CODEWITHSUNDEEP
pythonarrays
Ruham
Ruham

Reputation: 759

check for consecutive values membership in arrays

There are several similar questions around that suggest using any() with sets, however mine is a bit different in a way that I want to get the most effective and Pythonic way of forming a new array out of existing arrays based on membership.

I have the following array [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6], [2, 4, 5, 6, 9, 1, 4, 5, 6, 4, 5, 6], [1, 2, 9, 4, 5, 6, 7, 7, 8, 5]]

What I need is a way to form a new object {[4, 5, 6] : 6} where [4, 5, 6] is the key, and 6 is the number of times the key sequence appears in the array above.

I have achieved the result by making a function with simple for loops, but I don't think it is the most Pythonic and efficient way of achieving it. I was thinking of using map() and filter() as well.

What would be the most efficient and Pythonic way of achieving the result?

Upvotes: 0

Views: 59

Answers (2)

Alain T.
Alain T.

Reputation: 42143

You could use Counter from collection and initialize a Counter object with all the 3 number tuples in the various lists. You will then be able to get the count of any 3 number series instantly:

lists = [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6], [2, 4, 5, 6, 9, 1, 4, 5, 6, 4, 5, 6], [1, 2, 9, 4, 5, 6, 7, 7, 8, 5]]

from collections import Counter
counts = Counter( (a,b,c) for lst in lists for a,b,c in zip(lst,lst[1:],lst[2:]) )

print(counts[(4,5,6)]) # 6

counts is a Counter dictionary that will contain:

{ (4, 5, 6): 6,  (5, 6, 7): 2,  (1, 2, 3): 1, 
  (2, 3, 4): 1,  (3, 4, 5): 1,  (6, 7, 8): 1, 
  (7, 8, 9): 1,  (8, 9, 10): 1, (9, 10, 4): 1, 
  (10, 4, 5): 1, (2, 4, 5): 1,  (5, 6, 9): 1, 
  (6, 9, 1): 1,  (9, 1, 4): 1,  (1, 4, 5): 1, 
  (5, 6, 4): 1,  (6, 4, 5): 1,  (1, 2, 9): 1,
  (2, 9, 4): 1,  (9, 4, 5): 1,  (6, 7, 7): 1, 
  (7, 7, 8): 1,  (7, 8, 5): 1 }

If you only need that single series' count you can use the list comprehension directly:

len([s for lst in lists for s in zip(lst,lst[1:],lst[2:]) if s==(4,5,6)])

Upvotes: 1

Mohamed Sherif
Mohamed Sherif

Reputation: 11

best way to maintain a mapping of elements in an array to a value is a hash table. as @JacobIRR mentioned {[4, 5, 6] : 6} is not valid because you cant have multiple keys associated to a value. i cant give you an exact solution to your problem as you didnt mention how your function works. if your algorithm dosent depend on a list as a key,you can try {6 :[4, 5, 6]} will be a valid solution as HashMap doesn't allow duplicate keys but allows duplicate values.

Upvotes: 1

Related Questions