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perlacueto Aug 29, 2021
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lab_tuple_set_dict [Perla Cueto]
perlacueto Aug 29, 2021
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392 changes: 392 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": [
"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": 2,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"<class 'tuple'>\n"
]
}
],
"source": [
"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": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('I', 'r', 'o', 'n', 'a', 'c', 'k')\n"
]
}
],
"source": [
"ti=['r','o','n','a','c','k']\n",
"\n",
"lista_1=list(tup)\n",
"\n",
"lista_2=tuple(lista_1+ti)\n",
"\n",
"print(lista_2)\n",
"\n"
]
},
{
"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": [
"parte_1=list(lista_2[0:4])\n",
"\n",
"parte_2=list(lista_2[4:])\n",
"\n",
"parte_3=[\"h\"]\n",
"\n",
"lista_4=tuple(parte_1+parte_3+parte_2)\n",
"\n",
"print(lista_4)"
]
},
{
"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": [
"tup1=lista_4[0:4]\n",
"\n",
"tup2=lista_4[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": 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": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"True\n"
]
}
],
"source": [
"x=len(tup1)\n",
"y=len(tup2)\n",
"\n",
"if (x + y ==len(tup3)):\n",
" \n",
" print(True)\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### What is the index number of `\"h\"` in `tup3`?"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"4\n"
]
}
],
"source": [
"for i in range(len(tup3)):\n",
" \n",
" if tup3[i] == 'h':\n",
" \n",
" print(i)"
]
},
{
"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": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"True a esa letra si esta\n",
"False b esa letra no esta\n",
"True c esa letra si esta\n",
"False d esa letra no esta\n",
"False e esa letra no esta\n"
]
}
],
"source": [
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"\n",
"\n",
"\n",
"for i in letters:\n",
" \n",
" x=0\n",
" \n",
" for k in tup3:\n",
" \n",
" if k == i:\n",
" \n",
" x+=1 \n",
" \n",
" break\n",
" \n",
" else:\n",
" \n",
" x=x\n",
" if x>0:\n",
" \n",
" print(True,i,'esa letra si esta')\n",
" \n",
" else:\n",
" \n",
" print(False,i, 'esa letra no esta')\n",
" \n",
" \n",
" \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": 10,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a esta 1 veces\n",
"b esta 0 veces\n",
"c esta 1 veces\n",
"d esta 0 veces\n",
"e esta 0 veces\n"
]
}
],
"source": [
"letters = [\"a\", \"b\", \"c\", \"d\", \"e\"]\n",
"\n",
"\n",
"\n",
"for i in letters:\n",
" \n",
" x=0\n",
" \n",
" for k in tup3:\n",
" \n",
" if k == i:\n",
" \n",
" x+=1 \n",
" \n",
" break\n",
" \n",
" else:\n",
" \n",
" x=x\n",
" if x>0:\n",
" \n",
" print(i, 'esta',x, 'veces')\n",
" \n",
" else:\n",
" \n",
" print(i, 'esta',x, 'veces')\n",
" \n",
" \n",
" \n"
]
}
],
"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": "3.9.6"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
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