[英]Comparing a tuple of a pair to a list of tuple pairs
So I am working on this programming homework and I am currently stuck on comparing the pairs of values in one tuple to the tuple pairs in a list. 因此,我正在从事此编程作业,目前我坚持将一个元组中的值对与列表中的元组对进行比较。
The pairs are basically x and y coordinates and I need to find the closest one from the list to the tuple pair. 这些对基本上是x和y坐标,我需要从列表中找到最接近元组对的一个。 As an example, given the point
(-4, 3)
and the list [(10, 6), (1, 7), (6, 3), (1, 9)]
, the closest one would be (1, 7)
. 例如,给定点
(-4, 3)
和列表[(10, 6), (1, 7), (6, 3), (1, 9)]
,最接近的将是(1, 7)
。
These numbers always changes with a randoming part of the programming but the the above is defined as a function. 这些数字总是随着编程的随机化部分而变化,但是以上被定义为函数。 Here is the whole thing:
这是整个事情:
def nearest(point, more_points):
'''
Finds the nearest point to the base point
'''
(x, y) = point
for i, j in more_points:
a = math.fabs(x - i)
tempI = i
b = math.fabs(y - j)
tempJ = j
tempA, tempB = a , b
if min(tempA) < a:
point = ()
my_points = []
c = 0
lpoint = list(point)
while c < 2:
lpoint.append(random.randrange(-5,5,1)) # generate the "point"
c += 1
tpoint = tuple(lpoint)
c = 0
colx = [] # x points
coly = [] # y points
# generate the points
while c < 4:
colx.append(random.randint(0,10))
coly.append(random.randint(0,10))
c += 1
my_point = list(zip(colx,coly))
print(my_point)
the_nearest = nearest(tpoint,my_point)
print(the_nearest)
What I am trying to do is to take the x,y in the point and then take the "other" point and get the difference and then use that to find the "nearest" but I lost and I am stuck. 我想做的就是先取x,y,然后取“其他”点并求出差值,然后用它来找到“最近”,但我输了,被困住了。 The focus is on the user defined function.
重点是用户定义的功能。
Assuming the following function calculate the distance within 2 points: 假设以下函数计算2点内的距离:
def distance(point_a, point_b):
"""Returns the distance between two points."""
x0, y0 = point_a
x1, y1 = point_b
return math.fabs(x0 - x1) + math.fabs(y0 - y1)
You can iterate over all points and find the minimum distance: 您可以遍历所有点并找到最小距离:
def nearest(point, all_points):
closest_point, best_distance = None, float("inf")
for other_point in all_points:
d = distance(point, other_point)
if d < best_distance:
closest_point, best_distance = other_point, d
return closest_point
Although I could come up with a more pythonic approach: 尽管我可以提出一种更Python化的方法:
def nearest(point, all_points):
"""Returns the closest point in all_points from the first parameter."""
distance_from_point = functools.partial(distance, point)
return min(all_points, key=distance_from_point)
The overall idea of the above solution is to build a partial function. 上述解决方案的总体思路是构建部分功能。 This partial function takes a single parameter and returns the distance to the point given as parameter.
此部分函数采用单个参数,并将距离返回到作为参数给定的点。 This is could be re-written as
lambda other_point: distance(point, other_point)
but this is prettier. 可以将其重写为
lambda other_point: distance(point, other_point)
但这更漂亮。
Note the above function will raise ValueError
if you call it with an empty list of points: nearest(point, [])
. 请注意,如果您使用一个空的点列表进行调用,上述函数将引发
ValueError
: nearest(point, [])
。 You can add an if for this case if necessary. 如果需要,您可以为此添加一个if。
Use min()
with a key
function: 将
min()
与key
函数一起使用:
#!/usr/bin/env python3
import math
from functools import partial
def distance_between_points(a, b):
ax, ay = a
bx, by = b
return math.sqrt(pow(ax - bx, 2) + pow(ay - by, 2))
def nearest(point, more_points):
'''
Finds the nearest point to the base point
'''
distance_to_point = partial(distance_between_points, point)
return min(more_points, key=distance_to_point)
point = (-4, 3)
my_points = [(10, 6), (1, 7), (6, 3), (1, 9)]
n = nearest(point, my_points)
print(n)
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