Find object in list that has attribute equal to some value (that meets any condition)

Question

I've got a list of objects. I want to find one (first or whatever) object in this list that has an attribute (or method result - whatever) equal to value.

What's the best way to find it?

Here's a test case:

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

import random

value = 5

test_list = [Test(random.randint(0,100)) for x in range(1000)]

# that I would do in Pascal, I don't believe it's anywhere near 'Pythonic'
for x in test_list:
    if x.value == value:
        print "i found it!"
        break

I think using generators and reduce() won't make any difference because it still would be iterating through the list.

ps.: Equation to value is just an example. Of course, we want to get an element that meets any condition.

Answer 1

next((x for x in test_list if x.value == value), None)

This gets the first item from the list that matches the condition, and returns None if no item matches. It's my preferred single-expression form.

However,

for x in test_list:
    if x.value == value:
        print("i found it!")
        break

The naive loop-break version, is perfectly Pythonic -- it's concise, clear, and efficient. To make it match the behavior of the one-liner:

for x in test_list:
    if x.value == value:
        print("i found it!")
        break
else:
    x = None

This will assign None to x if you don't break out of the loop.

Answer 2

Since it has not been mentioned just for completion. The good ol' filter to filter your to be filtered elements.

Functional programming ftw.

####### Set Up #######
class X:

    def __init__(self, val):
        self.val = val

elem = 5

my_unfiltered_list = [X(1), X(2), X(3), X(4), X(5), X(5), X(6)]

####### Set Up #######

### Filter one liner ### filter(lambda x: condition(x), some_list)
my_filter_iter = filter(lambda x: x.val == elem, my_unfiltered_list)
### Returns a flippin' iterator at least in Python 3.5 and that's what I'm on

print(next(my_filter_iter).val)
print(next(my_filter_iter).val)
print(next(my_filter_iter).val)

### [1, 2, 3, 4, 5, 5, 6] Will Return: ###
# 5
# 5
# Traceback (most recent call last):
#   File "C:\Users\mousavin\workspace\Scripts\test.py", line 22, in <module>
#     print(next(my_filter_iter).value)
# StopIteration


# You can do that None stuff or whatever at this point, if you don't like exceptions.

I know that generally in python list comprehensions are preferred or at least that is what I read, but I don't see the issue to be honest. Of course Python is not an FP language, but Map / Reduce / Filter are perfectly readable and are the most standard of standard use cases in functional programming.

So there you go. Know thy functional programming.

filter condition list

It won't get any easier than this:

next(filter(lambda x: x.val == value,  my_unfiltered_list)) # Optionally: next(..., None) or some other default value to prevent Exceptions

Answer 3

A simple example: We have the following array

li = [{"id":1,"name":"ronaldo"},{"id":2,"name":"messi"}]

Now, we want to find the object in the array that has id equal to 1

1. Use method next with list comprehension

next(x for x in li if x["id"] == 1 )

2. Use list comprehension and return first item

[x for x in li if x["id"] == 1 ][0]

3. Custom Function

def find(arr , id):
    for x in arr:
        if x["id"] == id:
            return x
find(li , 1)

Output all the above methods is {'id': 1, 'name': 'ronaldo'}

Answer 4

You could do something like this

dict = [{
   "id": 1,
   "name": "Doom Hammer"
 },
 {
    "id": 2,
    "name": "Rings ov Saturn"
 }
]

for x in dict:
  if x["id"] == 2:
    print(x["name"])

Thats what i use to find the objects in a long array of objects.

Answer 5

Old question but I use this quite frequently (for version 3.8). It's a bit of syntactic salt, but it has the advantage over the top answer in that you could retrieve a list of results (if there are multiple) by simply removing the [0] and it still defaults to None if nothing is found. For any other condition, simply change the x.value==value to what ever you're looking for.

_[0] if (_:=[x for x in test_list if x.value==value]) else None

Answer 6

You could also implement rich comparison via __eq__ method for your Test class and use in operator. Not sure if this is the best stand-alone way, but in case if you need to compare Test instances based on value somewhere else, this could be useful.

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

    def __eq__(self, other):
        """To implement 'in' operator"""
        # Comparing with int (assuming "value" is int)
        if isinstance(other, int):
            return self.value == other
        # Comparing with another Test object
        elif isinstance(other, Test):
            return self.value == other.value

import random

value = 5

test_list = [Test(random.randint(0,100)) for x in range(1000)]

if value in test_list:
    print "i found it"

Answer 7

I just ran into a similar problem and devised a small optimization for the case where no object in the list meets the requirement.(for my use-case this resulted in major performance improvement):

Along with the list test_list, I keep an additional set test_value_set which consists of values of the list that I need to filter on. So here the else part of agf's solution becomes very-fast.