Develop Reference

How can I split concatenated strings that contain no delimiters in python?

Let's say I have a list of concatenated firstname + lastname combinations like this:
["samsmith","sallyfrank","jamesandrews"]
I also have lists possible_firstnames and possible_lastnames.
If I want to split those full name strings based on values that appear in possible_firstnames and possible_lastnames, what is the best way of doing so?
My initial strategy was to compare characters between full name strings and each possible_firstnames/possible_lastnames value one by one, where I would split the full name string on discovery of a match. However, I realize that I would encounter a problem if, for example, "Sal" was included as a possible first name (my code would try to turn "sallyfrank" into "Sal Lyfrank" etc).
My next step would be to crosscheck what remains in the string after "sal" to values in possible_lastnames before finalizing the split, but this is starting to approach the convoluted and so I am left wondering if there is perhaps a much simpler option that I have been overlooking from the very beginning?
The language that I am working in is Python.

If you are getting similar names, like sam, samantha and saman, put them in reverse order so that the shortest is last
full_names = ["samsmith","sallyfrank","jamesandrews", "samanthasang", "samantorres"]
first_name = ["sally","james", "samantha", "saman", "sam"]
matches = []
for name in full_names:
for first in first_name:
if name.startswith(first):
matches.append(f'{first} {name[len(first):]}')
break
print(*matches, sep='\n')
Result
sam smith
sally frank
james andrews
samantha sang
saman torres
This won't pick out a name like Sam Antony. It would show this as *Saman Tony", in which case, your last name idea would work.
It also won't pick out Sam Anthanei. This could be Samantha Nei, Saman Thanei or Sam Anthanei if all three surnames were in your surname list.

Is this what u wanted
names = ["samsmith","sallyfrank","jamesandrews"]
pos_fname = ["sally","james"]
pos_lname = ["smith","frank"]
matches = []
for i in names:
for n in pos_fname:
if i.startswith(n):
break
else:
continue
for n in pos_lname:
if i.endswith(n):
matches.append(f"{i[:-len(n)].upper()} {n.upper()}")
break
else:
continue
print(matches)

Is there a way to combine multiple strings using Regex?

Having an issue with Regex and not really understanding its usefulness right now.
Trying to extrapolate data from a file. file consists of first name, last name, grade
File:
Peter Jenkins: A
Robert Right: B
Kim Long: C
Jim Jim: B
Opening file code:
##Regex Code r'([A-Za-z]+)(: B)
regcode = r'([A-Za-z]+)(: B)'
answer=re.findall(regcode,file)
return answer
The expected result is first name last name. The given result is last name and letter grade. How do I just get the first name and last name for all B grades?

Since you must use regex for this task, here's a simple regex solution that returns the full name:
'(.*): B'
Which works in this case because:
(.*) returns all text up to a match of : B
Click here to see my test and matching output. I recommend this site for your regex testing needs.

You can do it without regex:
students = '''Peter Jenkins: A
Robert Right: B
Kim Long: C
Jim Jim: B'''
for x in students.split('\n'):
string = x.split(': ')
if string[1] == 'B':
print(string[0])
# Robert Right
# Jim Jim
or
[x[0:-3] for x in students.split('\n') if x[-1] == 'B']

If a regex solution is required (I perosnally like the solution of Roman Zhak more), put inside a group what you are interested in, i.e. the first name and the second name. Follows colon and B:
import re
file = """
Peter Jenkins: A
Robert Right: B
Kim Long: C
Jim Jim: B
"""
regcode = r'([A-Za-z]+) ([A-Za-z]+): B'
answer=re.findall(regcode,file,re.)
print(answer) # [('Robert', 'Right'), ('Jim', 'Jim')]

Add a capturing group ('()') to your expression. Everything outside the group will be ignored, even if it matches the expression.
re.findall('(\w+\s+\w+):\s+B', file)
#['Robert Right', 'Jim Jim']
'\w' is any alphanumeric character, '\s' is any space-like character.
You can add two groups, one for the first name and one for the last name:
re.findall('(\w+)\s+(\w+):\s+B', data)
#[('Robert', 'Right'), ('Jim', 'Jim')]
The latter will not work if there are more than two names on one line.

