Python: Retrieve Tuples From Set Based on First Value of Tuple

Suppose I have a set, s that looks like this:

s = set([(1,2), (1,4), (2,6)])

I want to retrieve all tuples in my set that have first element 1. Usually I'd have to give a full tuple, something like:

(1,2) in s

In this case, I want to retrieve all tuples of the form (1,_) where _ can be any number.

Any thoughts on how to do this?

Edit: To be clear, I wanted to do this using a set because I want O(1) time. I understand I could just iterate through a list and collect the tuples that have their first element as 1, but am looking for something faster.

I thought of using a separate set for the first elements and a second set for the second elements but not sure how I'd use that to retrieve efficiently.

Thanks!

Asked by: Alford359 | Posted: 27-01-2022

Answer 1

Using list comprehension.

s = set([(1,2), (1,4), (2,6)])
print(set([i for i in s if i[0] == 1]))    #Check if first value in tuple is 1

Output:

set([(1, 2), (1, 4)])

Answered by: Joyce387 | Posted: 28-02-2022

Answer 2

Instead of s = set([(1,2), (1,4), (2,6)]) use d = {(1,):[(2,),(4,)], ((2,):[(6,)]} Then you can generate list of tuples starting with 1 in worst case O(k) where k is max number of tuples starting with one particular value (which is hopefully much better than O(n)).

To do the lookup:

[(1,) + x for x in d[(1,)]]

Answered by: Kirsten192 | Posted: 28-02-2022

Answer 3

A simple list comprehension:

[t for t in s if t[0] == 1]

Result:

>>> s = set([(1,2), (1,4), (2,6)])
>>> [t for t in s if t[0] == 1]
[(1, 2), (1, 4)]
>>>

Answered by: Thomas750 | Posted: 28-02-2022

Answer 4

Why not store your tuples in a numpy array?

import numpy as np

s = np.array([[1,2], [1,4], [2,6]])

s[np.where(s[:,0] == 1)]

gives you

array([[1,2], [1,4]])

Answered by: Stuart879 | Posted: 28-02-2022

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