Develop Reference

Error when observing on uniform with sampled parameters in PyMC

I’m new to PyMC and am trying to model a situation where you are rolling marbles at a wall and trying to find the block. The data is only for the values where the marble hits the block.
I’m first sampling an x location and size and then calculating a point from those with a uniform, but I’m getting an error.
import pymc3 as pm
import theano.tensor as tt
basic_model = pm.Model()
with basic_model:
# We are assuming independence of these.
x = pm.Uniform("x", lower=1, upper=30)
l = pm.Uniform("l", lower=1, upper=30)
lower = pm.Deterministic('lower', x-0.5*l)
upper = pm.Deterministic('upper', x+0.5*l)
point_x = pm.Uniform('point_x', lower=lower, upper=upper, observed=x_vals)
pm.sample()
With the error:
SamplingError: Initial evaluation of model at starting point failed!
Starting values:
{'x_interval__': array(0.), 'l_interval__': array(0.)}
Initial evaluation results:
x_interval__ -1.39
l_interval__ -1.39
point_x -inf
Name: Log-probability of test_point, dtype: float64
Clearly the issue is with point_x. I’m guessing the error has to do with the fact that the observed data may potentially fall outside the lower-upper range depending on the value of x and l sampled. But how might I fix this?

The sampler doesn't know how to handle starting off in an invalid region of the parameter space. A quick and dirty fix is to provide testval arguments that ensure the sampling begins in a logically valid solution. For example, we know the minimum block must have:
l_0 = np.max(x_vals) - np.min(x_vals)
x_0 = np.min(x_vals) + 0.5*l_0
and could use those:
x = pm.Uniform("x", lower=1, upper=30, testval=x_0)
l = pm.Uniform("l", lower=1, upper=30, testval=l_0)
Also, the nature of this model leads to many rejections due to impossibility, so you may want to use Metropolis for sampling, which almost always needs more steps and tuning
pm.sample(tune=10000, draws=10000, step=pm.Metropolis())
Alternative Models
Otherwise, consider reparameterizing so that only valid solutions are in the parameter space. One approach would be to sample l and then use that to constrain x. Something like:
other_model = pm.Model()
x_min = np.min(x_vals)
x_max = np.max(x_vals)
l_0 = x_max - x_min
with other_model:
# these have logical constraints from the data
l = pm.Uniform("l", lower=l_0, upper=30)
x = pm.Uniform("x", lower=x_max - 0.5*l, upper=x_min + 0.5*l)
lower = pm.Deterministic('lower', x - 0.5*l)
upper = pm.Deterministic('upper', x + 0.5*l)
point_x = pm.Uniform('point_x', lower=lower, upper=upper, observed=x_vals)
res = pm.sample(step=pm.NUTS(), return_inferencedata=True)
Another approach would be to sample lower and upper directly, and compute the x and l as deterministic variables from those.

"Exception: #error: Solution not Found" on and off while solving the same problem

I am solving multiple optimization problems with the Gekko optimization suite. I generate random numbers for three variables (book3.csv) and import them to python as initial values and parameters. I run optimization solver and collect the results. It works well. However, when I tried this about 20 times to check, I received an error message of Exception: #error: Solution not Found several times. It stops at the same location: ...gekko.py line 2174, in solve raise Exception(response)
I collected the value of control variables, u, at each round of optimization, i, at each time points, j, and saved it at a dataframe, utot, in order to track all the decisions made at each time for each round. I noticed that the last column (at the last time points) of utot has some positive values, which I think shouldn't be. I intentionally put Pe0=0, so that u=0 is optimum at all time points, but still the last column of utot has some positive value. I am not sure what am I missing here.
from gekko import GEKKO
import pandas as pd
import numpy as np
from numpy import random
file_path='./Book3.csv'
d=pd.read_csv("Book3.csv")
dftemp = pd.DataFrame(data=d)
na=len(dftemp)
# time points
n=51
year=50
# constants
Pa0 = 1
Pe0 = 0
C = 10
r = 0.05
k=50
shift=100
ll=0.5*random.rand(n)
index = range(0, na)
columns = range(0, n)
columns1 = range(0, 1)
#Dataframe to collect results
Atot=pd.DataFrame(index=index,columns=columns)
Etot=pd.DataFrame(index=index,columns=columns)
utot=pd.DataFrame(index=index,columns=columns)
PVtot=pd.DataFrame(index=index,columns=columns1)
for i in range(0,na):
# create GEKKO model
m = GEKKO(remote=False)
m.time = np.linspace(0,year,n)
t=m.time
A0=dftemp.loc[i][0]
h=dftemp.loc[i][1]
emax=dftemp.loc[i][2]
u = m.MV(value=0,lb=0, ub=emax)
u.STATUS = 1
u.DCOST = 0
A = m.SV(value=A0)
E = m.SV(value=0)
t = m.Param(value=m.time)
Pe = m.Var(value=Pe0)
d = m.Var(value=1)
l = m.Param(value=ll)
# Equation
m.Equation(A.dt()==-u)
m.Equation(E.dt()==u)
m.Equation(Pe.dt()==-Pe/k)
m.Equation(d==m.exp(-t*r))
m.Equation(A>=0)
# Objective (Utility)
J = m.Var(value=0)
# Final objective
Jf = m.FV()
Jf.STATUS = 1
m.Connection(Jf,J,pos2='end')
m.Equation(J.dt() == m.log(A+E*(1-l))*h*Pa0-r*C*u+E*Pe+shift)*d
# maximize U
m.Maximize(Jf)
# options
m.options.IMODE = 6 # optimal control
m.options.NODES = 3 # collocation nodes
m.options.SOLVER = 3 # solver (IPOPT)
# solve optimization problem
m.solve()
# print profit
print('Optimal Profit: ' + str(Jf.value[0]))
for j in range(0,n):
Atot.loc[i][j]=A.value[j]
Etot.loc[i][j]=E.value[j]
utot.loc[i][j]=u.value[j]
PVtot.loc[i][0]=Jf.value[0]
print(PVtot.sum())

