Python 返回给定两点的直线方程的方法
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Method to return the equation of a straight line given two points
提问by reggaelizard
I have a class Point, consisting of a point with x and y coordinates, and I have to write a method that computes and returns the equation of a straight line joining a Pointobject and another Pointobject that is passed as an argument (my_point.get_straight_line(my_point2). I know how to calculate that on paper with y-y1= m(x-x1) and I already have a method my_point.slope(my_point2)to compute m, but I can't really wrap my head around how to translate the equation to Python. Here's the entire class:
我有一个类Point，由一个带有 x 和 y 坐标的点组成，我必须编写一个方法来计算并返回连接一个Point对象和另一个Point作为参数传递的对象的直线的方程（my_point.get_straight_line(my_point2)。我知道如何在纸上用 yy 1= m(xx 1)my_point.slope(my_point2)计算m，我已经有了计算的方法，但我无法真正理解如何将方程转换为 Python。这是整个课程：
class Point:
def __init__(self,initx,inity):
self.x = initx
self.y = inity
def getx(self):
return self.x
def gety(self):
return self.y
def negx(self):
return -(self.x)
def negy(self):
return -(self.y)
def __str__(self):
return 'x=' + str(self.x) + ', y=' + str(self.y)
def halfway(self,target):
midx = (self.x + target.x) / 2
midy = (self.y + target.y) / 2
return Point(midx, midy)
def distance(self,target):
xdiff = target.x - self.x
ydiff = target.y - self.y
dist = math.sqrt(xdiff**2 + ydiff**2)
return dist
def reflect_x(self):
return Point(self.negx(),self.y)
def reflect_y(self):
return Point(self.x,self.negy())
def reflect_x_y(self):
return Point(self.negx(),self.negy())
def slope_from_origin(self):
if self.x == 0:
return None
else:
return self.y / self.x
def slope(self,target):
if target.x == self.x:
return None
else:
m = (target.y - self.y) / (target.x - self.x)
return m
Any help is appreciated.
任何帮助表示赞赏。
EDIT:I figured it out with an equation that computes cand then just returns it in a string along with self.slope(target)! This turned out to be way less complicated than I thought.
编辑：我用一个计算公式计算出来c，然后将它与self.slope(target)! 结果证明这比我想象的要简单得多。
def get_line_to(self,target):
c = -(self.slope(target)*self.x - self.y)
return 'y = ' + str(self.slope(target)) + 'x + ' + str(c)
回答by Joran Beasley
from numpy import ones,vstack
from numpy.linalg import lstsq
points = [(1,5),(3,4)]
x_coords, y_coords = zip(*points)
A = vstack([x_coords,ones(len(x_coords))]).T
m, c = lstsq(A, y_coords)[0]
print("Line Solution is y = {m}x + {c}".format(m=m,c=c))
but really your method should be fine ...
但真的你的方法应该没问题......
回答by Hugh Bothwell
I cleaned it up a bit; see what you think.
我清理了一下；看看你的想法。
def slope(dx, dy):
return (dy / dx) if dx else None
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
return '({}, {})'.format(self.x, self.y)
def __repr__(self):
return 'Point({}, {})'.format(self.x, self.y)
def halfway(self, target):
midx = (self.x + target.x) / 2
midy = (self.y + target.y) / 2
return Point(midx, midy)
def distance(self, target):
dx = target.x - self.x
dy = target.y - self.y
return (dx*dx + dy*dy) ** 0.5
def reflect_x(self):
return Point(-self.x,self.y)
def reflect_y(self):
return Point(self.x,-self.y)
def reflect_x_y(self):
return Point(-self.x, -self.y)
def slope_from_origin(self):
return slope(self.x, self.y)
def slope(self, target):
return slope(target.x - self.x, target.y - self.y)
def y_int(self, target): # <= here's the magic
return self.y - self.slope(target)*self.x
def line_equation(self, target):
slope = self.slope(target)
y_int = self.y_int(target)
if y_int < 0:
y_int = -y_int
sign = '-'
else:
sign = '+'
return 'y = {}x {} {}'.format(slope, sign, y_int)
def line_function(self, target):
slope = self.slope(target)
y_int = self.y_int(target)
def fn(x):
return slope*x + y_int
return fn
and here are some use examples:
以下是一些使用示例：
a = Point(2., 2.)
b = Point(4., 3.)
print(a) # => (2.0, 2.0)
print(repr(b)) # => Point(4.0, 3.0)
print(a.halfway(b)) # => (3.0, 2.5)
print(a.slope(b)) # => 0.5
print(a.y_int(b)) # => 1.0
print(a.line_equation(b)) # => y = 0.5x + 1.0
line = a.line_function(b)
print(line(x=6.)) # => 4.0
回答by funk
Let's assume that we have the following points:
让我们假设我们有以下几点：
P0: ( x0 = 100, y0 = 240 )
P1: ( x1 = 400, y1 = 265 )
P0: ( x0 = 100, y0 = 240 )
P1: ( x1 = 400, y1 = 265 )
We can compute the coefficients of the line y = a*x + b that connects the two points using the polyfitmethod from numpy.
我们可以使用numpy 中的polyfit方法计算连接两点的直线 y = a*x + b 的系数。
import numpy as np
import matplotlib.pyplot as plt
# Define the known points
x = [100, 400]
y = [240, 265]
# Calculate the coefficients. This line answers the initial question.
coefficients = np.polyfit(x, y, 1)
# Print the findings
print 'a =', coefficients[0]
print 'b =', coefficients[1]
# Let's compute the values of the line...
polynomial = np.poly1d(coefficients)
x_axis = np.linspace(0,500,100)
y_axis = polynomial(x_axis)
# ...and plot the points and the line
plt.plot(x_axis, y_axis)
plt.plot( x[0], y[0], 'go' )
plt.plot( x[1], y[1], 'go' )
plt.grid('on')
plt.show()
a = 0.0833333333333
a = 0.08333333333333
b = 231.666666667
b = 231.666666667
For installing numpy: http://docs.scipy.org/doc/numpy/user/install.html
回答by ironmann350
I think you are making pretty advanced code, but you are making it complicated. Here is a function that can do that:
我认为您正在编写非常高级的代码，但您使其变得复杂。这是一个可以做到这一点的函数：
from decimal import Decimal
def lin_equ(l1, l2):
"""Line encoded as l=(x,y)."""
m = Decimal((l2[1] - l1[1])) / Decimal(l2[0] - l1[0])
c = (l2[1] - (m * l2[0]))
return m, c
# Example Usage:
lin_equ((-40, 30,), (20, 45))
# Result: (Decimal('0.25'), Decimal('40.00'))
回答by Pavan Shetkar
class Line(object):
def __init__(self,coor1,coor2):
self.coor1 = coor1
self.coor2 = coor2
def distance(self):
x1,y1 = self.coor1
x2,y2 = self.coor2
return ((x2-x1)**2+(y2-y1)**2)**0.5
def slope(self):
x1,x2 = self.coor1
y1,y2 = self.coor2
return (float(y2-y1))/(x2-x1)
