Clearly an 1D Poisson equation with a constant source has an unique solution even if both Dirichlet and Newmann boundary conditions are on the same side. However I can't think of a way to solve this in FiPy. Please advise. I tried:
from fipy import CellVariable, Grid1D, DiffusionTerm
import matplotlib.pyplot as plt
L = 1.
nx = 20
dx = L/nx
mesh = Grid1D(nx=nx, dx=dx)
D = 1.
valueLeft = 1.
gradValue = 0.
source = 1.
var = CellVariable(mesh=mesh)
var.constrain(valueLeft, where=mesh.facesLeft)
var.faceGrad.constrain(gradValue, where=mesh.facesLeft)
eq = 0.0 == DiffusionTerm(coeff=D) + source
eq.solve(var=var)
plt.plot(var.value);
This is not setting gradValue at facesLeft. However, it can set the needed gradValue at facesRight.
Having both a Dirichlet and a Neumann on the same face in FV turns the nature of the problem from a boundary value problem into an initial value problem. In that sense the problem becomes over-specified as the right hand side boundary condition is still required. There may be ways to handle it with FD/FV with some hacks. However, FiPy certainly isn't set up to handle this type of problem.
Related
I'm looking at a global netcdf file. I want to set all land points that are within the 60-75 deg N band to zero but keep the ocean points in that band as nan. As a second step, I want to keep the values on the land points from 60-75 but set all other land points to zero. Ocean values are NaNs. I just don't get my xarray script to do that - here is what I tried
import numpy as np
import matplotlib.pyplot as plt
ds = xr.open_dataset('ifle.nc')
ds['Shrub_total'] = ds['Shrub']
shrub_total = ds.Shrub_total
tundra = shrub_total.where((shrub_total!=np.nan)&(shrub_total.Lat>60)&
(shrub_total.Lat<75), 0)
shrub = shrub_total.where((shrub_total!=np.nan)&(shrub_total.Lat<60)&
(shrub_total.Lat>75), 0)
ds['Tundra'] = tundra
ds['Shrub'] = shrub
fig, axes = plt.subplots(ncols=2,figsize=(12,3))
ds['Shrub_total'].isel(Time=0).plot(ax=axes[0])
ds['Tundra'].isel(Time=0).plot(ax=axes[1])
ds['Shrub'].isel(Time=0).plot(ax=axes[2])
plt.show()
This is what it looks like
The left panel is the original data, for the middle one at least I managed to keep the data I wanted - but instead of the two massive violet blocks I wanted to keep the map with all values outside the selected area set to zero. The right panel was intended to be the 'inverse' of the middle one but I completely failed there. It feels like this should be such an easy thing to do but I just can't figure it out!
This appeared to be mostly an issue with the logical side, as well as the method used to deal with the NaNs.
The below seems to work for me:
tundra = shrub_total.where((np.isnan(shrub_total)==True)|
((shrub_total.Lat>60)&(shrub_total.Lat<75)), 0)
shrub = shrub_total.where((np.isnan(shrub_total)==True)|
((shrub_total.Lat<60)|(shrub_total.Lat>75)), 0)
I changed the shrub logical to an OR statement (we want to mask either less than 60 or more than 75 - it's not possible for somewhere to be both!).
We use np.isnan()==True rather than ()!=np.nan. I am unsure about why we can't treat this the way you did... This necessitated further changes to the logic.
Note, I do not use python so this may be very hacky, and I'm sure someone else will have a much more elegant and knowledgeable answer but it intrigued me so I attempted it :)
I am using GEKKO to estimate the parameters of a differential equation and I have bounded one of the variables between 0 and 1. However, when I solve the ODE, I get values outside of the bounds for this variable, so I was wondering if somebody knew how GEKKO finds the solution, as this might help me resolve the issue.
Here is the code I use to fit the data. This gives me a solution x and u where x is between 0 and 1.
However, afterwards, I try to solve the ODE using scipy.integrate.solve_ivp, with the initial value of u that I got, and the solution I get for u is not between this bounds. Since it should be unique, I am wondering what is the process that GEKKO follows to find the solution (does it proyect the values to the bound or how does it deal with this?) Any comment is very appreciated.
