Check if list of objects contain an object with a certain attribute value

Question

I want to check if my list of objects contain an object with a certain attribute value.

class Test:
    def __init__(self, name):
        self.name = name

# in main()
l = []
l.append(Test("t1"))
l.append(Test("t2"))
l.append(Test("t2"))

I want a way of checking if list contains an object with name "t1" for example. How can it be done? I found https://stackoverflow.com/a/598415/292291,

[x for x in myList if x.n == 30]               # list of all matches
any(x.n == 30 for x in myList)                 # if there is any matches
[i for i,x in enumerate(myList) if x.n == 30]  # indices of all matches

def first(iterable, default=None):
    for item in iterable:
        return item
    return default

first(x for x in myList if x.n == 30)          # the first match, if any

I don’t want to go through the whole list every time; I just need to know if there’s 1 instance which matches. Will first(...) or any(...) or something else do that?

Asked By: Jiew Meng

||

Source

Answer 1

As you can easily see from the documentation, the any() function short-circuits an returns True as soon as a match has been found.

any(x.name == "t2" for x in l)

Answered By: Sven Marnach

Answer 2

Another built-in function next() can be used for this job. It stops at the first instance where the condition is True much like any().

next((True for x in l if x.name=='t2'), False)

Also, next() can return the object itself where the condition is True (so behaves like first() function in the OP).

next((x for x in l if x.name == 't2'), None)

Answered By: cottontail

Check if list of objects contain an object with a certain attribute value

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