Try setting m.options.TIME_SHIFT=0 to avoid advancing in time if the initial conditions should remain the same. The default is m.options.TIME_SHIFT=1 where all of the prior values are shifted left and the second horizon point becomes the first time point. This is the default for applications that are running in real-time to a physical process. Each cycle, the simulation or optimal control problem advances one time step ahead.
The last point in the horizon sometimes has no effect on the objective function so the solver isn't forced to move it to a particular value. This can be checked by fixing the last value with m.fix_final(utot,100) and see if this changes the objective function value. If the objective function value doesn't change then the solver will not find a unique solution for this value. If you'd like to make it unique then include an additional objective such as m.Minimize(1e-5 * utot**2).
When submitting questions, please include all of the files needed to reproduce the behavior. Here are some tips on creating a Minimal, Reproducible example.

Efficient computation of a loop of integrals in Python

I was wondering how to speed up the following code in where I compute a probability function which involves numerical integrals and then I compute some confidence margins.
Some possibilities that I have thought about are Numba or vectorization of the code
EDIT:
I have made minor modifications because there was a mistake. I am looking for some modifications that provide major time improvements (I know that there are some minor changes that would provide some minor time improvements, such as repeated functions, but I am not concerned about them)
The code is:
# -*- coding: utf-8 -*-
"""
Created on Tue Jan 26 17:05:46 2021
#author: Ignacio
"""
import numpy as np
from scipy.integrate import simps
def pdf(V,alfa_points):
alfa=np.linspace(0,2*np.pi,alfa_points)
return simps(1/np.sqrt(2*np.pi)/np.sqrt(sigma_R2)*np.exp(-(V*np.cos(alfa)-eR)**2/2/sigma_R2)*1/np.sqrt(2*np.pi)/np.sqrt(sigma_I2)*np.exp(-(V*np.sin(alfa)-eI)**2/2/sigma_I2),alfa)
def find_nearest(array,value):
array=np.asarray(array)
idx = (np.abs(array-value)).argmin()
return array[idx]
N = 20
n=np.linspace(0,N-1,N)
d=1
sigma_An=0.1
sigma_Pn=0.2
An=np.ones(N)
Pn=np.zeros(N)
Vs=np.linspace(0,30,1000)
inc=np.max(Vs)/len(Vs)
th=np.linspace(0,np.pi/2,250)
R=np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I=np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
PDF=np.zeros(len(Vs))
for vv in range(len(Vs)):
PDF[vv]=pdf(Vs[vv],100)
total=simps(PDF,Vs)
values=np.cumsum(PDF)*inc/total
xval_05=find_nearest(values,0.05)
fmin[tt]=Vs[values==xval_05]
xval_95=find_nearest(values,0.95)
fmax[tt]=Vs[values==xval_95]

