I want to fit some data to obtain the minimum (both x and y), but when I use UnivariateSpline it will generate an error: 'ValueError: x must be increasing if s > 0'
This is my code:
import matplotlib.pyplot as plt
from scipy.interpolate import UnivariateSpline
import numpy as np
import pandas as pd
list_dist=pd.read_csv('C:/.../dist.txt')
dist=list_dist['Distanze']
ar_dist=np.array(dist)
list_energy=pd.read_csv('C:/.../energy.txt')
energy=list_energy['Energie']
ar_energy=np.array(energy)
x = ar_dist
y = ar_energy
print(x)
print(y)
s = UnivariateSpline(x, y)
xs = np.linspace(0, 10, 10000)
ys = s(xs)
ys_min=np.min(ys)
ys_min_pos=np.argmin(ys)
xs_min=xs[ys_min_pos]
print('UnivariateSpline: ', '\n', 'dist_min: ',xs_min , '\t', 'en_min: ', '\t',ys_min)
plt.plot(x, y, 'o')
plt.plot(xs, ys)
plt.show()
Related
I'm trying to create a vector field and some curve, I've created a vector field as shown
import matplotlib.pyplot as plt
import numpy as np
x,y = np.meshgrid(np.arange(-3,3,.35),np.arange(-3,3,.35))
u = x
v = y
plt.quiver(x, y, u, v, color = 'black')
plt.show()
But I want to add the curve $y=x^2$ in the same plot, how could I do that?
I've tryeid to add plt.plot and the curve but the result is weird.
You probably want to keep the y-axis limit as was in the mesh grid. plt.ylim is helpful in that case
import matplotlib.pyplot as plt
import numpy as np
x, y = np.meshgrid(np.arange(-3, 3, .35), np.arange(-3, 3, .35))
u = x
v = y
plt.quiver(x, y, u, v, color = 'black')
x = np.linspace(-3, 3, 100)
ylim = plt.ylim()
plt.plot(x, x**2)
plt.ylim(ylim)
plt.show()
Output:
I have a function f(x,y) where t is a parameter. I'm trying to plot the function where t = 1 for x and y values ranging from -5 to 5. The plot doesn't render.
import sympy as sp
import numpy as np
import matplotlib.pyplot as plt
%matplotlib notebook
C = sv.CoordSys3D("")
x, y, z = C.base_scalars()
t = sp.symbols("t")
f = sp.sin(2*sp.pi*t)*sp.exp(-(x-3*sp.sin(sp.pi*t))**2 -(y-3*sp.cos(sp.pi*t))**2)
fig = plt.figure(figsize=(6, 6))
ax = fig.add_subplot(projection='3d')
X = np.linspace(-5,5,100)
Y = np.linspace(-5,5,100)
xvals, yvals = np.meshgrid(X,Y)
zvals = sp.lambdify((x,y),f.subs(t,1),"numpy")(xvals,yvals)
ax.plot_surface(xvals,yvals,zvals)
plt.show()
I get an error 'int' object has no attribute 'ndim' which I don't know how to solve.
The problem is that when you execute f.subs(t,1) it returns a number (zero in this case). So, f=0 is the expression that you are going to lambdify. Let's see the function generated by lambdify:
import inspect
print(inspect.getsource(sp.lambdify((x,y),f.subs(t,1),"numpy")))
# def _lambdifygenerated(Dummy_25, Dummy_24):
# return 0
So, no matter the values and shape of xvals and yvals, that numerical function will always return 0, which is an integer number.
However, ax.plot_surface requires zvals to have the same shape as xvals or yval. Luckily, we can easily fix that with a simple if statement:
import sympy as sp
import sympy.vector as sv
import numpy as np
import matplotlib.pyplot as plt
C = sv.CoordSys3D("")
x, y, z = C.base_scalars()
t = sp.symbols("t")
f = sp.sin(2*sp.pi*t)*sp.exp(-(x-3*sp.sin(sp.pi*t))**2 -(y-3*sp.cos(sp.pi*t))**2)
fig = plt.figure(figsize=(6, 6))
ax = fig.add_subplot(projection='3d')
X = np.linspace(-5,5,100)
Y = np.linspace(-5,5,100)
xvals, yvals = np.meshgrid(X,Y)
zvals = sp.lambdify((x,y),f.subs(t,1),"numpy")(xvals,yvals)
# if zvals is just a number, create a proper matrix
if not isinstance(zvals, np.ndarray):
zvals = zvals * np.ones_like(xvals)
ax.plot_surface(xvals,yvals,zvals)
plt.show()
The fact that this doesn't render is bug in lambdify that it doesn't work well for constant expressions.
