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    "(List_comprehensions)=\n",
    "# List comprehensions\n",
    "``` {index} List comprehension (Python)\n",
    "```\n",
    "List comprehensions are 'shortcuts' in creating lists using loops within one statement. The main advantage is that they are shorter to write. The syntax is:\n",
    "\n",
    "```python\n",
    "    some_list = [expression for item in iterable_object if condition]\n",
    "```\n",
    "    \n",
    "This will generate a list with an expression based on items in 'iterable_object', if the condition is met. The `if` condition is optional."
   ]
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  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[1, 4, 7, 10, 13, 16, 19]\n",
      "[1, 64, 343, 1000, 2197, 4096, 6859]\n",
      "[1, 4, 7, 10, 13, 16, 19]\n",
      "[1, 64, 343, 1000, 2197, 4096, 6859]\n"
     ]
    }
   ],
   "source": [
    "# List comprehensions\n",
    "\n",
    "normal_numbers = [number for number in range(1, 20, 3)]\n",
    "print(normal_numbers)\n",
    "\n",
    "cubic_numbers = [number**3 for number in range(1, 20, 3)]\n",
    "print(cubic_numbers)\n",
    "\n",
    "# These are equivalent to \n",
    "\n",
    "normal_numbers = []\n",
    "cubic_numbers = []\n",
    "\n",
    "for number in range(1, 20, 3):\n",
    "    normal_numbers.append(number)\n",
    "    cubic_numbers.append(number**3)\n",
    "    \n",
    "print(normal_numbers)\n",
    "print(cubic_numbers)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The expression in list comprehension can be anything, the same value for each item, a string...:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[0, 0, 0, 0, 0]\n",
      "['text', 'text', 'text', 'text', 'text']\n"
     ]
    }
   ],
   "source": [
    "zeros_list = [ 0 for i in range(5)]\n",
    "text_list = [\"text\" for i in range(5)]\n",
    "\n",
    "print(zeros_list)\n",
    "print(text_list)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Using `if` condition:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]\n",
      "[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]\n"
     ]
    }
   ],
   "source": [
    "# List comprehension with if statement\n",
    "# number%2 == 1 means to search for numbers whose \n",
    "# remainder from division by 2 is 1 (odd numbers)\n",
    "\n",
    "odd_numbers = [ number for number in range(20) if number%2 == 1]\n",
    "print(odd_numbers)\n",
    "\n",
    "# Equivalent to\n",
    "\n",
    "odd_numbers = []\n",
    "for number in range(20):\n",
    "    if number%2 == 1:\n",
    "        odd_numbers.append(number)\n",
    "        \n",
    "print(odd_numbers)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "(list_comprehensions_exercises)=\n",
    "## Exercises:\n",
    "_____________\n",
    "* **Create a list comprehension** to find Celsius temperatures for Fahrenheit temperatures between $20^\\circ$F and $80^\\circ$F at $5^\\circ$F increments. The Fahrenheit-Celsius conversion formula is:\n",
    "\n",
    "\n$$ T_{Celsius} = \\frac{5}{9}*(T_{Fahrenheit}-32). $$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "````{admonition} Answer\n",
    ":class: dropdown\n",
    "\n",
    "```python\n",
    "temp_C = [ 5./9.*(i-32) for i in range(20, 81, 5)]\n",
    "\n",
    "print(temp_C)\n",
    "```\n",
    "\n",
    "````"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* **Create a list** of powers of two $2^x$ where $x$ is positive, odd, not divisible by 3 and smaller than 40. Iterate through that list and print the last integer of each element. Numbers tend to show up patterns in such analysis, can you spot one?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "````{admonition} Answer\n",
    ":class: dropdown\n",
    "\n",
    "```python\n",
    "l = [2**x for x in range(40) if x % 2 == 1 and not(x % 3 == 0)]\n",
    "\n",
    "for i in l:\n",
    "    print(i%10)  \n",
    "```\n",
    "\n",
    "````"
   ]
  }
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