Here is an MVCE. If you run it you can see that with GEKKO, I get a solution between the bounds and then, when I solve the ODE with solve_ivp, I don't get the same solution. Can you explain why this happens and how can I deal with it? I want to use solve_ivp to predict the next values.
from scipy.integrate import solve_ivp
from gekko import GEKKO
import matplotlib.pyplot as plt
time=[0.0, 0.11784511784511785, 0.18855218855218855,\
0.2356902356902357]
m = GEKKO(remote=False)
m.time= [0.0, 0.11784511784511785, 0.18855218855218855,\
0.2356902356902357]
x_data= [0.0003777630481280617, 0.002024573836061331,\
0.0008954383363035536, 0.005331749410182463]
x = m.CV(value=x_data, lb=0); x.FSTATUS = 1 # fit to measurement
x.SPLO = 0
sigma = m.FV(value=0.5, lb= 0, ub=1); sigma.STATUS=1
d = m.Param(0.05)
k = m.Param(0.001)
b = m.Param(0.5)
r = m.FV(value=0.5, lb= 0); r.STATUS=1
m_param = m.Param(1)
u = m.Var(value=0.1, lb=0, ub=1)
m.free(u)
a = m.Param(0.999)
Kmax= m.Param(100000)
m.Equations([x.dt()==x*(r*(1-a*u**2)*(1-x/(Kmax*(1-a*u**2)))-\
m_param/(k+b*u)-d), u.dt() == \
sigma*((-2*a*(b**2)*r*(u**3)+4*a*b*k*r*(u**2)\
+2*a*(k**2)*r*u-b*m_param)/((b*u+k)**2))])
m.options.IMODE = 5 # dynamic estimation
m.options.NODES = 5 # collocation nodes
m.options.EV_TYPE = 1 # linear error (2 for squared)
m.solve(disp=False, debug=False) # display solver output
def model_case_3(t, z, r, k, b, Kmax, sigma):
m=1
a=0.999
x, u= z
dxdt = x*(r*(1-a*u**2)*(1-x/(Kmax*(1-a*u**2)))-m/(k+b*u)-0.05)
dudt = sigma*((-2*a*(b**2)*r*(u**3)+4*a*b*k*r*(u**2)\
+2*a*(k**2)*r*u-b*m)/((b*u+k)**2))
return [dxdt, dudt]
sol = solve_ivp(fun=model_case_3, t_span=[0.0, 0.2356902356902357],\
y0=[0.0003777630481280617, u.value[0]],\
t_eval=[0.0, 0.11784511784511785, 0.18855218855218855,\
0.2356902356902357], \
args=(r.value[0], 0.001, 0.5,1000000 , sigma.value[0]))
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10,3), constrained_layout=True)
ax1.set_title('x')
ax1.plot(time, x.value, time, sol['y'][0])
ax2.set_title('u')
ax2.plot(time, u.value, time, sol['y'][1])
plt.show()
It is not an issue with the version of Gekko as I have Gekko 0.2.8, so I am wondering if it has anything to do with the initialization of variables. I run the example I posted on spyder (I was using google colab) and I got the correct solution, but when I run the rest of the cases I got again negative values for u (solving with solve_ivp), which is quite strange.
You can add a bound to a variable when it is created by setting lb (lower bound) and ub (upper bound).
z = m.Var(lb=0,ub=10)
After you create the variable, the bound is adjusted with .lower and .upper.
z.LOWER = 1
z.UPPER = 9
Here is an example problem that shows the use of bounds where x is constrained to be greater than 0.5.
from gekko import GEKKO
t_data = [0, 0.1, 0.2, 0.4, 0.8, 1]
x_data = [2.0, 1.6, 1.2, 0.7, 0.3, 0.15]
m = GEKKO(remote=False)
m.time = t_data
x = m.CV(value=x_data,lb=0.5,ub=3); x.FSTATUS = 1 # fit to measurement
k = m.FV(); k.STATUS = 1 # adjustable parameter
m.Equation(x.dt()== -k * x) # differential equation
m.options.IMODE = 5 # dynamic estimation
m.options.NODES = 5 # collocation nodes
m.solve(disp=False) # display solver output
k = k.value[0]; print(k)
A plot of the results shows that the bounds are enforced but the model prediction does not fit the data because of the lower bound constraint (x>=0.5).
