I'm brand new to sympy, and what seems like it should be a simple example doesn't work for me. I'm running this in a fresh Google colab.
import sympy
from sympy.plotting import plot3d
x, y = sympy.symbols('x y')
expr = sympy.Min(x, y)
print(expr)
expr.subs(x, .2).subs(y, .5)
So far so good. But now when I try to plot this using plot3d, I get an error:
plot3d(expr, (x, -1, 1), (y, -1, 1))
results in TypeError: only integer scalar arrays can be converted to a scalar index. More strangely, when I change the expression to Min(x, x) it plots as expected. What am I missing?
Related
I am new to sympy, and I cannot understand why the result of the following piece of code does not results in f(x)=0
from sympy import *
f = Function('f')
x = Symbol('x')
simplify(Eq(f(x)+1,1))
When SymPy rewrites x + x as 2*x that is automatic rewriting. Not everything is automatic, however, as you have seen. If you want to know what value of f(x) makes that Equality true, you can solve for it:
>>> solve(Eq(f(x) + 1, 1), f(x))
[0]
I am trying to make a simple example in SymPy to compute some coefficients and then use them in a sum of legendre polynomials. Finally plot it. Very simple but can not make it work. I want to use it for my electromagnetism course. I get errors in both the attempts below:
%matplotlib inline
from sympy import *
x, y, z = symbols('x y z')
k, m, n = symbols('k m n', integer=True)
f, step, potential = symbols('f step potential', cls=Function)
var('n x')
A=SeqFormula(2*(2*m+1)*Integral(legendre(2*m+1,x),(x,0,1)).doit(),(m,0,oo)).doit()
Sum(A.coeff(m).doit()*legendre(2*m+1,x),(m,0,10)).doit()
B=Tuple.fromiter(2*(2*n+1)*Integral(legendre(2*n+1,x),(x,0,1)).doit() for n in range(50))
Sum(B[m]*legendre(2*m+1,x),(m,0,10)).doit()
Here is a part of an script in Mathematica of what I would like to replicate:
Nn = 50;
Array[A, Nn]
For[i = 0, i <= Nn, i++, A[i + 1] = Integrate[LegendreP[2*i + 1, x]*(2*(2*i + 1) + 1), {x, 0, 1}]];
Step = Sum[A[n + 1]*LegendreP[2*n + 1, #], {n, 0, Nn}] & Plot[Step[x], {x, -1, 1}]
I think the structure you were searching for with A is Python's lambda.
A = lambda m: 2*(2*m+1)*Integral(legendre(2*m+1, x), (x, 0, 1))
f = Sum(A(m)*legendre(2*m+1, x), (m, 0, 10)).doit()
plot(f, (x, -1, 1))
The key point is that m has to be explicit in order for integration to happen; SymPy does not know a general formula for integrating legendre(n, x). So, the integration here is attempted only when A is called with a concrete value of m, like A(0), A(1), etc.
I'm having problems defining a metric using sympy's diffgeom package, where the metric depends on a function f(x,y).
I get the error ValueError: Can't calculate 1st derivative wrt x.
import sympy as sym
from sympy.diffgeom import Manifold, Patch, CoordSystem
m = Manifold("M",2)
patch = Patch("P",m)
cartesian = CoordSystem("cartesian", patch, ["x", "y"])
x, y = cartesian.coord_functions()
f = sym.Function('f')
xi = sym.Symbol('xi')
g = sym.Matrix([
[ xi + sym.diff(f,x)**2 , sym.diff(f,x) * sym.diff(f,y) ],
[ sym.diff(f,x) * sym.diff(f,y) , xi + sym.diff(f,y)**2 ]
])
I have a feeling it's because of how x and y are defined, but I haven't been able to figure it out.
Indeed, differentiation with respect to these x and y (objects of class sympy.diffgeom.diffgeom.BaseScalarField) is not supported. You can see this by accessing the internal property _diff_wrt which indicates if something can be a thing with respect to which to differentiate.
>>> x._diff_wrt
False
Do those derivative (with respect to a scalar field) make mathematical sense here? I'm not sure.
An additional issue is that SymPy does not differentiate functions, so
f = Function('f')
diff(f, x)
is always an error. SymPy can differentiate expressions, e.g., diff(f(x, y), x).
