Find tuple structure containing an unknown value inside a list

Question

Say I have list of tuples:

list = [(1,5), (1,7), (2,3)] 

Is there a way in Python to write something like

if (1, *) in list: do things 

where * means "I don’t care about this value"? So we are checking if there is a tuple with 1 at the first position and with whatever value on the second one.

As far as I know there are special mechanisms in other languages, but I just don’t know the name of this particular problem. So is there similar behavior in Python?

P.S.: I know that I can use list comprehensions here. I am just interested in this particular mechanism.

Answer 1

You can use the any() function:

if any(t[0] == 1 for t in yourlist): 

This efficiently tests and exits early if 1 is found in the first position of a tuple.

Answer 2

A placeholder object like you're asking for isn't supported natively, but you can make something like that yourself:

class Any(object):     def __eq__(self, other):         return True ANYTHING = Any()  lst = [(1,5), (1,7), (2,3)] 

The __eq__ method defines how two objects test for equality. (See https://docs.python.org/3/reference/datamodel.html for details.) Here, ANYTHING will always test positive for equality with any object. (Unless that object also overrode __eq__ in a way to return False.)

The in operator merely calls __eq__ for each element in your list. I.e. a in b does something like:

for elem in b:     if elem == a:         return True 

This means that, if you say (1, ANYTHING) in lst, Python will first compare (1, ANYTHING) to the first element in lst. (Tuples, in turn, define __eq__ to return True if all its elements' __eq__ return True. I.e. (x, y) == (a, b) is equivalent to x==a and y==b, or x.__eq__(a) and y.__eq__(b).)

Hence, (1, ANYTHING) in lst will return True, while (3, ANYTHING) in lst will return False.

Also, note that I renamed your list lst instead of list to prevent name clashes with the Python built-in list.