This version's speedup: 31x
A simple profiling (%%prun) reveals that most of the time is spent in simps.
You are in control of the integration done in pdf(): for example, you can use the trapeze method instead of Simpson with negligible numerical difference if you increase a bit the resolution of alpha. In fact, the higher resolution obtained by a higher sampling of alpha more than makes up for the difference between simps and trapeze (see picture at the bottom as for why). This is by far the highest speedup. We go one bit further by implementing the trapeze method ourselves instead of using scipy, since it is so simple. This alone yields marginal gain, but opens the door for a more drastic optimization (below, about pdf2D.
Also, the remaining simps(PDF, ...) goes faster when it knows that the dx step is constant, so we can just say so instead of passing the whole alpha array.
You can avoid doing the loop to compute PDF and use np.vectorize(pdf) directly on Vs, or better (as in the code below), do a 2-D version of that calculation.
There are some other minor things (such as using an index directly fmin[tt] = Vs[closest(values, 0.05)] instead of finding the index, returning the value, and then using a boolean mask for where values == xval_05), or taking all the constants (including alpha) outside functions and avoid recalculating every time.
This above gives us a 5.2x improvement. There is a number of things I don't understand in your code, e.g. why having An (ones) and Pn (zeros)?
But, importantly, another ~6x speedup comes from the observation that, since we are implementing our own trapeze method by using numpy primitives, we can actually do it in 2D in one go for the whole PDF.
The final speed up of the code below is 31x. I believe that a better understanding of "the big picture" of what you want to do would yield additional, perhaps substantial, speed gains.
Modified code:
import numpy as np
from scipy.integrate import simps
alpha_points = 200 # more points as we'll use trapeze not simps
alpha = np.linspace(0, 2*np.pi, alpha_points)
cosalpha = np.cos(alpha)
sinalpha = np.sin(alpha)
d_alpha = np.mean(np.diff(alpha)) # constant dx
coeff = 1 / np.sqrt(2*np.pi)
Vs=np.linspace(0,30,1000)
d_Vs = np.mean(np.diff(Vs)) # constant dx
inc=np.max(Vs)/len(Vs)
def f2D(Vs, eR, sigma_R2, eI, sigma_I2):
a = coeff / np.sqrt(sigma_R2)
b = coeff / np.sqrt(sigma_I2)
y = a * np.exp(-(np.outer(cosalpha, Vs) - eR)**2 / 2 / sigma_R2) * b * np.exp(-(np.outer(sinalpha, Vs) - eI)**2 / 2 / sigma_I2)
return y
def pdf2D(Vs, eR, sigma_R2, eI, sigma_I2):
y = f2D(Vs, eR, sigma_R2, eI, sigma_I2)
s = y.sum(axis=0) - (y[0] + y[-1]) / 2 # our own impl of trapeze, on 2D y
return s * d_alpha
def closest(a, val):
return np.abs(a - val).argmin()
N = 20
n = np.linspace(0,N-1,N)
d = 1
sigma_An = 0.1
sigma_Pn = 0.2
An=np.ones(N)
Pn=np.zeros(N)
th = np.linspace(0,np.pi/2,250)
R = np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I = np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
PDF=pdf2D(Vs, eR, sigma_R2, eI, sigma_I2)
total = simps(PDF, dx=d_Vs)
values = np.cumsum(PDF) * inc / total
fmin[tt] = Vs[closest(values, 0.05)]
fmax[tt] = Vs[closest(values, 0.95)]
Note: most of the fmin and fmax are np.allclose() compared with the original function, but some of them have a small error: after some digging, it turns out that the implementation here is more precise as that function f() can be pretty abrupt, and more alpha points actually help (and more than compensate the minuscule lack of precision due to using trapeze instead of Simpson).
For example, at index tt=244, vv=400:

Considering several methods, the one that provides the largest time improvement is the Numba method. The method proposed by Pierre is very interesting and it does not require to install other packages, which is an asset.
However, in the examples that I have computed, the time improvement is not as large as with the numba example, specially when the points in th grows to a few tenths of thousands (which is my actual case). I post here the Numba code just in case someone is interested:
import numpy as np
from numba import njit
#njit
def margins(val_min,val_max):
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
Vs=np.linspace(0,30,1000)
inc=np.max(Vs)/len(Vs)
integration_points=200
PDF=np.zeros(len(Vs))
for vv in range(len(Vs)):
PDF[vv]=np.trapz(1/np.sqrt(2*np.pi)/np.sqrt(sigma_R2)*np.exp(-(Vs[vv]*np.cos(np.linspace(0,2*np.pi,integration_points))-eR)**2/2/sigma_R2)*1/np.sqrt(2*np.pi)/np.sqrt(sigma_I2)*np.exp(-(Vs[vv]*np.sin(np.linspace(0,2*np.pi,integration_points))-eI)**2/2/sigma_I2),np.linspace(0,2*np.pi,integration_points))
total=np.trapz(PDF,Vs)
values=np.cumsum(PDF)*inc/total
idx = (np.abs(values-val_min)).argmin()
xval_05=values[idx]
fmin[tt]=Vs[np.where(values==xval_05)[0][0]]
idx = (np.abs(values-val_max)).argmin()
xval_95=values[idx]
fmax[tt]=Vs[np.where(values==xval_95)[0][0]]
return fmin,fmax
N = 20
n=np.linspace(0,N-1,N)
d=1
sigma_An=1/2**6
sigma_Pn=2*np.pi/2**6
An=np.ones(N)
Pn=np.zeros(N)
th=np.linspace(0,np.pi/2,250)
R=np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I=np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
F=R+1j*I
Fa=np.abs(F)/np.max(np.abs(F))
fmin, fmax = margins(0.05,0.95)

