基本python程序

Question

我试图完成该程序，但答案是错误的，我无法精确确定究竟是什么。

问题：给定两条直线（y = mx + b）的方程，确定两条直线是平行的，相同的还是相交的。计算并输出交点。

我的代码：

equation_1 =raw_input("Please enter the equation of your 1st line(y=mx+b): ") 
equation_2 =raw_input("Please enter the equation of your 2nd line(y=mx+b): ") 

plus_1 = equation_1.find('+') 
plus_2 = equation_2.find('+') 

x_1 = equation_1.find('x') 
x_2 = equation_2.find('x') 

equalsign_1 = equation_1.find('=') 
equalsign_2 = equation_2.find('=') 

b1 = equation_1[x_1+1:] 
b2 = equation_2[x_2+1:] 

m1 = equation_1[equalsign_1+1:x_1] 
m2 = equation_2[equalsign_2+1:x_2] 

if m1==m2 and b1!=b2: 
    print "Your equations are parallel. " 

elif m1==m2 and b1==b2: 
    print "Your equations are the same. " 

else: 
    equation_intersect_y = float(b2)-float(b1) 
    equation_intersect_x = float(m2)-float(m1) # equation_intersect_x = float(m1)-float(m2) 

    poi_x = float(equation_intersect_y)/float(equation_intersect_x) 
    poi_y = float(b1)*float(poi_x)+float(m1)` 

Answer 1

时使用的计算poi_x是错误的等式。此外，您的代码用于计算poi_y的公式已互换m和b。这里有一个稍微修改代码，应该帮助：

#! /usr/bin/env python 
equation_1 ="y=2x+3" 
equation_2 ="y=-0.5x+7" 

plus_1 = equation_1.find('+') 
plus_2 = equation_2.find('+') 

x_1 = equation_1.find('x') 
x_2 = equation_2.find('x') 

equalsign_1 = equation_1.find('=') 
equalsign_2 = equation_2.find('=') 

b1 = float(equation_1[x_1+1:]) 
b2 = float(equation_2[x_2+1:]) 

m1 = float(equation_1[equalsign_1+1:x_1]) 
m2 = float(equation_2[equalsign_2+1:x_2]) 

print m1,b1,m2,b2 

if m1==m2 and b1!=b2: 
    print "Your equations are parallel. " 

elif m1==m2 and b1==b2: 
    print "Your equations are the same. " 

else: 
    equation_intersect_y = b2 - b1 
    equation_intersect_x = m1 - m2 

    poi_x = equation_intersect_y/equation_intersect_x 
    poi_y = m1*poi_x+b1 

    print poi_x, poi_y 

输出是：

2.0 3.0 -0.5 7.0 
1.6 6.2 

这里是一个稍微好一点的代码，减少重复：

#! /usr/bin/env python 
def parse_equation_string(eq_string): 
    x_pos = eq_string.find('x') 
    equal_pos = eq_string.find('=') 

    b = float(eq_string[x_pos+1:]) 
    m = float(eq_string[equal_pos+1:x_pos]) 
    return m, b 

def get_point_of_intersection(line1, line2): 
    m1, b1 = line1 
    m2, b2 = line2 

    if m1==m2 and b1!=b2: 
     return "The lines are parallel. " 

    elif m1==m2 and b1==b2: 
     return "The lines are the same. " 

    else: 
     equation_intersect_y = b2 - b1 
     equation_intersect_x = m1 - m2 

     poi_x = equation_intersect_y/equation_intersect_x 
     poi_y = m1*poi_x+b1 

     return poi_x, poi_y 

equation_1 = "y=2x+3" 
equation_2 = "y=-0.5x+7" 

line_1 = parse_equation_string(equation_1) 
line_2 = parse_equation_string(equation_2) 

print line_1, line_2 
print get_point_of_intersection(line_1, line_2) 

输出是：

(2.0, 3.0) (-0.5, 7.0) 
(1.6, 6.2) 

Answer 2

不应该

b1 = equation_1[x_1+1:] 
b2 = equation_2[x_2+1:] 

要

b1 = equation_1[plus_1+1:] 
b2 = equation_2[plus_2+1:] 

或者

b1 = equation_1[x_1+2:] 
b2 = equation_2[x_2+2:] 

此外，我认为

m1 = equation_1[equalsign_1+1:x_1] 
m2 = equation_2[equalsign_2+1:x_2] 

应该

m1 = equation_1[equalsign_1+1:x_1-1] 
m2 = equation_2[equalsign_2+1:x_2-1] 

Answer 3

第一对的建议：

打印back（之前的第一个“if”语句）执行操作之前的方程给用户：

print "Equation 1 y={}x+{}".format(m1, b1) 
print "Equation 2 y={}x+{}".format(m2, b2) 

重或字符串函数分裂和条可以比字符串索引更容易：

m1 = equation_1.split('=')[1].split('x')[0] 
b1 = equation_1.split('=')[1].split('+')[1] 

给程序之前较硬的某些简单的测试情况： 1：Y = 0X + -3 2：Y = 1X + 0 相交于X = 3

1: y=-1x+0 
2: y=0x+3 
intersects at X = -3 


1: y=2x+2 
2: y=-2x+0 
intersects at X = -0.5 

现在，所有剩下的就是代数：

不要用手第一硬盒：

假设他们不平行或相同： 发现的地步，X1 = X2和Y1 = （m1）* x1 + b1 = y1 = y2 =（m2）* x2 + b2 找到X，其中两个Y都相等：因此：
但在感兴趣点（X）x1 = x2 重写：（m1 + m2）* X = b2 -b1 重写：X =（b2 - b1）/（m1 + m2）

现在我们可以看到这与您的equation_intersect x公式不匹配。