matching content creating new column

Hello I have a dataset where I want to match my keyword with the location. The problem I am having is the location "Afghanistan" or "Kabul" or "Helmund" I have in my dataset appears in over 150 combinations including spelling mistakes, capitalization and having the city or town attached to its name. What I want to do is create a separate column that returns the value 1 if any of these characters "afg" or "Afg" or "kab" or "helm" or "are contained in the location. I am not sure if upper or lower case makes a difference.
For instance there are hundreds of location combinations like so: Jegdalak, Afghanistan, Afghanistan,Ghazni♥, Kabul/Afghanistan,
I have tried this code and it is good if it matches the phrase exactly but there is too much variation to write every exception down
keywords= ['Afghanistan','Kabul','Herat','Jalalabad','Kandahar','Mazar-i-Sharif', 'Kunduz', 'Lashkargah', 'mazar', 'afghanistan','kabul','herat','jalalabad','kandahar']
#how to make a column that shows rows with a certain keyword..
def keyword_solution(value):
strings = value.split()
if any(word in strings for word in keywords):
return 1
else:
return 0
taleban_2['keyword_solution'] = taleban_2['location'].apply(keyword_solution)
# below will return the 1 values
taleban_2[taleban_2['keyword_solution'].isin(['1'])].head(5)
Just need to replace this logic where all results will be put into column "keyword_solution" that matches either "Afg" or "afg" or "kab" or "Kab" or "kund" or "Kund"

Given the following:
Sentences from the New York Times
Remove all non-alphanumeric characters
Change everything to lowercase, thereby removing the need for different word variations
Split the sentence into a list or set. I used set because of the long sentences.
Add to the keywords list as needed
Matching words from two lists
'afgh' in ['afghanistan']: False
'afgh' in 'afghanistan': True
Therefore, the list comprehension searches for each keyword, in each word of word_list.
[True if word in y else False for y in x for word in keywords]
This allows the list of keywords to be shorter (i.e. given afgh, afghanistan is not required)
import re
import pandas as pd
keywords= ['jalalabad',
'kunduz',
'lashkargah',
'mazar',
'herat',
'mazar',
'afgh',
'kab',
'kand']
df = pd.DataFrame({'sentences': ['The Taliban have wanted the United States to pull troops out of Afghanistan Turkey has wanted the Americans out of northern Syria and North Korea has wanted them to at least stop military exercises with South Korea.',
'President Trump has now to some extent at least obliged all three — but without getting much of anything in return. The self-styled dealmaker has given up the leverage of the United States’ military presence in multiple places around the world without negotiating concessions from those cheering for American forces to leave.',
'For a president who has repeatedly promised to get America out of foreign wars, the decisions reflect a broader conviction that bringing troops home — or at least moving them out of hot spots — is more important than haggling for advantage. In his view, decades of overseas military adventurism has only cost the country enormous blood and treasure, and waiting for deals would prolong a national disaster.',
'The top American commander in Afghanistan, Gen. Austin S. Miller, said Monday that the size of the force in the country had dropped by 2,000 over the last year, down to somewhere between 13,000 and 12,000.',
'“The U.S. follows its interests everywhere, and once it doesn’t reach those interests, it leaves the area,” Khairullah Khairkhwa, a senior Taliban negotiator, said in an interview posted on the group’s website recently. “The best example of that is the abandoning of the Kurds in Syria. It’s clear the Kabul administration will face the same fate.”',
'afghan']})
# substitute non-alphanumeric characters
df['sentences'] = df['sentences'].apply(lambda x: re.sub('[\W_]+', ' ', x))
# create a new column with a list of all the words
df['word_list'] = df['sentences'].apply(lambda x: set(x.lower().split()))
# check the list against the keywords
df['location'] = df.word_list.apply(lambda x: any([True if word in y else False for y in x for word in keywords]))
# final
print(df.location)
0 True
1 False
2 False
3 True
4 True
5 True
Name: location, dtype: bool

Regex to match strings in quotes that contain only 3 or less capitalized words

I've searched and searched, but can't find an any relief for my regex woes.
I wrote the following dummy sentence:
Watch Joe Smith Jr. and Saul "Canelo" Alvarez fight Oscar de la Hoya and Genaddy Triple-G Golovkin for the WBO belt GGG. Canelo Alvarez and Floyd 'Money' Mayweather fight in Atlantic City, New Jersey. Conor MacGregor will be there along with Adonis Superman Stevenson and Mr. Sugar Ray Robinson. "Here Goes a String". 'Money Mayweather'. "this is not a-string", "this is not A string", "This IS a" "Three Word String".
I'm looking for a regular expression that will return the following when used in Python 3.6:
Canelo, Money, Money Mayweather, Three Word String
The regex that has gotten me the closest is:
(["'])[A-Z](\\?.)*?\1
I want it to only match strings of 3 capitalized words or less immediately surrounded by single or double quotes. Unfortunately, so far it seem to match any string in quotes, no matter what the length, no matter what the content, as long is it begins with a capital letter.
I've put a lot of time into trying to hack through it myself, but I've hit a wall. Can anyone with stronger regex kung-fu give me an idea of where I'm going wrong here?

