Develop Reference

Parse a very large text file with Python?

So, the file has about 57,000 book titles, author names and a ETEXT No. I am trying to parse the file to only get the ETEXT NOs
The File is like this:
TITLE and AUTHOR ETEXT NO.
Aspects of plant life; with special reference to the British flora,      56900
by Robert Lloyd Praeger
The Vicar of Morwenstow, by Sabine Baring-Gould 56899
[Subtitle: Being a Life of Robert Stephen Hawker, M.A.]
Raamatun tutkisteluja IV, mennessä Charles T. Russell 56898
[Subtitle: Harmagedonin taistelu]
[Language: Finnish]
Raamatun tutkisteluja III, mennessä Charles T. Russell 56897
[Subtitle: Tulkoon valtakuntasi]
[Language: Finnish]
Tom Thatcher's Fortune, by Horatio Alger, Jr. 56896
A Yankee Flier in the Far East, by Al Avery 56895
and George Rutherford Montgomery
[Illustrator: Paul Laune]
Nancy Brandon's Mystery, by Lillian Garis 56894
Nervous Ills, by Boris Sidis 56893
[Subtitle: Their Cause and Cure]
Pensées sans langage, par Francis Picabia 56892
[Language: French]
Helon's Pilgrimage to Jerusalem, Volume 2 of 2, by Frederick Strauss 56891
[Subtitle: A picture of Judaism, in the century
which preceded the advent of our Savior]
Fra Tommaso Campanella, Vol. 1, di Luigi Amabile 56890
[Subtitle: la sua congiura, i suoi processi e la sua pazzia]
[Language: Italian]
The Blue Star, by Fletcher Pratt 56889
Importanza e risultati degli incrociamenti in avicoltura, 56888
di Teodoro Pascal
[Language: Italian]
And this is what I tried:
def search_by_etext():
fhand = open('GUTINDEX.ALL')
print("Search by ETEXT:")
for line in fhand:
if not line.startswith(" [") and not line.startswith("~"):
if not line.startswith(" ") and not line.startswith("TITLE"):
words = line.rstrip()
words = line.lstrip()
words = words[-7:]
print (words)
search_by_etext()
Well the code mostly works. However for some lines it gives me part of title or other things. Like:
This kind of output(), containing 'decott' which is a part of author name and shouldn't be here.
2
For this:
The Bashful Earthquake, by Oliver Herford                                56765
[Subtitle: and Other Fables and Verses]
The House of Orchids and Other Poems, by George Sterling                 56764
North Italian Folk, by Alice Vansittart Strettel Carr                    56763
 and Randolph Caldecott
[Subtitle: Sketches of Town and Country Life]
Wild Life in New Zealand. Part 1, Mammalia, by George M. Thomson 56762
[Subtitle: New Zealand Board of Science and Art, Manual No. 2]
Universal Brotherhood, Volume 13, No. 10, January 1899, by Various 56761
De drie steden: Lourdes, door Émile Zola 56760
[Language: Dutch]
Another example:
4
For
Rhandensche Jongens, door Jan Lens 56702
[Illustrator: Tjeerd Bottema]
[Language: Dutch]
The Story of The Woman's Party, by Inez Haynes Irwin 56701
Mormon Doctrine Plain and Simple, by Charles W. Penrose 56700
[Subtitle: Or Leaves from the Tree of Life]
The Stone Axe of Burkamukk, by Mary Grant Bruce 56699
[Illustrator: J. Macfarlane]
The Latter-Day Prophet, by George Q. Cannon 56698
[Subtitle: History of Joseph Smith Written for Young People]
Here: Life] shouldn't be there. Lines starting with blank space has been parsed out with this:
if not line.startswith(" [") and not line.startswith("~"):
But Still I am getting those off values in my output results.

