebounds in the defence by player was a staggering 3.5
I want to find all the bold words and after that extract them.
I managed to build a script that does the extraction but it only works for exact matches.
I was thinking of using difflib.SequenceMatcher but I got stuck.
You can use regex:
import re
#Find [defence(s)][space][rebound(s)][space][any word]
re.findall('defensive[\w]* rebound[\w]* [\w]+', s)
#Find [rebound(s)][space][any word][space][any word][space][any word]
re.findall('rebound[\w]* [\w]+ [\w]+ [\w]+', s)
findall returns a list of matches
If all your matches are in the same form of bold words you can extract them with:
re.findall('rebound[ \w]*defence', s)
re.findall('defensive[\w]* rebound[\w]*[ rate]*', s)
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.