我该如何解决这个错误？值错误和类型错误。滑块

Question

问题：

TypeError：float() 参数必须是字符串或数字，而不是“函数”

ValueError：x 和 y 必须具有相同的第一个维度，但形状为 (10,) 和 (1,)

第 32 行 grap_plot, = mplt.plot(x,F,'r')

总结

我有一个 sympy 函数（“U”），U 取决于 x 和 t，它们是符号变量，然后我使用 lambdify 从 sympy 切换到 numpy，因为我想使用取决于 t 的滑块绘制 U，但我不能

from sympy import *
import numpy as np
from sympy.plotting import plot
import matplotlib.pyplot as mplt
from matplotlib.widgets import Slider
init_printing()
( x, t, n) = symbols('x t n' , positive = True)
Ro= Rational(10,1)**-3 #[Kg/m]
T= Rational(49) #[N]
L = Rational(1) #[m]
c = sqrt(T/Ro)
f = x*(x-1)**2
b = simplify((2/L)*integrate(f*sin(n*pi*x/L),(x,0,L)))
U = expand(summation(b*sin(n*pi*x/L)*cos(c*n*pi*t/L),(n,1,10)))
U0 = simplify(U.subs({t:0}))
mplt.style.use('seaborn-whitegrid')
p = plot(f,(x,0,L))
F = (lambdify(['x','t'], U ,'numpy'))
F0 = (lambdify('x', U0,'numpy'))
x = np.arange(0, L, 0.1)
fig = mplt.figure(figsize=(8,3))
grap_ax = mplt.axes([0.1, 0.2, 0.8, 0.65])
slider_ax = mplt.axes([0.1, 0.05, 0.8, 0.05])
mplt.axes(grap_ax) 
mplt.title('Movimiento de la cuerda')
grap_plot, = mplt.plot(x,F,'r')
mplt.xlim(0, L)
mplt.ylim(-0.15, 0.15)
t_slider = Slider(slider_ax,'t',1, 10,0)
def update(t):
    grap_plot.set_ydata(F) 
    fig.canvas.draw_idle()     
t_slider.on_changed(update)
mplt.show()

我尝试了很多，但我无法修复它，请帮助，谢谢。

Answer 1

sympy documentation 表明lambdify 的返回是一个函数。因此，当您根据范围绘制此图时，matplotlib 会感到困惑，因为它需要两个等长的数组类型。然后您必须使用您的 lambda 生成点。

plt.plot(x, F(x), 'r', label = 'Fixed function')

这将修复这两个错误 - 您将传递两个长度相等且类型相同的数组。

Answer 2

我解决了 2 个错误，但我还有一个错误，当我运行代码时一切正常，但当我使用滑块时，出现类型错误。

from sympy import *
import numpy as np
from sympy.plotting import plot
import matplotlib.pyplot as mplt
from matplotlib.widgets import Slider
init_printing()
( x, t, n) = symbols('x t n' , positive = True)
Ro= Rational(10,1)**-3 #[Kg/m]
T= Rational(49) #[N]
L = Rational(1) #[m]
c = sqrt(T/Ro)
f = x*(x-1)**2
b = simplify((2/L)*integrate(f*sin(n*pi*x/L),(x,0,L)))
U = expand(summation(b*sin(n*pi*x/L)*cos(c*n*pi*t/L),(n,1,10)))
u = b*sin(n*pi*x/L)*cos(c*n*pi*t/L)
freq = nfloat(n*c/(2*L),6)
Peri = nfloat(1/freq,3)
for i in range (1,4):
    init_printing()
    display(u.subs({n:i}))
    display(freq.subs({n:i}))
    display(Peri.subs({n:i}))
mplt.style.use('seaborn-whitegrid')
p = plot(f,(x,0,L))
F = lambdify(['x','t'], U ,'numpy')
x = np.arange(0,1,0.1)
t_init = 0.
fig = mplt.figure(figsize=(8,3))
grap_ax = mplt.axes([0.1, 0.2, 0.8, 0.65])
slider_ax = mplt.axes([0.1, 0.05, 0.8, 0.05])
mplt.axes(grap_ax) 
mplt.title('Movimiento de la cuerda')
grap_plot, = mplt.plot(x,F(x,t_init),'r')
mplt.xlim(0, 1)
mplt.ylim(-0.15, 0.15)
t_slider = Slider(slider_ax,'t',0., 10., valinit=t_init)
def update(t):
    grap_plot.set_ydata(F(x,t_slider)) 
    fig.canvas.draw_idle()     
t_slider.on_changed(update)
mplt.show()

TypeError： 不支持的操作数类型：'float' 和 'Slider'

Answer 3

一切都很好，谢谢

from sympy import *
import numpy as np
from sympy.plotting import plot
import matplotlib.pyplot as mplt
from matplotlib.widgets import Slider
init_printing()
( x, t, n) = symbols('x t n' , positive = True)
Ro= Rational(10,1)**-3 #[Kg/m]
T= Rational(49) #[N]
L = Rational(1) #[m]
c = sqrt(T/Ro)
f = x*(x-1)**2
b = simplify((2/L)*integrate(f*sin(n*pi*x/L),(x,0,L)))
U = expand(summation(b*sin(n*pi*x/L)*cos(c*n*pi*t/L),(n,1,10)))
u = b*sin(n*pi*x/L)*cos(c*n*pi*t/L)
freq = nfloat(n*c/(2*L),6)
Peri = nfloat(1/freq,3)
for i in range (1,4):
    init_printing()
    display(u.subs({n:i}))
    display(freq.subs({n:i}))
    display(Peri.subs({n:i}))
mplt.style.use('seaborn-whitegrid')
p = plot(f,(x,0,L))

F = lambdify(['x','t'], U ,'numpy')
x = np.arange(0,1,0.1)
t_init = 0
fig = mplt.figure(figsize=(8,3))
grap_ax = mplt.axes([0.1, 0.2, 0.8, 0.65])
slider_ax = mplt.axes([0.1, 0.05, 0.8, 0.05])
mplt.axes(grap_ax) 
mplt.title('Movimiento de la cuerda')
grap_plot, = mplt.plot(x,F(x,t_init),'r')
mplt.xlim(0, 1)
mplt.ylim(-0.15, 0.15)
t_slider = Slider(slider_ax,'t',0, 10, valinit=t_init,valfmt="%i")
def update(t):
    grap_plot.set_ydata(F(x,t)) 
    fig.canvas.draw_idle()     
t_slider.on_changed(update)
mplt.show()