Searching a list of objects in Python

Question

Let\'s assume I\'m creating a simple class to work similar to a C-style struct, to just hold data elements. I\'m trying to figure out how to search a list of objects for ob

Answer 1

You should add a __eq__ and a __hash__ method to your Data class, it could check if the __dict__ attributes are equal (same properties) and then if their values are equal, too.

If you did that, you can use

test = Data()
test.n = 5

found = test in myList

The in keyword checks if test is in myList.

If you only want to a a n property in Data you could use:

class Data(object):
    __slots__ = ['n']
    def __init__(self, n):
        self.n = n
    def __eq__(self, other):
        if not isinstance(other, Data):
            return False
        if self.n != other.n:
            return False
        return True
    def __hash__(self):
        return self.n

    myList = [ Data(1), Data(2), Data(3) ]
    Data(2) in myList  #==> True
    Data(5) in myList  #==> False

Answer 2

Simple, Elegant, and Powerful:

A generator expression in conjuction with a builtin… (python 2.5+)

any(x for x in mylist if x.n == 10)

Uses the Python any() builtin, which is defined as follows:

any(iterable) -> Return True if any element of the iterable is true. Equivalent to:

def any(iterable):
    for element in iterable:
        if element:
            return True
    return False

Answer 3

Just for completeness, let's not forget the Simplest Thing That Could Possibly Work:

for i in list:
  if i.n == 5:
     # do something with it
     print "YAY! Found one!"

Answer 4

Consider using a dictionary:

myDict = {}

for i in range(20):
    myDict[i] = i * i

print(5 in myDict)

Answer 5

Another way you could do it is using the next() function.

matched_obj = next(x for x in list if x.n == 10)

Answer 6

You can get a list of all matching elements with a list comprehension:

[x for x in myList if x.n == 30]  # list of all elements with .n==30

If you simply want to determine if the list contains any element that matches and do it (relatively) efficiently, you can do

def contains(list, filter):
    for x in list:
        if filter(x):
            return True
    return False

if contains(myList, lambda x: x.n == 3)  # True if any element has .n==3
    # do stuff