Stepsize control of dopri5 integrator

I am trying to solve a simple example with the dopri5 integrator in scipy.integrate.ode. As the documentation states
This is an explicit runge-kutta method of order (4)5 due to Dormand & Prince (with stepsize control and dense output).
this should work. So here is my example:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def MassSpring_with_force(t, state):
""" Simple 1DOF dynamics model: m ddx(t) + k x(t) = f(t)"""
# unpack the state vector
x = state[0]
xd = state[1]
# these are our constants
k = 2.5 # Newtons per metre
m = 1.5 # Kilograms
# force
f = force(t)
# compute acceleration xdd
xdd = ( ( -k*x + f) / m )
# return the two state derivatives
return [xd, xdd]
def force(t):
""" Excitation force """
f0 = 1 # force amplitude [N]
freq = 20 # frequency[Hz]
omega = 2 * np.pi *freq # angular frequency [rad/s]
return f0 * np.sin(omega*t)
# Time range
t_start = 0
t_final = 1
# Main program
state_ode_f = ode(MassSpring_with_force)
state_ode_f.set_integrator('dopri5', rtol=1e-6, nsteps=500,
first_step=1e-6, max_step=1e-3)
state2 = [0.0, 0.0] # initial conditions
state_ode_f.set_initial_value(state2, 0)
sol = np.array([[t_start, state2[0], state2[1]]], dtype=float)
print("Time\t\t Timestep\t dx\t\t ddx\t\t state_ode_f.successful()")
while state_ode_f.t < (t_final):
state_ode_f.integrate(t_final, step=True)
sol = np.append(sol, [[state_ode_f.t, state_ode_f.y[0], state_ode_f.y[1]]], axis=0)
print("{0:0.8f}\t {1:0.4e} \t{2:10.3e}\t {3:0.3e}\t {4}".format(
state_ode_f.t, sol[-1, 0]- sol[-2, 0], state_ode_f.y[0], state_ode_f.y[1], state_ode_f.successful()))
The result I get is:
Time Timestep dx ddx state_ode_f.successful()
0.49763822 4.9764e-01 2.475e-03 -8.258e-04 False
0.99863822 5.0100e-01 3.955e-03 -3.754e-03 False
1.00000000 1.3618e-03 3.950e-03 -3.840e-03 False
with a warning:
c:\python34\lib\site-packages\scipy\integrate_ode.py:1018: UserWarning: dopri5: larger nmax is needed
self.messages.get(idid, 'Unexpected idid=%s' % idid))
The result is incorect. If I run the same code with vode integrator, I get the expected result.
Edit
A similar issue is described here:
Using adaptive step sizes with scipy.integrate.ode
The suggested solution recommends setting nsteps=1, which solves the ODE correctly and with step-size control. However the integrator returns state_ode_f.successful() as False.

No, there is nothing wrong. You are telling the integrator to perform an integration step to t_final and it performs that step. Internal steps of the integrator are not reported.
The sensible thing to do is to give the desired sampling points as input of the algorithm, set for example dt=0.1 and use
state_ode_f.integrate( min(state_ode_f.t+dt, t_final) )
There is no single-step method in dopri5, only vode has it defined, see the source code https://github.com/scipy/scipy/blob/v0.14.0/scipy/integrate/_ode.py#L376, this could account for the observed differences.
As you found in Using adaptive step sizes with scipy.integrate.ode, one can force single-step behavior by setting the iteration bound nsteps=1. This will produce a warning every time, so one has to suppress these specific warnings to see a sensible result.
You should not use a parameter (which is a constant for the integration interval) for a time-dependent force. Use inside MassSpring_with_force the evaluation f=force(t). Possibly you could pass the function handle of force as parameter.