Try to use this one: (["'])((?:[A-Z][a-z]+ ?){1,3})\1
(["']) - opening quote
([A-Z][a-z]+ ?){1,3} - Capitalized word repeating 1 to 3 times separated by space
[A-Z] - capital char (word begining char)
[a-z]+ - non-capital chars (end of word)
_? - space separator of capitalized words (_ is a space), ? for single word w/o ending space
{1,3} - 1 to 3 times
\1 - closing quote, same as opening
Group 2 is what you want.
Match 1
Full match 29-37 `"Canelo"`
Group 1. 29-30 `"`
Group 2. 30-36 `Canelo`
Match 2
Full match 146-153 `'Money'`
Group 1. 146-147 `'`
Group 2. 147-152 `Money`
Match 3
Full match 318-336 `'Money Mayweather'`
Group 1. 318-319 `'`
Group 2. 319-335 `Money Mayweather`
Match 4
Full match 398-417 `"Three Word String"`
Group 1. 398-399 `"`
Group 2. 399-416 `Three Word String`
RegEx101 Demo: https://regex101.com/r/VMuVae/4

Working with the text you've provided, I would try to use regular expression lookaround to get the words surrounded by quotes and then apply some conditions on those matches to determine which ones meet your criterion. The following is what I would do:
[p for p in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt) if all(x.istitle() for x in p.split(' ')) and len(p.split(' ')) <= 3]
txt is the text you've provided here. The output is the following:
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']
Cleaner:
matches = []
for m in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt):
if all(x.istitle() for x in m.split(' ')) and len(m.split(' ')) <= 3:
matches.append(m)
print(matches)
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']

Here's my go at it: ([\"'])(([A-Z][^ ]*? ?){1,3})\1

Regex Python [python-2.7]

I'm working on a Python program that sifts through a .txt file to find the genus and species name. The lines are formatted like this (yes, the equals signs are consistently around the common name):
1. =Common Name= Genus Species some other words that I don't want.
2. =Common Name= Genus Species some other words that I don't want.
I can't seem to figure out a regex that will work to match only the genus and species and not the common name. I know the equals signs (=) will probably help in some way but I cannot think of how to use them.
Edit: Some real data:
1. =Western grebe.= ÆCHMOPHORUS OCCIDENTALIS. Rare migrant; western species, chiefly interior regions of North America.
2. =Holboell's grebe.= COLYMBUS HOLBOELLII. Rare migrant; breeds far north; range, all of North America.
3. =Horned grebe.= COLYMBUS AURITUS. Rare migrant; range, almost the same as the last.
4. =American eared grebe.= COLYMBUS NIGRICOLLIS CALIFORNICUS. Summer resident; rare in eastern, common in western Colorado; breeds from plains to 8,000 feet; partial to alkali lakes; western species.

You probably don't need regex for this one. If the order of the words you need and the count of the words is always the same, you can just split each line into list of substrings and get the third (genus) and the fourth (species) element of that list. The code will probably look like that:
myfile = open('myfilename.txt', 'r')
for line in myfile.readlines():
words = line.split()
genus, species = words[2], words[3]
It just looks a little more "pythonic" to me.
If common name can consist of multiple words, then suggested code will return an incorrect result. To get the right result in this case too, you can use this code:
myfile = open('myfilename.txt', 'r')
for line in myfile.readlines():
words = line.split('=')[2].split() # If the program returns wrong results, try changing the index from 2 to 1 or 3. What number is the right one depends on whether there can be any symbols before the first "=".
genus, species = words[0], words[1]

If it is enough to capture words in groups (and you dont't wont direct match) you can try with:
(?=\d\.\s*=[^=]+=\s(?:(?P<genus>\w+)\s(?P<species>\w+)))
DEMO
the desired values will be in groups <genus> and <species>. The whole regex is a positive lookbehind, so it match a zero point position on a beginning of string, but it captures some content into groups.
(?=\d\.\s*=[^=]+=\s - decimal folowed by some content between equal
signs and space,
(?:(?P<genus>\w+)\s(?P<species>\w+))) - capture first word to genus
groups, and second word do species groups,

You can try something like:
import re
txt='1. =Common Name= Genus Species some other words that I don\'t want.'
re1='.*?' # Non-greedy match on filler
re2='(?:[a-z][a-z]+)' # Uninteresting: word
re3='.*?' # Non-greedy match on filler
re4='(?:[a-z][a-z]+)' # Uninteresting: word
re5='.*?' # Non-greedy match on filler
re6='((?:[a-z][a-z]+))' # Word 1
re7='.*?' # Non-greedy match on filler
re8='((?:[a-z][a-z]+))' # Word 2
rg = re.compile(re1+re2+re3+re4+re5+re6+re7+re8,re.IGNORECASE|re.DOTALL)
m = rg.search(txt)
if m:
word1=m.group(1)
word2=m.group(2)
print "("+word1+")"+"("+word2+")"+"\n"
In your test input as shown in txt, this will print
(Genus)(Species)
You can you this awesome site to help do regexes like this!
Hope this helps