Simple solution: regexps to the rescue !
import re
with open("etext.txt") as f:
for line in f:
match = re.search(r" (\d+)$", line.strip())
if match:
print(match.group(1))
the regular expression (\d+)$ will match "at least one space followed by 1 or more digits at the end of the string", and capture only the "one or more digits" group.
You can eventually improve the regexp - ie if you know all etext codes are exactly 5 digits long, you can change the regexp to (\d{5})$.
This works with the example text you posted. If it doesn't properly work on your own file then we need enough of the real data to find out what you really have.

It could be that those extra lines that are not being filtered out start with whitespace other than a " " char, like a tab for example. As a minimal change that might work, try filtering out lines that start with any whitespace rather than specifically a space char?
To check for whitespace in general rather than a space char, you'll need to use regular expressions. Try if not re.match(r'^\s', line) and ...

Python: replace \n \r \t in a list excluding those starting \n\n and ends with \n\r\n\t

My List:
['\n\r\n\tThis article is about sweet bananas. For the genus to which
banana plants belong, see Musa (genus).\n\r\n\tFor starchier bananas
used in cooking, see Cooking banana. For other uses, see Banana
(disambiguation)\n\r\n\tMusa species are native to tropical Indomalaya
and Australia, and are likely to have been first domesticated in Papua
New Guinea.\n\r\n\tThey are grown in 135
countries.\n\n\n\r\n\tWorldwide, there is no sharp distinction between
"bananas" and "plantains".\n\nDescription\n\r\n\tThe banana plant is
the largest herbaceous flowering plant.\n\r\n\tAll the above-ground
parts of a banana plant grow from a structure usually called a
"corm".\n\nEtymology\n\r\n\tThe word banana is thought to be of West
African origin, possibly from the Wolof word banaana, and passed into
English via Spanish or Portuguese.\n']
Example code:
import requests
from bs4 import BeautifulSoup
import re
re=requests.get('http://www.abcde.com/banana')
soup=BeautifulSoup(re.text.encode('utf-8'), "html.parser")
title_tag = soup.select_one('.page_article_title')
print(title_tag.text)
list=[]
for tag in soup.select('.page_article_content'):
list.append(tag.text)
#list=([c.replace('\n', '') for c in list])
#list=([c.replace('\r', '') for c in list])
#list=([c.replace('\t', '') for c in list])
print(list)
After I scraping a web page, I need to do data cleansing. I want to replace all the "\r", "\n", "\t" as "", but I found that I have subtitle in this, if I do this, subtitles and sentences are going to mix together.
Every subtitle always starts with \n\n and ends with \n\r\n\t, is it possible that I can do something to distinguish them in this list like \aEtymology\a. It's not going to work if I replace \n\n and \n\r\n\t separately to \a first cause other parts might have the same elements like this \n\n\r and it will become like \a\r. Thanks in advance!

Approach
Replace the subtitles to a custom string, <subtitles> in the list
Replace the \n, \r, \t etc. in the list
Replace the custom string with the actual subtitle
Code
l=['\n\r\n\tThis article is about sweet bananas. For the genus to which banana plants belong, see Musa (genus).\n\r\n\tFor starchier bananas used in cooking, see Cooking banana. For other uses, see Banana (disambiguation)\n\r\n\tMusa species are native to tropical Indomalaya and Australia, and are likely to have been first domesticated in Papua New Guinea.\n\r\n\tThey are grown in 135 countries.\n\n\n\r\n\tWorldwide, there is no sharp distinction between "bananas" and "plantains".\n\nDescription\n\r\n\tThe banana plant is the largest herbaceous flowering plant.\n\r\n\tAll the above-ground parts of a banana plant grow from a structure usually called a "corm".\n\nEtymology\n\r\n\tThe word banana is thought to be of West African origin, possibly from the Wolof word banaana, and passed into English via Spanish or Portuguese.\n']
import re
regex=re.findall("\n\n.*.\n\r\n\t",l[0])
print(regex)
for x in regex:
l = [r.replace(x,"<subtitles>") for r in l]
rep = ['\n','\t','\r']
for y in rep:
l = [r.replace(y, '') for r in l]
for x in regex:
l = [r.replace('<subtitles>', x, 1) for r in l]
print(l)
Output
['\n\nDescription\n\r\n\t', '\n\nEtymology\n\r\n\t']
['This article is about sweet bananas. For the genus to which banana plants belong, see Musa (genus).For starchier bananas used in cooking, see Cooking banana. For other uses, see Banana (disambiguation)Musa species are native to tropical Indomalaya and Australia, and are likely to have been first domesticated in Papua New Guinea.They are grown in 135 countries.Worldwide, there is no sharp distinction between "bananas" and "plantains".\n\nDescription\n\r\n\tThe banana plant is the largest herbaceous flowering plant.All the above-ground parts of a banana plant grow from a structure usually called a "corm".\n\nEtymology\n\r\n\tThe word banana is thought to be of West African origin, possibly from the Wolof word banaana, and passed into English via Spanish or Portuguese.']

