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尺寸，但形状为（10，）和（1，）

第32行grap_plot，= mplt.plot（x，F，'r'）

摘要

我有一个符号函数（“ U”），U依赖于x和t（它们是符号变量），然后我使用lambdify从sympy切换为numpy，因为我想使用依赖于t的滑块来绘制U，但是我不能

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()

我已经尝试了很多，但是无法解决，请提供一些帮助，谢谢。

Answer 1

sympy documentation表明plt.plot(x, F(x), 'r', label = 'Fixed function')的返回是一个函数。当您针对范围绘制此内容时，由于期望两个相等长度的数组类型，因此matplotlib会感到困惑。然后，您必须使用Lambda生成点。

{{1}}

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，但是还有另一个错误，当我运行代码时一切都很好，但是当我使用滑块时，出现了Type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()

如何解决此错误？ ValueError和TypeError。滑杆

3 个答案: