Matplotlib，python，如何求一条直线与x轴相交的点

Question

我一直在研究一些代码来求解二次方程并将其绘制在图形上，我想在直线与 x 轴的交点处得到一个点，并显示点的坐标在他们下面。

代码现在是这样的：

这就是我想要它做的：

代码如下：

import matplotlib

matplotlib.use('TkAgg')
import matplotlib.pyplot as plt
import numpy as np
import math


def int_input(prompt):
    while True:
        try:
            variable_name = int(input(prompt))
            return variable_name
        except ValueError:
            print("Please enter a whole number (in digits)")


def float_input(prompt):
    while True:
        try:
            variable_name = float(input(prompt))
            return variable_name
        except ValueError:
            print("Please enter a numeric value (in digits)")


def yes_input(prompt):
    while True:
        variable_name = input(prompt).lower()
        if variable_name in ["y", "yes"]:
            return "yes"
        elif variable_name in ["n", "no"]:
            return "no"
        else:
            print("""Please enter either "y" or "n". """)


print("Quadratic Solver")

while True:

    print("Input below the values of a, b and c from an equation that is of the form : ax² + bx + c = 0")
    a = float_input('a: ')

    b = float_input('b: ')

    c = float_input('c: ')

    # calculate the discriminant
    d = (b ** 2) - (4 * a * c)

    # find two solutions
    try:
        solution1 = (-b - math.sqrt(d)) / (2 * a)
    except ValueError:
        solution1 = "none"

    try:
        solution2 = (-b + math.sqrt(d)) / (2 * a)
    except ValueError:
        solution2 = "none"

    if solution1 == solution2:
        if solution1 == "none":
            print("There are no solutions.")
        else:
            print("There is one solution: x = {0}".format(solution1))

    elif solution1 != solution2:
        print('There are two solutions, x can be either {0} or {1}'.format(solution1, solution2))

    graph = yes_input("Do you want to plot that on a graph? (y/n):  ")
    if graph == "yes":
        x = np.linspace(-6, 6, 50)

        fig = plt.figure(figsize=(7, 4))

        y4 = a * x ** 2 + b * x + c
        plt.plot(x, y4, 'g:', label='Degree 2')

        # Add features to our figure
        plt.legend()
        plt.grid(True, linestyle=':')
        plt.xlim([-6, 6])
        plt.ylim([-4, 4])

        plt.title(f'A graph of {a}x² + {b}x + {c} = 0')
        plt.xlabel('x-axis')
        plt.ylabel('y-axis')
        plt.get_current_fig_manager().window.wm_geometry("+0+0")
        fig.canvas.manager.window.attributes('-topmost', 1)
        # Show plot
        plt.show()
    new_question = yes_input('New question? (y/n): ')
    if new_question == "no":
        break

我正在使用 windows OS 并正在使用 IntelliJ IDEA 编辑我的 python 代码。

Answer 1

要在二次方程的图形中添加点，您可以使用带有圆圈标记的 plt.plot 函数，并使用 plt.text 打印点的坐标。

根据您的代码查看我的解决方案。它不是最干净的，但很容易理解（第 86-96 行）。

import matplotlib

matplotlib.use('TkAgg')
import matplotlib.pyplot as plt
import numpy as np
import math

def int_input(prompt):
    while True:
        try:
            variable_name = int(input(prompt))
            return variable_name
        except ValueError:
            print("Please enter a whole number (in digits)")


def float_input(prompt):
    while True:
        try:
            variable_name = float(input(prompt))
            return variable_name
        except ValueError:
            print("Please enter a numeric value (in digits)")


def yes_input(prompt):
    while True:
        variable_name = input(prompt).lower()
        if variable_name in ["y", "yes"]:
            return "yes"
        elif variable_name in ["n", "no"]:
            return "no"
        else:
            print("""Please enter either "y" or "n". """)


print("Quadratic Solver")

while True:

    print("Input below the values of a, b and c from an equation that is of the form : ax² + bx + c = 0")
    a = float_input('a: ')

    b = float_input('b: ')

    c = float_input('c: ')

    # calculate the discriminant
    d = (b ** 2) - (4 * a * c)

    # find two solutions
    try:
        solution1 = (-b - math.sqrt(d)) / (2 * a)
    except ValueError:
        solution1 = "none"

    try:
        solution2 = (-b + math.sqrt(d)) / (2 * a)
    except ValueError:
        solution2 = "none"

    if solution1 == solution2:
        if solution1 == "none":
            print("There are no solutions.")
        else:
            print("There is one solution: x = {0}".format(solution1))

    elif solution1 != solution2:
        print('There are two solutions, x can be either {0} or {1}'.format(solution1, solution2))

    graph = yes_input("Do you want to plot that on a graph? (y/n):  ")
    if graph == "yes":
        x = np.linspace(-6, 6, 50)

        fig = plt.figure(figsize=(7, 4))

        y4 = a * x ** 2 + b * x + c
        plt.plot(x, y4, 'g:', label='Degree 2')

        # Add features to our figure
        plt.legend()
        plt.grid(True, linestyle=':')
        plt.xlim([-6, 6])
        plt.ylim([-4, 4])

        # Add dots
        offset = 0.3

        if solution1 == solution2:
            if solution1 != "none":
                plt.plot(solution1, 0 , marker="o", markersize=10, color="black", linestyle="None")
                plt.text(solution1, 0-offset, f'({solution1}, {0})', ha='center', va='center')
        elif solution1 != solution2:
            plt.plot([solution1, solution2], [0, 0], marker="o", markersize=10, color="black", linestyle="None")
            plt.text(solution1, 0-offset, f'({solution1}, {0})', ha='center', va='center')
            plt.text(solution2, 0-offset, f'({solution2}, {0})', ha='center', va='center')


        plt.title(f'A graph of {a}x² + {b}x + {c} = 0')
        plt.xlabel('x-axis')
        plt.ylabel('y-axis')
        plt.get_current_fig_manager().window.wm_geometry("+0+0")
        fig.canvas.manager.window.attributes('-topmost', 1)
        # Show plot
        plt.show()
    new_question = yes_input('New question? (y/n): ')
    if new_question == "no":
        break

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