import re
print([re.sub(r'[\n\r\t]', '', c) for c in list])
I think you may use regex

You can do this by using regular expressions:
import re
subtitle = re.compile(r'\n\n(\w+)\n\r\n\t')
new_list = [subtitle.sub(r"\a\g<1>\a", l) for l in li]
\g<1> is a backreference to the (\w+) in the first regex. It lets you reuse what ever is in there.

Pattern of regular expressions while using Look Behind or Look Ahead Functions to find a match

I am trying to split a sentence correctly bases on normal grammatical rules in python.
The sentence I want to split is
s = """Mr. Smith bought cheapsite.com for 1.5 million dollars,
i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a
probability of .9 it isn't."""
The expected output is
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with a probability of .9 it isn't.
To achieve this I am using regular , after a lot of searching I came upon the following regex which does the trick.The new_str was jut to remove some \n from 's'
m = re.split(r'(?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)\s',new_str)
for i in m:
print (i)
Mr. Smith bought cheapsite.com for 1.5 million dollars,i.e. he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with aprobability of .9 it isn't.
So the way I understand the reg ex is that we are first selecting
1) All the characters like i.e
2) From the filtered spaces from the first selection ,we select those characters
which dont have words like Mr. Mrs. etc
3) From the filtered 2nd step we select only those subjects where we have either dot or question and are preceded by a space.
So I tried to change the order as below
1) Filter out all the titles first.
2) From the filtered step select those that are preceded by space
3) remove all phrases like i.e
but when I do that the blank after is also split
m = re.split(r'(?<![A-Z][a-z]\.)(?<=\.|\?)\s(?<!\w\.\w.)',new_str)
for i in m:
print (i)
Mr. Smith bought cheapsite.com for 1.5 million dollars,i.e.
he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with aprobability of .9 it isn't.
Shouldn't the last step in the modified procedure be capable in identifying phrases like i.e ,why is it failing to detect it ?

First, the last . in (?<!\w\.\w.) looks suspicious, if you need to match a literal dot with it, escape it ((?<!\w\.\w\.)).
Coming back to the question, when you use r'(?<![A-Z][a-z]\.)(?<=\.|\?)\s(?<!\w\.\w.)' regex, the last negative lookbehind checks if the position after a whitespace is not preceded with a word char, dot, word char, any char (since the . is unescaped). This condition is true, because there are a dot, e, another . and a space before that position.
To make the lookbehind work that same way as when it was before \s, put the \s into the lookbehind pattern, too:
(?<![A-Z][a-z]\.)(?<=\.|\?)\s(?<!\w\.\w.\s)
See the regex demo
Another enhancement can be using a character class in the second lookbehind: (?<=\.|\?) -> (?<=[.?]).

