Develop Reference

Replace equal sub-strings with different words

I have a string:
s = '96 ST MARY ST'
Now the first occurrence of 'ST' is Saint, and the second occurrence is Street i.e. Saint Mary Street.
I want to replace the first ST with Saint, and the second ST with Street. For this I tried to use find() and rfind():
# index of ST
ind = s.find('ST')
s[ind:(ind+2)] = 'Saint'
# index of last ST
ind2 = s.rfind('ST')
s[ind2:(ind2+2)] = 'Street'
TypeError: 'str' object does not support item assignment
I don't know how to get around this.
Is there a way to extract these sub-strings somehow and replace them?

Two replacement:
s = s.replace("ST", "Saint", 1).replace("ST", "Street", 1)

You might be OK with using re.sub along with its count parameter, to target the first replacement:
s = '96 ST MARY ST'
print(s)
out = re.sub(r'\bST\b', 'Saint', s, 1)
print(out)
out = re.sub(r'\bST\b', 'Street', s)
print(out)
This prints:
96 ST MARY ST
96 Saint MARY ST
96 Street MARY Street
However, while the above coincidentally works for your exact sample input, there are many edge cases where this would fail. It assumes that Saint comes before Street, and this may not always be the case, nor may there always be only two occurrences of ST.

A simple way, assuming there are always 2 occurences of 'ST':
p1, p2, _ = s.split('ST')
res = f"{p1}Saint{p2}Street"
If your input strings happen to be more complexes, you should go for regex (as Tim Biegeleisen proposed)

How to find a word inside a list and search backwards from that word?

How can I search for a specific word inside a list? Once I have searched for that specific word, how can I search backwards from there? In my example, I need to search for the city, and then search backwards to find the street type (ex: Rd, St, Ave, etc.)
I first allow a user to input an address, like 123 Fakeville St SW San Francisco CA 90215:
searchWord = 'San Francisco'
searchWord = searchWord.upper()
address = raw_input("Type an address: ").upper()
Once the address is entered, I split it using address = address.split(), which results in:
['123', 'Fakeville', 'St', 'SW', 'San Francisco', 'CA', '90215']
I then search for city in the list:
for items in address:
if searchWord in items:
print searchWord
But I'm not sure how to count backwards to find the street type (ex: St).

You can use list.index method to search the index of an item in a list.
There is no list.rindex method to search backward.
You need to use:
rev_idx = len(my_list) - my_list[::-1].index(item) - 1
I don't really understand what’s your aim, be I can explain how to search backward the "St" string in the address list of strings:
address = ['123', 'Fakeville', 'St', 'SW', 'San Francisco', 'CA', '90215']
town_idx = address.index('San Francisco')
print(town_idx)
# You'll get: 4
before = address[:town_idx]
st_index = len(before) - before[::-1].index("St") - 1
print(st_index)
# You'll get: 2

for items in address:
if searchWord in items:
for each in reversed(address[0:address.index(searchWord)]):
if each == 'St':
print each
Once you find the city, reverse traverse the list using reversed

Remove numbers conditionally?

I'm sorry if the title isn't very descriptive. I don't exactly know how to sum up my problem in a few words.
Here's my issue. I'm cleaning addresses and some of them are causing some issues.
I have a list of delimiters (avenue, street, road, place, etc etc etc) named patterns.
Let's say I have this address for example: SUITE 1603 200 PARK AVENUE SOUTH NEW YORK
I would like the output to be SUITE 200 PARK AVENUE SOUTH NEW YORK
Is there any way I could somehow look to see if there are 2 batches of numbers (in this case 1603 and 200) before one of my patterns and if so, strip the first batch of numbers from my string? i.e remove 1603 and keep 200.
Update: I've added this line to my code:
address = re.sub("\d+", "", address) however it's currently removing all the numbers. I thought that by putting ,1 after address it would only remove the first occurrence but that wasn't the case