Your real problem though is that the expression you are trying to plot is just zero:
In [5]: f
Out[5]:
2 2
- (x_ - 3⋅sin(π⋅t)) - (y_ - 3⋅cos(π⋅t))
ℯ ⋅sin(2⋅π⋅t)
In [6]: f.subs(t, 1)
Out[6]: 0
In order to create a 3d plot using plot_surface and wireframe I wrote this (looking here around)
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import rc
from matplotlib.ticker import MultipleLocator
import matplotlib.ticker as mticker
import numpy as np
from matplotlib.ticker import FormatStrFormatter
def log_tick_formatter(val, pos=None):
return f"10$^{{{int(val)}}}$"
data=np.genfromtxt('jpdfomegal2_90.dat')
x_len= len(np.unique(data[:, 0]))
y_len= len(np.unique(data[:, 1]))
X = data[:, 0].reshape(x_len, y_len)
Y = data[:, 1].reshape(x_len, y_len)
Z = data[:, 2].reshape(x_len, y_len)
#identify lowest non-negative Z value Zmin>0
Zmin = np.where(Z > 0, Z, np.inf).min()
Zmax = Z.max()
#and substitute zero with a slightly lower value than Zmin
Z[Z==0] = 0.9 * Zmin
#log transformation because the conversion in 3D
#does not work well in matplotlib
Zlog = np.log10(Z)
rc('font',family='palatino')
rc('font',size=18)
fig = plt.figure(figsize=(12,8))
#ax = fig.add_subplot(projection='3d')
ax = Axes3D(fig)
ax.set_xlim3d(0,15)
ax.set_zlim3d(np.floor(np.log10(Zmin))-1, np.ceil(np.log10(10)))
ax.zaxis.set_major_formatter(mticker.FuncFormatter(log_tick_formatter))
ax.zaxis.set_major_locator(mticker.MaxNLocator(integer=True))
rc('font',family='palatino')
rc('font',size=18)
tmp_planes = ax.zaxis._PLANES
ax.zaxis._PLANES = ( tmp_planes[2], tmp_planes[3],
tmp_planes[0], tmp_planes[1],
tmp_planes[4], tmp_planes[5])
ax.set_xlabel('$\omega^2 /<\omega^2>$')
ax.xaxis.labelpad = 10
ax.yaxis.labelpad = 10
ax.set_ylabel('cos$(\omega,\lambda^2)$')
ax.zaxis.set_rotate_label(False) # disable automatic rotation
ax.zaxis.labelpad = 10
ax.set_zlabel('')
ax.view_init(elev=17, azim=-60)
ax.grid(False)
ax.xaxis.pane.set_edgecolor('black')
ax.yaxis.pane.set_edgecolor('black')
ax.zaxis.pane.set_edgecolor('black')
ax.xaxis.pane.fill = False
ax.yaxis.pane.fill = False
ax.zaxis.pane.fill = False
ax.xaxis.set_major_locator(MultipleLocator(2))
ax.yaxis.set_major_locator(MultipleLocator(0.2))
ax.zaxis.set_major_locator(MultipleLocator(1))
#not sure this axis scaling routine is really necessary
scale_x = 1
scale_y = 1
scale_z = 0.8
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
ax.contour(X, Y, np.log10(Z), 4, lw=0.1, colors="k", linestyles="--", offset=np.floor(np.log10(Zmin))-1)#-7)
surf = ax.plot_surface(X, Y, np.log10(Z), cmap="binary", lw=0.1,alpha=0.5)
ax.plot_wireframe(X, Y, np.log10(Z),linewidth=1,color='k')
ax.contour(X, Y, np.log10(Z), 4, lw=0.1, colors="k", linestyles="solid")
fig.colorbar(surf, shrink=0.5, aspect=20)
plt.tight_layout()
plt.savefig('jpdf_lambda2_90.png', bbox_inches='tight')
plt.show()
the problem is related to the "minorticks" along zaxis .. I obtain this :
but I would have this format and ticks in the axis
Does somebody clarify how to obtain it and as well I did not find a way to use the log scale in pyplot 3d
There's an open bug on log-scaling in 3D plots, and it looks like there won't be a fix any time soon.
You can use a matplotlib.ticker.FixedLocator to add the z-axis minor ticks, as shown below.