Using adaptive step sizes with scipy.integrate.ode

The (brief) documentation for scipy.integrate.ode says that two methods (dopri5 and dop853) have stepsize control and dense output. Looking at the examples and the code itself, I can only see a very simple way to get output from an integrator. Namely, it looks like you just step the integrator forward by some fixed dt, get the function value(s) at that time, and repeat.
My problem has pretty variable timescales, so I'd like to just get the values at whatever time steps it needs to evaluate to achieve the required tolerances. That is, early on, things are changing slowly, so the output time steps can be big. But as things get interesting, the output time steps have to be smaller. I don't actually want dense output at equal intervals, I just want the time steps the adaptive function uses.
EDIT: Dense output
A related notion (almost the opposite) is "dense output", whereby the steps taken are as large as the stepper cares to take, but the values of the function are interpolated (usually with accuracy comparable to the accuracy of the stepper) to whatever you want. The fortran underlying scipy.integrate.ode is apparently capable of this, but ode does not have the interface. odeint, on the other hand, is based on a different code, and does evidently do dense output. (You can output every time your right-hand-side is called to see when that happens, and see that it has nothing to do with the output times.)
So I could still take advantage of adaptivity, as long as I could decide on the output time steps I want ahead of time. Unfortunately, for my favorite system, I don't even know what the approximate timescales are as functions of time, until I run the integration. So I'll have to combine the idea of taking one integrator step with this notion of dense output.
EDIT 2: Dense output again
Apparently, scipy 1.0.0 introduced support for dense output through a new interface. In particular, they recommend moving away from scipy.integrate.odeint and towards scipy.integrate.solve_ivp, which as a keyword dense_output. If set to True, the returned object has an attribute sol that you can call with an array of times, which then returns the integrated functions values at those times. That still doesn't solve the problem for this question, but it is useful in many cases.

Since SciPy 0.13.0,
The intermediate results from the dopri family of ODE solvers can
now be accessed by a solout callback function.
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
backend = 'dopri5'
# backend = 'dop853'
solver = ode(logistic).set_integrator(backend)
sol = []
def solout(t, y):
sol.append([t, *y])
solver.set_solout(solout)
solver.set_initial_value(y0, t0).set_f_params(r)
solver.integrate(t1)
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
Result:
The result seems to be slightly different from Tim D's, although they both use the same backend. I suspect this having to do with FSAL property of dopri5. In Tim's approach, I think the result k7 from the seventh stage is discarded, so k1 is calculated afresh.
Note: There's a known bug with set_solout not working if you set it after setting initial values. It was fixed as of SciPy 0.17.0.

I've been looking at this to try to get the same result. It turns out you can use a hack to get the step-by-step results by setting nsteps=1 in the ode instantiation. It will generate a UserWarning at every step (this can be caught and suppressed).
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
import warnings
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
#backend = 'vode'
backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend, nsteps=1)
solver.set_initial_value(y0, t0).set_f_params(r)
# suppress Fortran-printed warning
solver._integrator.iwork[2] = -1
sol = []
warnings.filterwarnings("ignore", category=UserWarning)
while solver.t < t1:
solver.integrate(t1, step=True)
sol.append([solver.t, solver.y])
warnings.resetwarnings()
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
result:

The integrate method accepts a boolean argument step that tells the method to return a single internal step. However, it appears that the 'dopri5' and 'dop853' solvers do not support it.
The following code shows how you can get the internal steps taken by the solver when the 'vode' solver is used:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
backend = 'vode'
#backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend)
solver.set_initial_value(y0, t0).set_f_params(r)
sol = []
while solver.successful() and solver.t < t1:
solver.integrate(t1, step=True)
sol.append([solver.t, solver.y])
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
Result:

FYI, although an answer has been accepted already, I should point out for the historical record that dense output and arbitrary sampling from anywhere along the computed trajectory is natively supported in PyDSTool. This also includes a record of all the adaptively-determined time steps used internally by the solver. This interfaces with both dopri853 and radau5 and auto-generates the C code necessary to interface with them rather than relying on (much slower) python function callbacks for the right-hand side definition. None of these features are natively or efficiently provided in any other python-focused solver, to my knowledge.

Here's another option that should also work with dopri5 and dop853. Basically, the solver will call the logistic() function as often as needed to calculate intermediate values so that's where we store the results:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
sol = []
def logistic(t, y, r):
sol.append([t, y])
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
# Maximum number of steps that the integrator is allowed
# to do along the whole interval [t0, t1].
N = 10000
#backend = 'vode'
backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend, nsteps=N)
solver.set_initial_value(y0, t0).set_f_params(r)
# Single call to solver.integrate()
solver.integrate(t1)
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()