If you want to apply this replacement only when one of your "separator" words is used, and only when there are two numbers, you can use a fancier regular expression.
import re
pattern = r"\d+ +(\d+ .*(STREET|AVENUE|ROAD|WHATEVER))"
input = "SUITE 1603 200 PARK AVENUE SOUTH NEW YORK"
output = re.sub(pattern, "\\1", input)
print(output) #SUITE 200 PARK AVENUE SOUTH NEW YORK

Your description of what you want to do isn't very clear, but if I understand correctly you want to is to delete the first occurrence of a number sequence?
You could do this without using a regex,
s = 'SUITE 1603 200 PARK AVENUE SOUTH NEW YORK'
l = s.split(' ')
for i, w in enumerate(l):
for c in w:
if c.isdigit():
del l[i]
break
print ' '.join(l)
Output: >>> SUITE 200 PARK AVENUE SOUTH NEW YORK

Rearrange position of words in string conditionally

I've spent the last few months developing a program that my company is using to clean and geocode addresses on a large scale (~5,000/day). It is functioning adequately well, however, there are certain address formats that I see daily that are causing issues for me.
Addresses with a format such as this park avenue 1 are causing issues with my geocoding. My thought process to tackle this issue is as follows:
Split the address into a list
Find the index of my delimiter word in the list. The delimiter words are words such as avenue, street, road, etc. I have a list of these delimiters called patterns.
Check to see if the word immediately following the delimiter is composed of digits with a length of 4 or less. If the number has a length of higher than 4 it is likely to be a zip code, which I do not need. If it's less than 4 it will most likely be the house number.
If the word meets the criteria that I explained in the previous step, I need to move it to the first position in the list.
Finally, I will put the list back together into a string.
Here is my initial attempt at putting my thoughts into code:
patterns ['my list of delimiters']
address = 'park avenue 1' # this is an example address
address = address.split(' ')
for pattern in patterns:
location = address.index(pattern) + 1
if address[location].isdigit() and len(address[location]) <= 4:
# here is where i'm getting a bit confused
# what would be a good way to go about moving the word to the first position in the list
address = ' '.join(address)
Any help would be appreciated. Thank you folks in advance.

Make the string address[location] into a list by wrapping it in brackets, then concatenate the other pieces.
address = [address[location]] + address[:location] + address[location+1:]
An example:
address = ['park', 'avenue', '1']
location = 2
address = [address[location]] + address[:location] + address[location+1:]
print(' '.join(address)) # => '1 park avenue'

Here's a modified version of your code. It uses simple list slicing to rearrange the parts of the address list.
Rather than using a for loop to search for a matching road type it uses set operations.
This code isn't perfect: it won't catch "numbers" like 12a, and it won't handle weird street names like "Avenue Road".
road_patterns = {'avenue', 'street', 'road', 'lane'}
def fix_address(address):
address_list = address.split()
road = road_patterns.intersection(address_list)
if len(road) == 0:
print("Can't find a road pattern in ", address_list)
elif len(road) > 1:
print("Ambiguous road pattern in ", address_list, road)
else:
road = road.pop()
index = address_list.index(road) + 1
if index < len(address_list):
number = address_list[index]
if number.isdigit() and len(number) <= 4:
address_list = [number] + address_list[:index] + address_list[index + 1:]
address = ' '.join(address_list)
return address
addresses = (
'42 tobacco road',
'park avenue 1 a',
'penny lane 17',
'nonum road 12345',
'strange street 23 london',
'baker street 221b',
'37 gasoline alley',
'83 avenue road',
)
for address in addresses:
fixed = fix_address(address)
print('{!r} -> {!r}'.format(address, fixed))
output
'42 tobacco road' -> '42 tobacco road'
'park avenue 1 a' -> '1 park avenue a'
'penny lane 17' -> '17 penny lane'
'nonum road 12345' -> 'nonum road 12345'
'strange street 23 london' -> '23 strange street london'
'baker street 221b' -> 'baker street 221b'
Can't find a road pattern in ['37', 'gasoline', 'alley']
'37 gasoline alley' -> '37 gasoline alley'
Ambiguous road pattern in ['83', 'avenue', 'road'] {'avenue', 'road'}
'83 avenue road' -> '83 avenue road'