Insert quotes in the string using index python

I want to insert quotes("") around the date and text in the string (which is in the file input.txt). Here is my input file:
created_at : October 9, article : ISTANBUL — Turkey is playing a risky game of chicken in its negotiations with NATO partners who want it to join combat operations against the Islamic State group — and it’s blowing back with violence in Turkish cities. As the Islamic militants rampage through Kurdish-held Syrian territory on Turkey’s border, Turkey says it won’t join the fight unless the U.S.-led coalition also goes after the government of Syrian President Bashar Assad.
created_at : October 9, article : President Obama chairs a special meeting of the U.N. Security Council last month. (Timothy A. Clary/AFP/Getty Images) When it comes to President Obama’s domestic agenda and his maneuvers to (try to) get things done, I get it. I understand what he’s up to, what he’s trying to accomplish, his ultimate endgame. But when it comes to his foreign policy, I have to admit to sometimes thinking “whut?” and agreeing with my colleague Ed Rogers’s assessment on the spate of books criticizing Obama’s foreign policy stewardship.
I want to put quotes around the date and text as follows:
created_at : "October 9", article : "ISTANBUL — Turkey is playing a risky game of chicken in its negotiations with NATO partners who want it to join combat operations against the Islamic State group — and it’s blowing back with violence in Turkish cities. As the Islamic militants rampage through Kurdish-held Syrian territory on Turkey’s border, Turkey says it won’t join the fight unless the U.S.-led coalition also goes after the government of Syrian President Bashar Assad".
created_at : "October 9", article : "President Obama chairs a special meeting of the U.N. Security Council last month. (Timothy A. Clary/AFP/Getty Images) When it comes to President Obama’s domestic agenda and his maneuvers to (try to) get things done, I get it. I understand what he’s up to, what he’s trying to accomplish, his ultimate endgame. But when it comes to his foreign policy, I have to admit to sometimes thinking “whut?” and agreeing with my colleague Ed Rogers’s assessment on the spate of books criticizing Obama’s foreign policy stewardship".
Here is my code which finds the index for comma(, after the date) and index for the article and then by using these, I want to insert quotes around the date. Also I want to insert quotes around the text, but how to do this?
f = open("input.txt", "r")
for line in f:
article_pos = line.find("article")
print article_pos
comma_pos = line.find(",")
print comma_pos

While you can do this with low-level operations like find and slicing, that's really not the easy or idiomatic way to do it.
First, I'll show you how to do it your way:
comma_pos = line.find(", ")
first_colon_pos = line.find(" : ")
second_colon_pos = line.find(" : ", comma_pos)
line = (line[:first_colon_pos+3] +
'"' + line[first_colon_pos+3:comma_pos] + '"' +
line[comma_pos:second_colon_pos+3] +
'"' + line[second_colon_pos+3:] + '"')
But you can more easily just split the line into bits, munge those bits, and join them back together:
dateline, article = line.split(', ', 1)
key, value = dateline.split(' : ')
dateline = '{} : "{}"'.format(key, value)
key, value = article.split(' : ')
article = '{} : "{}"'.format(key, value)
line = '{}, {}'.format(dateline, article)
And then you can take the repeated parts and refactor them into a simple function so you don't have to write the same thing twice (which may come in handy if you later need to write it four times).
It's even easier using a regular expression, but that might not be as easy to understand for a novice:
line = re.sub(r'(.*?:\s*)(.*?)(\s*,.*?:\s*)(.*)', r'\1"\2"\3"\4"', line)
This works by capturing everything up to the first : (and any spaces after it) in one group, then everything from there to the first comma in a second group, and so on:
(.*?:\s*)(.*?)(\s*,.*?:\s*)(.*)
Debuggex Demo
Notice that the regex has the advantage that I can say "any spaces after it" very simply, while with find or split I had to explicitly specify that there was exactly one space on either side of the colon and one after the comma, because searching for "0 or more spaces" is a lot harder without some way to express it like \s*.

You could also take a look at the regex library re.
E.g.
>>> import re
>>> print(re.sub(r'created_at:\s(.*), article:\s(.*)',
... r'created_at: "\1", article: "\2"',
... 'created_at: October 9, article: ...'))
created_at: "October 9", article: "..."
The first param to re.sub is the pattern you are trying to match. The parens () capture the matches and can be used in the second argument with \1. The third argument is the line of text.

