numpy tasks

core/numpy/numpy-basics.ipynb

@@ -981,6 +981,176 @@
     "to investigate every preceding dimension along our the last entry of our last axis, the same as `c[:, :, -1]`."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Numpy exercises\n",
+    "This block exists to add more practical exercises for the students. The exercises are not required but they can be very helpful for undestanding the subject. \n",
+    "\n",
+    "### Q1\n",
+    "Write a function that finds the sum of even diagonal elements of a square matrix. If there are no such elements, then print 0."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def np_diag_2k(a):\n",
+    "    # YOUR CODE\n",
+    "    return None"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# GIVEN CODE\n",
+    "a = np.random.randint(1, 10, size=(10, 10))\n",
+    "a"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "np_diag_2k(a)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "<details>\n",
+    "    <summary>Answer</summary>\n",
+    "\n",
+    "```python\n",
+    "def np_diag_2k(a):\n",
+    "    diag = a.diagonal()\n",
+    "    return np.sum(diag[diag % 2 == 0])\n",
+    "```\n",
+    "</details>"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Q2\n",
+    "\n",
+    "Write a function that, using a given sequence $\\{A_i\\}_{i=1}^n$, builds a sequence $S_n$, where $S_k = \\frac{A_1+ ... + A_k}{k}$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# ANSWER\n",
+    "def np_sec_av(A):\n",
+    "    # YOUR CODE\n",
+    "    return None"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# GIVEN CODE\n",
+    "import scipy.stats as sps\n",
+    "\n",
+    "A = sps.uniform.rvs(size=10**3)\n",
+    "\n",
+    "np_sec_av(A)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "<details>\n",
+    "    <summary>Answer</summary>\n",
+    "\n",
+    "```python\n",
+    "# ANSWER\n",
+    "def np_sec_av(A):\n",
+    "    return sum(A)/len(A)\n",
+    "```\n",
+    "</details>"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Q3\n",
+    "\n",
+    "A two-dimensional array $X$ is specified. For each row of the array X, the following transformation must be performed.\n",
+    "\n",
+    "Let the line x be given. It is necessary to build a new array, where all elements with odd indexes must be replaced with the number a (default value a=1). All elements with even indexes must be cubed. Then write down the elements in reverse order relative to their positions. At the end, you need to merge the array x with the transformed x and output it.\n",
+    "\n",
+    "Write a function that performs this transformation for each row of a two-dimensional array X. Array X should remain unchanged at the same time.\n",
+    "\n",
+    "Use the numpy library.\n",
+    "\n",
+    "Example:\n",
+    "$X = [[100,200,300,400,500]]$ -> $[[100, a,300,a,500]]$ -> $[[500^3, a,300^3,a,100^3]]$ -> glue -> $[[100,200,300,400,500,500^3,a,300^3,a,100^3]]$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# ANSWER\n",
+    "from copy import copy\n",
+    "\n",
+    "def transform(X, a=1):\n",
+    "    # YOUR CODE\n",
+    "    return None"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# GIVEN CODE\n",
+    "X = np.array([[100, 200, 300, 400, 500, 600], [200, 300, 500, 22, 11, 17]])\n",
+    "\n",
+    "S2 = transform(X)\n",
+    "print(S2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "<details>\n",
+    "    <summary>Answer</summary>\n",
+    "\n",
+    "```python\n",
+    "# ANSWER\n",
+    "def transform(X, a=1):\n",
+    "    Y = np.copy(X)\n",
+    "    Y[:,1::2] = a\n",
+    "    Y[:,0::2] **= 3\n",
+    "    return np.hstack((X, Y[:,::-1]))\n",
+    "```\n",
+    "</details>"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},