从对象列表中获取属性的最大值

Question

I have this list of objects which have ax and ay parameter (and some other stuff).

path.nodes = (
    <GSNode x=535.0 y=0.0 GSLINE GSSHARP>,
    <GSNode x=634.0 y=0.0 GSLINE GSSHARP>,
    <GSNode x=377.0 y=706.0 GSLINE GSSHARP>,
    <GSNode x=279.0 y=706.0 GSLINE GSSHARP>,
    <GSNode x=10.0 y=0.0 GSLINE GSSHARP>,
    <GSNode x=110.0 y=0.0 GSLINE GSSHARP>,
    <GSNode x=189.0 y=216.0 GSLINE GSSHARP>,
    <GSNode x=458.0 y=216.0 GSLINE GSSHARP>
)

I need to have the max y of this list. Though, I tried this:

print(max(path.nodes, key=y))

And I get this error:

NameError: name 'y' is not defined

I am kinda new to Python and the docs give me no clue. I think I am doing wrong with the keyword because if iterate through nodes like this:

for node in path.nodes:
    print(node.y)

I'll get the values of y. Could somebody provide me an explanation?

Answer 1

要获得最大值而不是整个对象，您可以使用生成器表达式：

print max(node.y for node in path.nodes)

Answer 2

There's a built-in to help with this case.

import operator

print max(path.nodes, key=operator.attrgetter('y'))

Alternatively:

print max(path.nodes, key=lambda item: item.y)

Edit: But Mark Byers' answer is most Pythonic.

print max(node.y for node in path.nodes)

Answer 3

There is an important difference for when to use the "Pythonic" style#1 versus lambda style#2:

max(node.y for node in path.nodes)    (style #1)

versus

max(path.nodes, key=lambda item: item.y)   (style #2)

If you look carefully you can see that style#1 returns the maximum value for the attribute "y" while style#2 returns the "node" that has maximum attribute "y". These two are not the same and code usage is important in case you want to iterate over the attribute values or iterate over the objects that holds that attribute.

Example:

class node():
    def __init__(self,x):
        self.x = x
        self.y = self.x + 10

node_lst = [node(1),node(2),node(3),node(4), node(5)]
print ([(e.x,e.y) for e in node_lst])

>>> [(1, 11), (2, 12), (3, 13), (4, 14), (5, 15)]

Now:

maxy = max(node.y for node in node_lst)
print(maxy)
>>> 15

max_node = max(node_lst, key=lambda node: node.y)
print(max_node.y)
>>> 15

Answer 4

from operator import attrgetter
print max(path.nodes, key=attrgetter("y"))

Answer 5

It's also possible to implement the __gt__ comparison operator for an object, and than use max without a key function:

class node:
    def __init__(self, y):
        self.y = y
    def __gt__(self, other):
        return self.y > other.y

and than something like:

ls = [node(3), node(5), node(11), node(0)]
print(max(ls).y)

is supposed to output 11 .

Answer 6

y isn't defined as a variable; it's an attribute of individual GSNode objects; you can't use it as a name on its own.

To access the individual attributes you can use something like key=lambda x: xy or attrgetter() from the operator module.

Answer 7

If y is a property attribute then you don't even need to import operator.attrgetter . You can use fget method instead:

my_node = max(path.nodes, key=Node.y.fget)

This will return the Node instance from where to get the max y value is just my_node.y

Answer 8

they already answered you, but if you want to get the object who has the max value:

max_val_object = lambda x: max(ob.value for ob in x)