Inserting ',' based on conditionals within a STR - Python

I am working with a very long list of street names that look like this:
1820 W 9000 SWest Jordan
455 S 500 ESalt Lake City
555 S 200 WBountiful
1000 N Green Valley PkwyHenderson
10100 W Tropicana AveLas Vegas
10305 S 1300 ESandy
10600 Southern Highlands PkwyLas Vegas
10616 S Eastern AveHenderson
111 Coors Blvd NWAlbuquerque
1170 E Gentile StLayton
1174 W 600 NSalt Lake City
1200 W Main StRiverton
....
....
I am trying to insert a ',' before the city name, which it appears is always after a lowercase character followed by NO SPACE and an UPPERCASE character.
So this is my thinking:
How do I write something that says, more or less,:
for cities in lst:
if [char] is lower and [nextchar] is UPPER:
[insert] ',' before UPPER

Following #Martijn's suggestion to take the last uppercase letter in a group, maybe:
import re
def fix(s):
return re.sub("([a-z]|[A-Z]+)([A-Z])",r"\1,\2", s)
which gives
>>> for line in lines:
... print fix(line)
...
1820 W 9000 S,West Jordan
455 S 500 E,Salt Lake City
555 S 200 W,Bountiful
1000 N Green Valley Pkwy,Henderson
10100 W Tropicana Ave,Las Vegas
10305 S 1300 E,Sandy
10600 Southern Highlands Pkwy,Las Vegas
10616 S Eastern Ave,Henderson
111 Coors Blvd NW,Albuquerque
1170 E Gentile St,Layton
1174 W 600 N,Salt Lake City
1200 W Main St,Riverton
[Disclaimer: I'm terrible with regexes.]

Something like this?
for big_index, cities in enumerate(lst):
for index, char in enumerate(cities):
if char == char.lower() and cities[index+1] != cities[index+1].lower():
lst[big_index] = cities[:index] + "," + cities[index:]
Disclaimer** Not tested. Since I don't have all your data, I won't attempt it, but this should give the output you're describing
**Edit: In fact, it doesn't look like your data follows these rules at all. Like the example in the comments, what about Coors Blvd NWAlbuquerque? Anyway, I'll keep the code here unless you change your question

cities = [re.sub(r'(?<=[a-z])(?=[A-Z])', ',', x) for x in cities]

This would be a solution without regex, basically implementing your logic:
new_list = []
for line in big_list:
for c in xrange(len(line)-1):
if 97 <= ord(line[c]) <= 122 and 65 <= ord(line[c+1]) <= 90:
line = line[:c+1]+","+line[c+1:]
break
new_list.append(line)
>>> new_list
['1820 W 9000 SWest Jordan', '455 S 500 ESalt Lake City', '555 S 200 WBountiful', '1000 N Green Valley Pkwy,Henderson', '10100 W Tropicana Ave,Las Vegas', '10305 S 1300 ESandy', '10600 Southern Highlands Pkwy,Las Vegas', '10616 S Eastern Ave,Henderson', '111 Coors Blvd NWAlbuquerque', '1170 E Gentile St,Layton', '1174 W 600 NSalt Lake City', '1200 W Main St,Riverton']
In case you're wondering what the ord function does: It translates a character into ASCII code. In ASCII, lower-case letters are bound to 97-122, so if ord(char) is in that range, it's a lower-case letter. Same goes for upper-case letters, except they're bound to 65-90.

Your core problem is the missing space between the street and the state portions. Assuming a solution where you have already split this string into component address parts (perhaps using #jazzpi 's solution), you can solve this secondary problem by building a collection of strings that match postal designations, such as ['Ave', 'E', 'Pkwy'] and so on, then look for matches to that collection on the left end of the state string.
Once you find a match, check to see if removing that sub-string leaves the state with an initial capital letter. If it leaves an initial capital letter intact, then you are free to truncate the substring and append the truncated street designation to the street portion.