Aside: diff can be used as a method of an expression, which in your case would result in shorter code, like f(x, y).diff(x).
I'm trying to setup sympy to calculate derivatives. When I test it with simple equation, I'm finding the same answer (equality is true between sympy calculation and my own calculation). However when I try with more complicated ones, when it doesnt work (I checked answers with wolfram alpha too).
Here is my code:
from __future__ import division
from sympy import simplify, cos, sin, expand
from sympy import *
x, y, z, t = symbols('x y z t')
k, m, n = symbols('k m n', integer=True)
f, g, h = symbols('f g h', cls=Function)
equation = (x**3*y-x*y**3)/(x**2+y**2)
equation2 = (x**4*y+4*x**2*y**3-y**5)/((x**2+y**2)**2)
pprint(equation)
print ""
pprint(equation2)
print diff(equation,x) == equation2
This is a common "gotcha" in Sympy. For creating symbolic equalities, you should use sympy.Eq and not = or == (see the tutorial). For your example,
Eq(equation.diff(x), equation2).simplify()
True
Note, as above, that you may have to call simplify() in order to see wheather the Eq object corresponds to True or False
How do I fit a curve on a barplot?
I have an equation, the diffusion equation, which has some unknown parameters, these parameters make the curve larger, taller, etc. On the other hand I have a barplot coming from a simulation. I would like to fit the curve on the barplot, and find the best parameters for the curve, how can I do that?
This is what I obtained by 'manual fitting', so basically I changed manually all the parameters for hours. However is there a way to do this with python?
To make it simple, imagine I have the following code:
import matplotlib.pyplot as plt
list1 = []
for i in range(-5,6):
list1.append(i)
width = 1/1.5
list2 = [0,0.2,0.6,3.5,8,10,8,3.5,0.6,0.2,0]
plt.bar(list1,list2,width)
plt.show()
T = 0.13
xx = np.arange(-6,6,0.01)
yy = 5*np.sqrt(np.pi)*np.exp(-((xx)**2)/(4*T))*scipy.special.erfc((xx)/(2*np.sqrt(T))) + np.exp(-((xx)**2)/(4*T))
plt.plot(xx,yy)
plt.show()
Clearly the fitting here would be pretty hard, but anyway, is there any function or such that allows me to find the best coefficients for the equation: (where T is known)
y = A*np.sqrt(np.pi*D)*np.exp(-((x-E)**2)/(4*D*T))*scipy.special.erfc((x-E)/(2*np.sqrt(D*T))) + 300*np.exp(-((x-E)**2)/(4*D*T))
EDIT: This is different from the already asked question and from the scipy documentation example. In the latter the 'xdata' is the same, while in my case it might and might not be. Furthermore I would also be able to plot this curve fitting, which isn't shown on the documentation. The height of the bars is not a function of the x's! So my xdata is not a function of my ydata, this is different from what is in the documentation.
To see what I mean try to change the code in the documentation a little bit, to fall into my example, try this:
def func(x,a,b,c):
return a * np.exp(-b * x) + c
xdata = np.linspace(0,4,50)
y = func(xdata, 2.5, 1.3, 0.5)
ydata = [1,6,3,4,6,7,8,5,7,0,9,8,2,3,4,5]
popt, pcov = curve_fit(func,xdata,ydata)
if you run this, it doesn't work. The reason is that I have 16 elements for the ydata and 50 for the function. This happens because y takes values from xdata, while ydata takes values from another set of x values, which is here unknown.
Thank you
I stand by my thinking that this question is a duplicate. Here is a brief example of the typical workflow using curve_fit. Let me know if you still think that your situation is different.
import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
# bar plot data
list1 = range(-5, 6)
list2 = [0, 0.2, 0.6, 3.5, 8, 10,
8, 3.5, 0.6, 0.2, 0]
width = 1/1.5
plt.bar(list1, list2, width, alpha=0.75)
# fit bar plot data using curve_fit
def func(x, a, b, c):
# a Gaussian distribution
return a * np.exp(-(x-b)**2/(2*c**2))
popt, pcov = curve_fit(func, list1, list2)
x = np.linspace(-5, 5, 100)
y = func(x, *popt)
plt.plot(x + width/2, y, c='g')