import numpy as np
import matplotlib.pyplot as plt # plot solution
plt.plot(m.time,x.value,'bo',\
label='Predicted (k='+str(np.round(k,2))+')')
plt.plot(m.time,x_data,'rx',label='Measured')
# plot exact solution
t = np.linspace(0,1); xe = 2*np.exp(-k*t)
plt.plot(t,xe,'k:',label='Exact Solution')
plt.legend()
plt.xlabel('Time'), plt.ylabel('Value')
plt.show()
Without the restrictive lower bound, the solver optimizes to best fit the points.
x = m.CV(value=x_data,lb=0.0,ub=3)
Response 1 to Question Edit
The only way that a variable (such as u) is outside of the bounds is if the solver did not report a successful solution. To report a successful solution, the solver must satisfy the Karush Kuhn Tucker conditions for optimality. I recommend that you check that it satisfied all of the equations by checking that m.options.APPSTATUS==1 after the m.solve() command. If you can include an MVCE (https://stackoverflow.com/help/minimal-reproducible-example) that has sample data so the script can run, we can help you check it.
Response 2 to Question Edit
Thanks for including a minimal reproducible example. Here are the results that I get with Gekko 0.2.8. If you are using an earlier version, I recommend that you upgrade with pip install gekko --upgrade.
The solver reports a successful solution.
EXIT: Optimal Solution Found.
The solution was found.
The final value of the objective function is 0.03164650667928192
---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 0.23339999999999997 sec
Objective : 0.0316473666078486
Successful solution
---------------------------------------------------
The constraints x>=0 and 0<=u<=1 are satisfied. Could it just be an issue with an older version of Gekko?
I am trying to create a t-distribution by taking the mean of many samples from a normal distribution (and then estimating the shape with kernel density estimation).
For some reason, I am getting pretty different results when I compare what I get with a proper t-distribution. I don't understand what is going wrong, so I think I am confused about something.
Here is the code:
import numpy as np
from scipy.stats import gaussian_kde
import matplotlib.pyplot as plt
import seaborn
inner_sample_size = 10
X = np.arange(-3, 3, 0.01)
results = [
np.mean(np.random.normal(size=inner_sample_size))
for _ in range(10000)
]
estimation = gaussian_kde(results)
plt.plot(X, estimation.evaluate(X))
t_samples = np.random.standard_t(inner_sample_size, 10000)
t_estimator = gaussian_kde(t_samples)
plt.plot(X, t_estimator.evaluate(X))
plt.ylabel("Probability density")
plt.show()
And here is the plot I get:
Where the orange line is numpy's own t-distribution, and the blue line is the one estimated by sampling.
Your assumption that the mean of Standard Normals has T distribution is incorrect. In fact, the mean of Standard Normals has Normal Distribution, which explains the shape of your blue graph. To generate one random variable T from a T distribution with k degrees of freedom, you first generate k+1 independent Standard Normals Z_i, i=0,...,k. You then compute
T = Z_0 / sqrt( sum(Z_i^2, i=1 to k)/k ).
The sum of squared Standard Normals sum(Z_i^2, i=1 to k) has Chi-Squared Distribution with k degrees of freedom, so if there is a pre-canned method to generate this, you should use it, since it's likely more efficient.
I have a bunch of points in a 2D space which all reside on a line (polygon). How can I compute the mean coordinate of these points on the line?
I don't mean the centroid of the points in the 2D space (as #rth initially proposed in his answer), but the mean location of the points along the line on which they reside. So basically, I could transform the line to a 1D axis, compute the mean location in 1D, and transform the location of the mean back into the 2D space.
Maybe these are exactly the necessary steps, but I think (or hope) that there is a function in numpy/scipy which allows me to do this in one step.
Edit: The approach you describe in the question is indeed probably the simplest way for solving this problem.
