在列表中查找具有等于某个值的属性（满足任何条件）的对象 [英] Find object in list that has attribute equal to some value (that meets any condition)

问题描述

我有对象列表。我想在此列表中找到一个具有等于值的属性（或方法结果 - 任何）的一个（第一个或任何）对象。

找到最好的方法是什么？

这里是测试用例：

  class Test：
 def __init __（self，value）：
 self.value = value 
 
 import随机
 
值= 5 
 
 test_list = [范围（1000）中的x的测试（random.randint（0,100））] 
 
＃我会在Pascal做的，我不相信在test_list中的Pythonic
附近没有任何地方：
如果x.value == value：
 print我发现了！  
 break 
  

我认为使用生成器和 reduce（） code>将不会有任何区别，因为它仍然是通过列表迭代。

ps .:等式为 value 只是一个例子。当然，我们希望得到满足任何条件的元素。

解决方案

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

与条件匹配的列表，如果没有匹配项则返回 None 。这是我首选的单一表达形式。<​​/ p>

但是，

  for x in test_list：
如果x.value ==值：
打印我找到了！ 
 break 
  

天真的循环版本，完美的是Pythonic - 简洁明了，高效。要使匹配一行的行为：

  for x in test_list：
 if x.value = = value：
 print我发现了！ 
 break 
 else：
x =无
  

code>无到 x 如果你不 break 循环。

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 isn't 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.

解决方案

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.

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