I didn't have your data, so I've plotted an arbitrary surface.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import rc
from matplotlib.ticker import MultipleLocator, FixedLocator
import matplotlib.ticker as mticker
import numpy as np
from matplotlib.ticker import FormatStrFormatter
def log_tick_formatter(val, pos=None):
return f"10$^{{{int(val)}}}$"
x = np.linspace(1,15,15)
y = np.linspace(0,1,15)
X, Y = np.meshgrid(x, y)
Z = 1 + X**2 * Y**2
#identify lowest non-negative Z value Zmin>0
Zmin = np.where(Z > 0, Z, np.inf).min()
Zmax = Z.max()
#and substitute zero with a slightly lower value than Zmin
Z[Z==0] = 0.9 * Zmin
rc('font',family='palatino')
rc('font',size=18)
fig = plt.figure(figsize=(12,8))
ax = Axes3D(fig, auto_add_to_figure=False)
fig.add_axes(ax)
ax.set_xlim3d(0,15)
ax.set_zlim3d(np.floor(np.log10(Zmin))-1, np.ceil(np.log10(Zmax)))
ax.zaxis.set_major_formatter(mticker.FuncFormatter(log_tick_formatter))
tmp_planes = ax.zaxis._PLANES
ax.zaxis._PLANES = ( tmp_planes[2], tmp_planes[3],
tmp_planes[0], tmp_planes[1],
tmp_planes[4], tmp_planes[5])
ax.set_xlabel('$\omega^2 /<\omega^2>$')
ax.xaxis.labelpad = 10
ax.yaxis.labelpad = 10
ax.set_ylabel('cos$(\omega,\lambda^2)$')
ax.zaxis.set_rotate_label(False) # disable automatic rotation
ax.zaxis.labelpad = 10
ax.set_zlabel('')
ax.view_init(elev=17, azim=-60)
ax.grid(False)
ax.xaxis.pane.set_edgecolor('black')
ax.yaxis.pane.set_edgecolor('black')
ax.zaxis.pane.set_edgecolor('black')
ax.xaxis.pane.fill = False
ax.yaxis.pane.fill = False
ax.zaxis.pane.fill = False
ax.xaxis.set_major_locator(MultipleLocator(2))
ax.yaxis.set_major_locator(MultipleLocator(0.2))
ax.zaxis.set_major_locator(MultipleLocator(1))
# Z minor ticks
zminorticks = []
zaxmin, zaxmax = ax.get_zlim()
for zorder in np.arange(np.floor(zaxmin),
np.ceil(zaxmax)):
zminorticks.extend(np.log10(np.linspace(2,9,8)) + zorder)
ax.zaxis.set_minor_locator(FixedLocator(zminorticks))
#not sure this axis scaling routine is really necessary
scale_x = 1
scale_y = 1
scale_z = 0.8
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
ax.contour(X, Y, np.log10(Z), 4, colors="k", linestyles="--", offset=np.floor(np.log10(Zmin))-1)#-7)
surf = ax.plot_surface(X, Y, np.log10(Z), cmap="binary", lw=0.1,alpha=0.5)
ax.plot_wireframe(X, Y, np.log10(Z),linewidth=1,color='k')
ax.contour(X, Y, np.log10(Z), 4, colors="k", linestyles="solid")
fig.colorbar(surf, shrink=0.5, aspect=20)
# get a warning that Axes3D is incompatible with tight_layout()
# plt.tight_layout()
# for saving
# fig.savefig('log3d.png')
plt.show()
I'm trying to create a plot in Python to illustrate how my sequence changes as n grows. I get a dimension error. How can I fix this?
My code:
import matplotlib.pyplot as plt
import numpy as np
x = np.zeros(101)
x[0] = 0
for n in range(0, 101):
x[n] = x[n-1] - n
if x[n] < 0:
x[n] = x[n-1] + n
y = set(x)
print(y)
i = np.linspace(0, 100)
plt.plot(y, i, 'g')
Error:
ValueError: x and y must have same first dimension, but have shapes (1,) and (50,)
The problem is you are using plt.plot with a set and because of that its saying first dimension has shape (1,). The x-argument in plt.plot should be a array-like or scalar you can read more in the documentation so you can convert it to a list:
y = list(set(x))
Also you need to make sure that the for each value in y there is a corresponding value in i (y and i need to be the same shape). So you need to set np.linspace to return len(y) values:
i = np.linspace(0, 100, len(y))
The code:
import matplotlib.pyplot as plt
import numpy as np
x = np.zeros(101)
x[0] = 0
for n in range(0,101):
x[n] = x[n-1] - n
if x[n]<0:
x[n] = x[n-1] + n
y = list(set(x))
i = np.linspace(0, 100, len(y))
plt.plot(y, i,'g')
plt.show()
Output:
import numpy as np
from matplotlib import pyplot as plt
data = np.random.normal(0,1,[100,3])
x = data[:,0]
y = data[:,1]
z = data[:,2]
plt.contour([x,y],z)
When I run this code with dummy data I get:
ValueError: Contour levels must be increasing
Do you have any idea what would this mean and how I could fix it?
plt.contour is a bit particular about its input, the z values must be on values on a rectangular 2D grid, see for example:
import matplotlib.pyplot as plt
import numpy as np
x = np.expand_dims(np.arange(1,11,1), axis=1)
y = np.expand_dims(np.arange(2,21,2), axis=0)
z = y * x
print(x.shape)
print(y.shape)
print(z.shape)
plt.figure()
plt.contour(z)
plt.show()
You can also provide x and y values for plt.contour by using np.meshgrid :
XX,YY = np.meshgrid(x,y)
plt.figure()
plt.contour(XX, YY, z)
plt.show()
If you have z-values with irregular values for x and y, you might use plt.tricontour, see the following example:
from matplotlib.tri import Triangulation
data = np.random.normal(0,1,[100,3])
x = data[:,0]
y = data[:,1]
#z = data[:,2]
z = x * y
tri = Triangulation(x,y)
plt.figure()
plt.tricontour(tri, z, )
plt.scatter(x,y, c=z)
plt.show()
Edit: from JohanC's comment i learned that this can be simplified without importing matplotlib.tri by:
plt.figure()
plt.tricontour(x,y,z)
plt.scatter(x,y, c=z)
plt.show()