Fix assert checks in online examples

Author: c-randall

docs/source/examples/robertson.ipynb

@@ -38,6 +38,8 @@
    "outputs": [],
    "source": [
     "import numpy as np\n",
+    "import numpy.testing as npt\n",
+    "\n",
     "import sksundae as sun\n",
     "import matplotlib.pyplot as plt\n",
     "\n",
@@ -76,9 +78,9 @@
     "soln.y[:,1] *= 1e4  # scale the y1 values for plotting\n",
     "\n",
     "plt.semilogx(soln.t, soln.y)\n",
-    "plt.legend(['y0', 'y1', 'y2'])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(\"concentration\");"
+    "plt.legend([\"y0\", \"y1\", \"y2\"])\n",
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"concentration, $c$\");"
    ]
   },
   {
@@ -114,9 +116,9 @@
     "soln.y[:,1] *= 1e4  # scale the y1 values for plotting\n",
     "\n",
     "plt.semilogx(soln.t, soln.y)\n",
-    "plt.legend(['y0', 'y1', 'y2'])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(\"concentration\");"
+    "plt.legend([\"y0\", \"y1\", \"y2\"])\n",
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"concentration, $c$\");"
    ]
   },
   {
@@ -126,7 +128,7 @@
     "## Step-wise solutions\n",
     "Solving step-by-step instead of across a full time span can be beneficial in some cases, especially for debugging. Therefore, a `step` method is also available in `IDA`. Before taking a step, the solver needs to know the initial conditions and time to determine the direction of integration for the following steps. Thus, before calling `step`, you should call `init_step`, as shown below. The initialization is handled automatically when using the `solve` method but must be done manually in a step-by-step approach. Since we are using the same solver instance that we initialized above, the `init_step` call will also provide a correction to `yp0` as specified by the `calc_initcond` option.\n",
     "\n",
-    "Below, we run `init_step` and compare the solution `soln_0` to the initial values from the full solve, from above. Afterward, we take a step evaluated at `soln.t[10]` using the solution object from the full solve, allowing us to compare the step-by-step solution to a portion of the full solution. We only check that the solutions are within some tolerance (`1e-6`) because the solver's internal steps may differ from those in the full solution, meaning the values will be close but may not be exactly the same."
+    "Below, we run `init_step` and compare the solution `soln_0` to the initial values from the full solve, from above. Afterward, we take a step evaluated at `soln.t[10]` using the solution object from the full solve, allowing us to compare the step-by-step solution to a portion of the full solution. We only check that the solutions are within some tolerance (`1e-7`) because the solver's internal steps may differ from those in the full solution, meaning the values will be close but may not be exactly the same."
    ]
   },
   {
@@ -140,14 +142,14 @@
     "soln_0 = solver.init_step(tspan[0], y0, yp0)\n",
     "print(soln_0)\n",
     "\n",
-    "assert soln_0.t == soln.t[0]\n",
-    "assert np.all(soln_0.y - soln.y[0] < 1e-6)\n",
+    "npt.assert_allclose(soln.t[0], soln_0.t)\n",
+    "npt.assert_allclose(soln.y[0], soln_0.y, atol=1e-7)\n",
     "\n",
     "soln_1 = solver.step(soln.t[10])\n",
     "print(soln_1)\n",
     "\n",
-    "assert soln_1.t == soln.t[10]\n",
-    "assert np.all(soln_1.y - soln.y[10] < 1e-6)"
+    "npt.assert_allclose(soln.t[10], soln_1.t)\n",
+    "npt.assert_allclose(soln.y[10], soln_1.y, atol=1e-7)"
    ]
   },
   {
@@ -190,9 +192,9 @@
     "soln.y_events[:,1] *= 1e4  # scale the y1 values for plotting\n",
     "\n",
     "plt.semilogx(soln.t, soln.y, '-', soln.t_events, soln.y_events, 'or')\n",
-    "plt.legend(['y0', 'y1', 'y2'])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(\"concentration\");\n",
+    "plt.legend([\"y0\", \"y1\", \"y2\"])\n",
+    "plt.xlabel(\"time, $t$\")\n",
+    "plt.ylabel(\"concentration, $c$\")\n",
     "\n",
     "for t in soln.t_events:\n",
     "    plt.axvline(t, linestyle='--', color='k')"
@@ -242,9 +244,9 @@
     "soln.y[:,1] *= 1e4  # scale the y1 values for plotting\n",
     "\n",
     "plt.semilogx(soln.t, soln.y)\n",
-    "plt.legend(['y0', 'y1', 'y2'])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(\"concentration\");"
+    "plt.legend([\"y0\", \"y1\", \"y2\"])\n",
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"concentration, $c$\");"
    ]
   },
   {
@@ -273,7 +275,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.12.5"
+   "version": "3.14.0"
   }
  },
  "nbformat": 4,

