查找給定的線（方程）是否能夠在python中成功地分離兩個點列表

Question

我遇到了一個問題，明確要求我不要使用 numpy 或 pandas

問題 ：

以元組列表的形式給出兩組數據點，例如

Red =[(R11,R12),(R21,R22),(R31,R32),(R41,R42),(R51,R52),..,(Rn1,Rn2)]
Blue=[(B11,B12),(B21,B22),(B31,B32),(B41,B42),(B51,B52),..,(Bm1,Bm2)]

和一組線方程（以字符串格式，即字符串列表）

Lines = [a1x+b1y+c1,a2x+b2y+c2,a3x+b3y+c3,a4x+b4y+c4,..,K lines]

注意：這里需要做字符串解析，得到x,y和intercept的系數。

您的任務是為給定的每一行打印“是”/“否”。 如果所有紅點都在直線的一側，藍點在直線的另一側，則應打印 YES，否則應打印 NO。

Ex:

Red= [(1,1),(2,1),(4,2),(2,4), (-1,4)]
Blue= [(-2,-1),(-1,-2),(-3,-2),(-3,-1),(1,-3)]
Lines=["1x+1y+0","1x-1y+0","1x+0y-3","0x+1y-0.5"]


Output:
YES
NO
NO
YES

在數學上，你取直線方程說 S

Suppose S(x)(y) = 1x+1y+0

Now take points (1,1) and (-6,-1)

S(1)(1) = 1(1)+ 1(1) = 2 >0

S(-6)(-1) = 1(-6)+(1)(-1) = -7 <0

因此，我們可以得出結論，(1,1) 和 (-6,-1) 位於直線 S 的不同側。

現在在給定的問題中，給定一個方程 S，所有紅色都應該在方程的一側，而藍色應該在另一側。

這里的問題是，我不確定如何使用 python 替換以字符串形式給出的等式中列表中點的值。

另外，我沒有為代碼提出一個邏輯（如何根據我們的要求使用循環）來解決上述問題以解決上述問題。

將不勝感激對這個問題的見解。 提前致謝。

Answer 1

假設你的字符串總是形式

ax+by+c

（按這個順序），你可以寫

import re

for line in Lines:
    a, b, c = [float(coef.strip()) for coef in re.split('x|y', line)]
    ...

Answer 2

使用 your_string.replace() 和 eval() 函數，用它的值替換 x 和 y 字符，eval() 將字符串作為方程執行

for element in red:
    equation=line.replace('x','*'+str(element[0]))
    equation=equation.replace('y','*'+str(element [1]))
    result=eval(equation)
    if result > 0:
        pass
    else:
        return "NO"
#code for blue
return "Yes"

Answer 3

Python3程序檢查兩點是否在同一側

def pointsAreOnSameSideOfLine(a, b, c, x1, y1, x2, y2): 
fx1 = 0 # Variable to store a * x1 + b * y1 - c 
fx2 = 0 # Variable to store a * x2 + b * y2 - c 
fx1 = a * x1 + b * y1 - c 
fx2 = a * x2 + b * y2 - c 

# If fx1 and fx2 have same sign 
if ((fx1 * fx2) > 0): 
    return True
return False

驅動程序代碼

a, b, c = 1, 1, 1
x1, y1 = 1, 1
x2, y2 = 2, 1
if (pointsAreOnSameSideOfLine(a, b, c, x1, y1, x2, y2)): 
    print("Yes") 
else: 
    print("No") 

Answer 4

@hemanth ravavarapu，re.split(x|y) 將基於 x 或 yie 分割線方程，如果你給空間，那么它將根據空間分割，這里也將在找到 x 或 y 的地方分割。

Answer 5

e1 = []
e2 = []
def points():
    for k in Lines:
        a=k[0];b=k[2]+k[3];c=k[5]+k[6]
        n = [];m=[]
        for i in Red:
            x1=i[0];y1=i[1]
            eq1 = int(a) * int(x1) + int(b) * int(y1)-int(c)
            n.append(eq1)
        e1.append(n)
        for j in Blue:
            x2=j[0];y2=j[1]
            eq2 = int(a) * int(x2) + int(b) * int(y2)-int(c)
            m.append(eq2)
        e2.append(m)
    print(e1)
    print('----------------------------------------------------------------------')
    print(e2)
    print('----------------------------------------------------------------------')
    p=[]
    for i,j in zip(e1,e2):
        q = []
        for k,l in zip(i,j):
            x=k*l
            if x<0:
                q.append(x)
        p.append(q)
    print(p)
    for i in p:
        if len(i)==5:
            print('yes')
        else:
            print('No')

Red= [(1,1),(2,1),(4,2),(2,4),(-1,4)]
Blue= [(-2,-1),(-1,-2),(-3,-2),(-3,-1),(1,-3)]
Lines=["1x+1y+0","1x-1y+0","1x+0y-3","0x+1y-0.5"]
points()

Answer 6

對於字符串解析：

def getCor(s):

    first = s.split('x')
    a = float(first[0])
    
    second = first[1].split('y')
    b = float(second[0])
    c = float(second[1])
    
    return (a,b,c)

getCor('24x+1y-0.5')

輸出 ：

(24.0, 1.0, -0.5)

Answer 7

import math

def getCor(s):
 
    first = s.split('x')
    a = float(first[0])
    
    second = first[1].split('y')
    b = float(second[0])
    c = float(second[1])
    
    return (a,b,c)


def getSide(t,p):
    
    cal = (t[0] * p[0]) + (t[1] * p[1]) + (t[2])

    if cal > 0:
        return 1
    elif cal < 0:
        return -1
    elif cal == 0:
        return 0
    return -2


def getAns(red,blue,line):
    
    sign_red = getSide(getCor(line),red[0])
    sign_blue = getSide(getCor(line),blue[0])
        
    for j in range(len(red)):
        if sign_red != getSide(getCor(line),red[j]):
            return 'no'

    for j in range(len(blue)):

        if sign_blue != getSide(getCor(line),blue[j]):
            return 'no'
    return 'Yes'


Red= [(1,1),(2,1),(4,2),(2,4), (-1,4)]   
Blue= [(-2,-1),(-1,-2),(-3,-2),(-3,-1),(1,-3)]
Lines=["1x+1y+0","1x-1y+0","1x+0y-3","0x+1y-0.5"]

for i in Lines:
    ans = getAns(Red, Blue, i)
    print(ans)

輸出：

Yes
no
no
Yes

Answer 8

import math
# write your python code here
# you can take the above example as sample input for your program to test
# it should work for any general input try not to hard code for only given input strings


# you can free to change all these codes/structure
def i_am_the_one(red,blue,line):
    # your code
    for i in red:
        eq=line.replace('x','*'+str(i[0]))
        eq=eq.replace('y','*'+str(i[1]))
        answer=eval(eq)
        if answer>0:
            pass
        else:
            return "NO"
        
    # Code for Blue
    for j in blue:
        eq1=line.replace('x','*'+str(j[0]))
        eq1=eq1.replace('y','*'+str(j[1]))
        answer1=eval(eq1)
        if answer1<0:
            pass
        else:
            return "NO"
    return "Yes"

Red= [(1,1),(2,1),(4,2),(2,4), (-1,4)]
Blue= [(-2,-1),(-1,-2),(-3,-2),(-3,-1),(1,-3)]
Lines=["1x+1y+0","1x-1y+0","1x+0y-3","0x+1y-0.5"]

for i in Lines:
    yes_or_no = i_am_the_one(Red, Blue, i)
    print(yes_or_no)