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agates696 Oct 20, 2020
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Áine Gates
agates696 Oct 20, 2020
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369 changes: 369 additions & 0 deletions challenge-1 Áine Gates.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Challenge 1: Tuples\n",
"\n",
"#### Do you know you can create tuples with only one element?\n",
"\n",
"**In the cell below, define a variable `tup` with a single element `\"I\"`.**\n",
"\n",
"*Hint: you need to add a comma (`,`) after the single element. The reason for this is that a tuple is defined as a series of comma-separated values. The parenthesis are not actually needed and are used simply for clarity. Therefore if we try to create a tuple without using a comma Python will simply interpret the element as a single element, be it a string, int or other types.*"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"tup = (\"I\",)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Print the type of `tup`. \n",
"\n",
"Make sure its type is correct (i.e. *tuple* instead of *str*)."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"tuple"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"type(tup)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Now try to append the following elements to `tup`. \n",
"\n",
"Are you able to do it? Explain.\n",
"\n",
"```\n",
"\"r\", \"o\", \"n\", \"h\", \"a\", \"c\", \"k\"\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"scrolled": true
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"['I', 'r', 'o', 'n', 'h', 'a', 'c', 'k']\n"
]
}
],
"source": [
"listed_tup = list(tup)\n",
"elements = [\"r\", \"o\", \"n\", \"h\", \"a\", \"c\", \"k\"]\n",
"full_list = listed_tup + elements\n",
"print(full_list)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"First make tuple a list. Create a list of the elements and then combine the two lists into one."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### How about re-assign a new value to an existing tuple?\n",
"\n",
"Re-assign the following elements to `tup`. Are you able to do it? Explain.\n",
"\n",
"```\n",
"\"I\", \"r\", \"o\", \"n\", \"h\", \"a\", \"c\", \"k\"\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n', 'h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"tup = tuple(full_list)\n",
"print(tup)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Yes, convert the list back to a tuple."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Split `tup` into `tup1` and `tup2` with 4 elements in each. \n",
"\n",
"`tup1` should be `(\"I\", \"r\", \"o\", \"n\")` and `tup2` should be `(\"h\", \"a\", \"c\", \"k\")`.\n",
"\n",
"*Hint: use positive index numbers for `tup1` assignment and use negative index numbers for `tup2` assignment. Positive index numbers count from the beginning whereas negative index numbers count from the end of the sequence.*\n",
"\n",
"Also print `tup1` and `tup2`."
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n')\n",
"('h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"list_ = list(tup)\n",
"tup1_list = list_[0:4]\n",
"tup2_list = list_[4:8]\n",
"tup1 = tuple(tup1_list)\n",
"tup2 = tuple(tup2_list)\n",
"print(tup1)\n",
"print(tup2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Add `tup1` and `tup2` into `tup3` using the `+` operator.\n",
"\n",
"Then print `tup3` and check if `tup3` equals to `tup`."
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n', 'h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"tup3 = tup1 + tup2\n",
"print(tup3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Count the number of elements in `tup1` and `tup2`. Then add the two counts together and check if the sum is the same as the number of elements in `tup3`."
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"tup1_len = len(tup1)\n",
"tup2_len = len(tup2)\n",
"sum_of_len_tups = (tup1_len + tup2_len)\n",
"len_tup3 = len(tup3)\n",
"len_tup3 == sum_of_len_tups"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### What is the index number of `\"h\"` in `tup3`?"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"'h'"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"tup3[-4]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Now, use a FOR loop to check whether each letter in the following list is present in `tup3`:\n",
"\n",
"```\n",
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"```\n",
"\n",
"For each letter you check, print `True` if it is present in `tup3` otherwise print `False`.\n",
"\n",
"*Hint: you only need to loop `letters`. You don't need to loop `tup3` because there is a Python operator `in` you can use. See [reference](https://stackoverflow.com/questions/17920147/how-to-check-if-a-tuple-contains-an-element-in-python).*"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"True\n",
"False\n",
"True\n",
"False\n",
"False\n"
]
}
],
"source": [
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"\n",
"for i in letters: \n",
" if i in tup3:\n",
" print(True)\n",
" else: \n",
" print(False)\n",
" \n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### How many times does each letter in `letters` appear in `tup3`?\n",
"\n",
"Print out the number of occurrence of each letter."
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"{'a': 1, 'b': 0, 'c': 1, 'd': 0, 'e': 0}\n"
]
}
],
"source": [
"appearances = {}\n",
"\n",
"for i in letters: \n",
" x = tup3.count(i)\n",
" appearances[i] = x\n",
"print(appearances)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**BONUS: use tuple unpacking together with the** `range()` **function to print all the integers from 10 to 20.**"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [],
"source": [
"# Your code here\n"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.8.6"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
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