Skip to content

lab-tuple-set-dict[Juan_Carlos_Gutiérrez_Valle] #41

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
juan1309-ctr wants to merge 1 commit into
base: master
Choose a base branch
from
Open

lab-tuple-set-dict[Juan_Carlos_Gutiérrez_Valle] #41

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
6 changes: 6 additions & 0 deletions your-code/.ipynb_checkpoints/Untitled-checkpoint.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,6 @@
{
"cells": [],
"metadata": {},
"nbformat": 4,
"nbformat_minor": 5
}
372 changes: 372 additions & 0 deletions your-code/.ipynb_checkpoints/challenge-1-checkpoint.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,372 @@
{
"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": 52,
"metadata": {},
"outputs": [],
"source": [
"tup= (\"I\",)\n"
]
},
{
"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": 53,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"tuple"
]
},
"execution_count": 53,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"type(tup)\n"
]
},
{
"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": 54,
"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;32mC:\\Users\\JUANCA~1\\AppData\\Local\\Temp/ipykernel_31256/2921012397.py\u001b[0m in \u001b[0;36m<module>\u001b[1;34m\u001b[0m\n\u001b[1;32m----> 1\u001b[1;33m \u001b[0mtup\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mappend\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m'r'\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 2\u001b[0m \u001b[1;31m#It is not possible to append those elements because a tuple is a inmutable object.\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": [
"tup.append('r')\n",
"#It is not possible to append those elements because a tuple is a inmutable object."
]
},
{
"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": 58,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"('I', 'r', 'o', 'n', 'h', 'a', 'c', 'k')"
]
},
"execution_count": 58,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"lista= list(tup)\n",
"\n",
"lista.append('r')\n",
"lista.append('o')\n",
"lista.append('n')\n",
"lista.append('h')\n",
"lista.append('a')\n",
"lista.append('c')\n",
"lista.append('k')\n",
"\n",
"tup=tuple(lista)\n",
"tup\n",
"\n",
"\n",
"\n",
"\n",
"#Se tiene que transformar la tupla en lista y, posteriormente, agregar uno por uno los elementos que pide el ejercicio mediante la función append. "
]
},
{
"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": 68,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n')\n",
"('h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"tup=list(tup)\n",
"tup\n",
"\n",
"tup1=tup[0:4]\n",
"tup2=tup[4:]\n",
"\n",
"tup1=tuple(tup1)\n",
"tup2=tuple(tup2)\n",
"\n",
"print(tup1)\n",
"print(tup2)\n",
"\n",
"\n"
]
},
{
"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": 70,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n', 'h', 'a', 'c', 'k')\n"
]
}
],
"source": [
"tup3= tup1+tup2\n",
"print(tup3)\n"
]
},
{
"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": 73,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Tienen el mismo número de elementos\n"
]
}
],
"source": [
"\n",
"a=len(tup1)\n",
"b=len(tup2)\n",
"c=len(tup3)\n",
"\n",
"if a+b==c:\n",
" print(\"Tienen el mismo número de elementos\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### What is the index number of `\"h\"` in `tup3`?"
]
},
{
"cell_type": "code",
"execution_count": 74,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"4"
]
},
"execution_count": 74,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"tup3.index('h')\n",
"#El ínide de h es el 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": 77,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a is true\n",
"b es false\n",
"c is true\n",
"d es false\n",
"e es false\n"
]
}
],
"source": [
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"\n",
"for i in letters:\n",
" if i in tup3:\n",
" print( i, 'is true')\n",
" else:\n",
" print(i, 'es 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": 88,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"la letra i se repite 1 veces\n",
"la letra i se repite 1 veces\n"
]
}
],
"source": [
"\n",
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"contador=0\n",
"\n",
"for i in letters:\n",
" num=tup3.count(i)\n",
" if num>0:\n",
" print('la letra i se repite', num, 'veces')"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.7"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
Loading