docs/source/examples/van_der_pol.ipynb

@@ -35,6 +35,8 @@
    "outputs": [],
    "source": [
     "import numpy as np\n",
+    "import numpy.testing as npt\n",
+    "\n",
     "import sksundae as sun\n",
     "import matplotlib.pyplot as plt\n",
     "\n",
@@ -69,8 +71,8 @@
     "print(soln)\n",
     "\n",
     "plt.plot(soln.t, soln.y[:,0])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(r\"$x$\");"
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"position, $x$\");"
    ]
   },
   {
@@ -80,7 +82,7 @@
     "## Step-wise solutions\n",
     "Solving step-by-step instead of across a full time span can be beneficial in some cases, especially for debugging. Therefore, a `step` method is also available in `CVODE`. Before taking a step, the solver needs to know the initial conditions and time to determine the direction of integration for the following steps. Thus, before calling `step`, you should call `init_step`, as shown below. The initialization is handled automatically when using the `solve` method but must be done manually in a step-by-step approach.\n",
     "\n",
-    "Below, we run `init_step` and compare the solution `soln_0` to the initial values from the full solve, from above. Afterward, we take a step evaluated at `soln.t[10]` using the solution object from the full solve, allowing us to compare the step-by-step solution to a portion of the full solution. We only check that the solutions are within some tolerance (`1e-6`) because the solver's internal steps may differ from those in the full solution, meaning the values will be close but may not be exactly the same."
+    "Below, we run `init_step` and compare the solution `soln_0` to the initial values from the full solve, from above. Afterward, we take a step evaluated at `soln.t[10]` using the solution object from the full solve, allowing us to compare the step-by-step solution to a portion of the full solution. We only check that the solutions are within some tolerance (`1e-7`) because the solver's internal steps may differ from those in the full solution, meaning the values will be close but may not be exactly the same."
    ]
   },
   {
@@ -92,14 +94,14 @@
     "soln_0 = solver.init_step(0, y0)\n",
     "print(soln_0)\n",
     "\n",
-    "assert soln_0.t == soln.t[0]\n",
-    "assert np.all(soln_0.y - soln.y[0] < 1e-6)\n",
+    "npt.assert_allclose(soln_0.t, soln.t[0])\n",
+    "npt.assert_allclose(soln_0.y, soln.y[0], atol=1e-7)\n",
     "\n",
     "soln_1 = solver.step(soln.t[10])\n",
     "print(soln_1)\n",
     "\n",
-    "assert soln_1.t == soln.t[10]\n",
-    "assert np.all(soln_1.y - soln.y[10] < 1e-6)"
+    "npt.assert_allclose(soln_1.t, soln.t[10])\n",
+    "npt.assert_allclose(soln_1.y, soln.y[10], atol=1e-7)"
    ]
   },
   {
@@ -142,8 +144,8 @@
     "print(soln)\n",
     "\n",
     "plt.plot(soln.t, soln.y[:,0], '-', soln.t_events, soln.y_events[:,0], 'or')\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(r\"$x$\");"
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"position, $x$\");"
    ]
   },
   {
@@ -181,8 +183,8 @@
     "print(soln)\n",
     "\n",
     "plt.plot(soln.t, soln.y[:,0])\n",
-    "plt.xlabel(r\"$t$\");\n",
-    "plt.ylabel(r\"$x$\");"
+    "plt.xlabel(\"time, $t$\");\n",
+    "plt.ylabel(\"position, $x$\");"
    ]
   },
   {
@@ -211,7 +213,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.12.4"
+   "version": "3.14.0"
   }
  },
  "nbformat": 4,