Here is an implementation that calculates the positions of vertices along the line in 1D, takes their mean, and finally calculates the corresponding 2D position with parametric interpolation,
import numpy as np
from scipy.interpolate import splprep, splev
vert = np.random.randn(1000, 2) # vertices definition here
# calculate the Euclidean distances between consecutive vertices
# equivalent to a for loop with
# dl[i] = ((vert[i+1, 0] - vert[i, 0])**2 + (vert[i+1,1] - vert[i,1])**2)**0.5
dl = (np.diff(vert, axis=0)**2).sum(axis=1)**0.5
# pad with 0, so dl.shape[0] == vert.shape[0] for convenience
dl = np.insert(dl, 0, 0.0)
l = np.cumsum(dl) # 1D coordinates along the line
l_mean = np.mean(l) # mean in the line coordinates
# calculate the coordinate of l_mean in 2D space
# with parametric B-spline interpolation
tck, _ = splprep(x=vert.T, u=l, k=3)
res = splev(l_mean, tck)
print(res)
Edit2: Assuming now that you have a high resolution set of points for your path vert_full and some approximate measurements vert_1, vert_2, etc, what you could do is the following.
Project each points of vert_1, etc. onto the exact path. Assuming that vert_full has much more datapoints than vert_1, we can simply look for the nearest neighbours of vert_1 in vert_full:
from scipy.spatial import cKDTree
tr = cKDTree(vert_full)
d, idx = tr.query(vert_1, k=1)
vert_1_proj = vert_full[idx] # this gives the projected corrdinates onto vert_full
# I have not actually run this, so it might require minor changes
Use the above mean calculation with the new vert_1_proj vector.
Meanwhile I've found the answer to my question, although using Shapely instead of Numpy.
from shapely.geometry import LineString, Point
# lists of points as (x,y) tuples
path_xy = [...]
points_xy = [...] # should be on or near path
path = LineString(path_xy) # create path object
pts = [Point(p) for p in points_xy] # create point objects
dist = [path.project(p) for p in pts] # distances along path
mean_dist = np.mean(dist) # mean distance along path
mean = path.interpolate(mean_dist) # mean point
mean_xy = (mean.x,mean.y)
This works perfectly!
(That's is also why I have to accept it as the answer, though I highly appreciate #rth's help!)
I'm currently writing line and contour plotting functions for my PyQuante quantum chemistry package using matplotlib. I have some great functions that evaluate basis sets along a (npts,3) array of points, e.g.
from somewhere import basisset, line
bfs = basisset(h2) # Generate a basis set
points = line((0,0,-5),(0,0,5)) # Create a line in 3d space
bfmesh = bfs.mesh(points)
for i in range(bfmesh.shape[1]):
plot(bfmesh[:,i])
This is fast because it evaluates all of the basis functions at once, and I got some great help from stackoverflow here and here to make them extra-nice.
I would now like to update this to do contour plotting as well. The slow way I've done this in the past is to create two one-d vectors using linspace(), mesh these into a 2D grid using meshgrid(), and then iterating over all xyz points and evaluating each one:
f = np.empty((50,50),dtype=float)
xvals = np.linspace(0,10)
yvals = np.linspace(0,20)
z = 0
for x in xvals:
for y in yvals:
f = bf(x,y,z)
X,Y = np.meshgrid(xvals,yvals)
contourplot(X,Y,f)
(this isn't real code -- may have done something dumb)
What I would like to do is to generate the mesh in more or less the same way I do in the contour plot example, "unravel" it to a (npts,3) list of points, evaluate the basis functions using my new fast routines, then "re-ravel" it back to X,Y matrices for plotting with contourplot.
The problem is that I don't have anything that I can simply call .ravel() on: I either have 1d meshes of xvals and yvals, the 2D versions X,Y, and the single z value.
Can anyone think of a nice, pythonic way to do this?
If you can express f as a function of X and Y, you could avoid the Python for-loops this way:
import matplotlib.pyplot as plt
import numpy as np
def bf(x, y):
return np.sin(np.sqrt(x**2+y**2))
xvals = np.linspace(0,10)
yvals = np.linspace(0,20)
X, Y = np.meshgrid(xvals,yvals)
f = bf(X,Y)
plt.contour(X,Y,f)
plt.show()
yields