FInd a US street address in text (preferably using Python regex)

Disclaimer: I read very carefully this thread:
Street Address search in a string - Python or Ruby
and many other resources.
Nothing works for me so far.
In some more details here is what I am looking for is:
The rules are relaxed and I definitely am not asking for a perfect code that covers all cases; just a few simple basic ones with assumptions that the address should be in the format:
a) Street number (1...N digits);
b) Street name : one or more words capitalized;
b-2) (optional) would be best if it could be prefixed with abbrev. "S.", "N.", "E.", "W."
c) (optional) unit/apartment/etc can be any (incl. empty) number of arbitrary characters
d) Street "type": one of ("st.", "ave.", "way");
e) City name : 1 or more Capitalized words;
f) (optional) state abbreviation (2 letters)
g) (optional) zip which is any 5 digits.
None of the above needs to be a valid thing (e.g. an existing city or zip).
I am trying expressions like these so far:
pat = re.compile(r'\d{1,4}( \w+){1,5}, (.*), ( \w+){1,5}, (AZ|CA|CO|NH), [0-9]{5}(-[0-9]{4})?', re.IGNORECASE)
>>> pat.search("123 East Virginia avenue, unit 123, San Ramondo, CA, 94444")
Don't work, and for me it's not easy to understand why. Specifically: how do I separate in my pattern a group of any words from one of specific words that should follow, like state abbrev. or street "type ("st., ave.)?
Anyhow: here is an example of what I am hoping to get:
Given
def ex_addr(text):
# does the re magic
# returns 1st address (all addresses?) or None if nothing found
for t in [
'The meeting will be held at 22 West Westin st., South Carolina, 12345 on Nov.-18',
'The meeting will be held at 22 West Westin street, SC, 12345 on Nov.-18',
'Hi there,\n How about meeting tomorr. #10am-sh in Chadds # 123 S. Vancouver ave. in Ottawa? \nThanks!!!',
'Hi there,\n How about meeting tomorr. #10am-sh in Chadds # 123 S. Vancouver avenue in Ottawa? \nThanks!!!',
'This was written in 1999 in Montreal',
"Cool cafe at 420 Funny Lane, Cupertino CA is way too cool",
"We're at a party at 12321 Mammoth Lane, Lexington MA 77777; Come have a beer!"
] print ex_addr(t)
I would like to get:
'22 West Westin st., South Carolina, 12345'
'22 West Westin street, SC, 12345'
'123 S. Vancouver ave. in Ottawa'
'123 S. Vancouver avenue in Ottawa'
None # for 'This was written in 1999 in Montreal',
"420 Funny Lane, Cupertino CA",
"12321 Mammoth Lane, Lexington MA 77777"
Could you please help?

I just ran across this in GitHub as I am having a similar problem. Appears to work and be more robust than your current solution.
https://github.com/madisonmay/CommonRegex
Looking at the code, the regex for street address accounts for many more scenarios. '\d{1,4} [\w\s]{1,20}(?:street|st|avenue|ave|road|rd|highway|hwy|square|sq|trail|trl|drive|dr|court|ct|parkway|pkwy|circle|cir|boulevard|blvd)\W?(?=\s|$)'

\d{1,4}( \w+){1,5}, (.*), ( \w+){1,5}, (AZ|CA|CO|NH), [0-9]{5}(-[0-9]{4})?
In this regex, you have one too many spaces (before ( \w+){1,5}, which already begins with one). Removing it, it matches your example.
I don't think you can assume that a "unit 123" or similar will be there, or there might be several ones (e.g. "building A, apt 3"). Note that in your initial regex, the . might match , which could lead to very long (and unwanted) matches.
You should probably accept several such groups with a limitation on the number (e.g. replace , (.*) with something like (, [^,]{1,20}){0,5}.
In any case, you will probably never get something 100% accurate that will accept any variation people might throw at them. Do lots of tests! Good luck.