I've a list of point with coordinates and another point.
我有一个带坐标和另一个点的点列表。
A sample from the list :
列表中的示例:
(45.1531912,5.7184742),(45.1531912,5.7184742),(45.1531113,5.7184544),(45.1525337,5.718298),(45.1525337,5.718298),
A point :
一个点 :
(45.1533837,5.7185242)
A function :
功能:
def dist(point1,point2)
....
return aDistance
Is there a python one liner (list-comprehension ?) to find a point in the list where a given function returns the minimal value for the list ?
是否有一个python one liner(list-comprehension?)来查找列表中给定函数返回列表最小值的点?
1 个解决方案
#1
8
The min()
function takes a key
argument already, no need for a list comprehension.
min()函数已经获取了一个关键参数,不需要列表推导。
Let's say you wanted to find the closest point in the list to the origin:
假设您想要找到列表中距离原点最近的点:
min(list_of_points, key=lambda p: distance(p, (0, 0)))
would find it (given a distance()
function that calculates the distance between two points).
会找到它(给定一个计算两点之间距离的距离()函数)。
From the documentation:
从文档:
The optional key argument specifies a one-argument ordering function like that used for
list.sort()
. The key argument, if supplied, must be in keyword form (for example,min(a,b,c,key=func)
).可选的key参数指定一个单参数排序函数,就像list.sort()一样。键参数(如果提供)必须采用关键字形式(例如,min(a,b,c,key = func))。
#1
8
The min()
function takes a key
argument already, no need for a list comprehension.
min()函数已经获取了一个关键参数,不需要列表推导。
Let's say you wanted to find the closest point in the list to the origin:
假设您想要找到列表中距离原点最近的点:
min(list_of_points, key=lambda p: distance(p, (0, 0)))
would find it (given a distance()
function that calculates the distance between two points).
会找到它(给定一个计算两点之间距离的距离()函数)。
From the documentation:
从文档:
The optional key argument specifies a one-argument ordering function like that used for
list.sort()
. The key argument, if supplied, must be in keyword form (for example,min(a,b,c,key=func)
).可选的key参数指定一个单参数排序函数,就像list.sort()一样。键参数(如果提供)必须采用关键字形式(例如,min(a,b,c,key = func))。