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378 changes: 378 additions & 0 deletions your-code/.ipynb_checkpoints/challenge-1-checkpoint.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.*"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"# Your code here\n",
"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": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"<class 'tuple'>\n"
]
}
],
"source": [
"# Your code here\n",
"print(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": {},
"outputs": [
{
"ename": "AttributeError",
"evalue": "'tuple' object has no attribute 'append'",
"output_type": "error",
"traceback": [
"\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
"\u001b[1;31mAttributeError\u001b[0m Traceback (most recent call last)",
"\u001b[1;32m<ipython-input-3-0d1fdd397c6d>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m\u001b[0m\n\u001b[0;32m 1\u001b[0m \u001b[1;31m# Your code here\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m----> 2\u001b[1;33m \u001b[0mtup\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mappend\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m[\u001b[0m\u001b[1;34m\"r\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"o\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"n\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"h\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"a\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"c\"\u001b[0m\u001b[1;33m,\u001b[0m \u001b[1;34m\"k\"\u001b[0m\u001b[1;33m]\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 3\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 4\u001b[0m \u001b[1;31m# Your explanation here\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
"\u001b[1;31mAttributeError\u001b[0m: 'tuple' object has no attribute 'append'"
]
}
],
"source": [
"# Your code here\n",
"tup.append([\"r\", \"o\", \"n\", \"h\", \"a\", \"c\", \"k\"])\n",
"\n",
"# Your explanation here\n",
"# Tuples son tipos de variables inmutables"
]
},
{
"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": [
"# Your code here\n",
"tup = (\"I\", \"r\", \"o\", \"n\", \"h\", \"a\", \"c\", \"k\")\n",
"print(tup)\n",
"\n",
"# Your explanation here\n",
"# ya que los tuples son inmitables, lo mejor forma para hacer una modificacion es l reasignacion"
]
},
{
"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": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n')\n",
"('h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"# Your code here\n",
"tup1 = tup[:4]\n",
"tup2 = tup[-4:]\n",
"\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": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n', 'h', 'a', 'c', 'k')\n",
"True\n"
]
}
],
"source": [
"# Your code here\n",
"tup3 = tup1 + tup2\n",
"\n",
"# chechk if tup3 == tup\n",
"print(tup3 == tup)"
]
},
{
"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": 11,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"True\n"
]
}
],
"source": [
"# Your code here\n",
"tup1_len = len(tup1)\n",
"tup2_len = len(tup2)\n",
"\n",
"print( tup1_len + tup2_len == len(tup))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### What is the index number of `\"h\"` in `tup3`?"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"h\n"
]
}
],
"source": [
"# Your code here\n",
"# Index 4\n",
"\n",
"print(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": 15,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a --> True\n",
"b --> False\n",
"c --> True\n",
"d --> False\n",
"e --> False\n"
]
}
],
"source": [
"# Your code here\n",
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"\n",
"for letter in letters:\n",
" print(letter, '-->', letter in tup3)"
]
},
{
"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": 27,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"------------\n",
"a --> 1\n",
"b --> 0\n",
"c --> 1\n",
"d --> 0\n",
"e --> 0\n",
"------------\n",
"a --> 1\n",
"b --> 0\n",
"c --> 1\n",
"d --> 0\n",
"e --> 0\n",
"------------\n",
"a --> 1\n",
"b --> 0\n",
"c --> 1\n",
"d --> 0\n",
"e --> 0\n"
]
}
],
"source": [
"# Your code here\n",
"\n",
"print('------------')\n",
"\n",
"for letter in letters:\n",
" print(letter, '-->', sum([letter == l for l in tup]))\n",
" \n",
"print('------------')\n",
"\n",
"for letter in letters:\n",
" c = 0\n",
" for l in tup:\n",
" if letter == l:\n",
" c += 1\n",
" print(letter, '-->', c)\n",
" \n",
"print('------------')\n",
"for letter in letters:\n",
" print(letter, '-->', tup.count(letter))"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"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": "3.8.8"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
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