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

Question

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

What's is the best way to find it?

Here's 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 list.

ps.: Equation to value is just an example. Of course we want to get element which 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.