Compare the numbers within different string using Python

Question

I have two types of addresses:

Unit 5, 123 Fake Street Drive
123 Fake St Dr, Unit 5

How can I use Python to compare the two addresses by the numbers?

For example:

Unit 5, 123 Fake Street Drive -> [5,123]
123 Fake St Dr, Unit 5 -> [123,5]

TRUE

123 Fake Street Drive -> [123]
123 Fake St Dr, Unit 5 -> [123,5]

FALSE

Unit 5, 155 Fake Street Drive -> [155,5]
123 Fake St Dr, Unit 5 -> [123,5]

FALSE

All I have now is:

if bool(set([int(s) for s in address.split() if s.isdigit()]) & set([int(s) for s in address2.split() if s.isdigit()])):

I want to find out if one list of numbers is the same as another list of numbers regardless of the order.

Just to make sure, but the if the addresses are Unit 1, 2 Street and 2 A completely different street, unit 1 the output should be False? — Aran-Fey
– Aran-Fey, Commented Mar 6, 2017 at 15:23
Try "1, 2".split() you'll get 1,: not all digits: fails. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Mar 6, 2017 at 15:23

Jean-François Fabre · Accepted Answer · 2017-03-06 16:18:03Z

3

You just have to build sets of extracted numbers and compare them with ==. set supports equality very well.

Another problem here is that str.split() won't work well for instance for 5,. So isdigit() fails and your sets aren't equal.

Let me suggest re.findall to find digits, put them into sets and compare, using \d+ or \b\d+\b to avoid digits inside words (like N2P for instance)

import re

address="Unit 5, 123 Fake Street Drive"
address2 = "123 Fake St Dr, Unit 5"

pattern = r"\b\d+\b"
print(set(re.findall(pattern,address))==set(re.findall(pattern,address2)))

This yields True, whereas if I change/add/remove one number from one of the lists above, I get False

As suggested in comments, that above fails if there are repeated numbers in one string, and not in the other: we could have a false positive since set clobbers duplicates.

If that's an issue, then replacing set with collections.Counter fixes that

collections.Counter(re.findall(pattern,address))==collections.Counter(re.findall(pattern,address2))

works too, Counter is a dictionary and compares to other dictionaries.

edited Mar 6, 2017 at 16:18

answered Mar 6, 2017 at 15:23

Jean-François Fabre♦

141k24 gold badges179 silver badges246 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Chris_Rands Over a year ago

{5} == {5,5} ; returns True (it should return False here), you'll need a Counter or multiset

Jean-François Fabre Over a year ago

you're right! OP example doesn't cover that. Good point. Edited with Counter, fits right in place.

Maciek Over a year ago

One more question, can I distinguish numbers that are not inside a word? For example, I don't want to capture numbers in postal codes such as "N2P"

John Bograd · Accepted Answer · 2017-03-06 15:37:41Z

0

I recommend that you use a sorted list, not a set. The set cannot distinguish between "Unit 1, 1 street X" and "1 Street Y" - but a sorted list will do.

answered Mar 6, 2017 at 15:37

John Bograd

11 bronze badge

Collectives™ on Stack Overflow

Compare the numbers within different string using Python

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related