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Fixes soyHenry#14
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‎03 - Flujos de Control/mis_soluciones_homework_3.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": []
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"1) Crear una variable que contenga un elemento del conjunto de números enteros y luego imprimir por pantalla si es mayor o menor a cero"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"-10 El numero es menor que cero\n"
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]
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}
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],
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"source": [
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"num = -10\n",
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"if (num > 0):\n",
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" print(num,' El numero es mayor a cero')\n",
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"elif (num < 0):\n",
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" print(num,' El numero es menor que cero')\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"2) Crear dos variables y un condicional que informe si son del mismo tipo de dato"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 9,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"No son del mismo tipo\n"
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]
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}
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],
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"source": [
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"var1 = '4'\n",
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"var2 = 2\n",
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"if (type (var1) == type (var2)):\n",
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" print( var1, 'y', var2, ' son del mismo tipo')\n",
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"else:\n",
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" print ('No son del mismo tipo')"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"3) Para los valores enteros del 1 al 20, imprimir por pantalla si es par o impar\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1 Es impar\n",
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"2 Es par\n",
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"3 Es impar\n",
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"4 Es par\n",
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"5 Es impar\n",
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"6 Es par\n",
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"7 Es impar\n",
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"8 Es par\n",
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"9 Es impar\n",
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"10 Es par\n",
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"11 Es impar\n",
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"12 Es par\n",
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"13 Es impar\n",
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"14 Es par\n",
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"15 Es impar\n",
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"16 Es par\n",
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"17 Es impar\n",
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"18 Es par\n",
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"19 Es impar\n",
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"20 Es par\n"
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]
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}
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],
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"source": [
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"for n in range(1,21):\n",
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" if (n % 2 == 0):\n",
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" print (n, ' Es par')\n",
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" else:\n",
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" print(n, ' Es impar')"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"4) En un ciclo for mostrar para los valores entre 0 y 5 el resultado de elevarlo a la potencia igual a 3"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 12,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"0 Elevado a la 3 es 0\n",
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"1 Elevado a la 3 es 1\n",
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"2 Elevado a la 3 es 8\n",
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"3 Elevado a la 3 es 27\n",
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"4 Elevado a la 3 es 64\n",
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"5 Elevado a la 3 es 125\n"
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]
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}
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],
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"source": [
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"for n in range(0,6):\n",
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" print (n, ' Elevado a la 3 es ', (n ** 3))"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"5) Crear una variable que contenga un número entero y realizar un ciclo for la misma cantidad de ciclos"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 13,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"0\n",
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"1\n",
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"2\n",
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"3\n",
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"4\n",
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"5\n",
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"6\n",
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"7\n",
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"8\n",
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"9\n"
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]
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}
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],
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"source": [
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"var = 10\n",
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"for var in range (var):\n",
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" print (var)\n",
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" var -= 1\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"6) Utilizar un ciclo while para realizar el factorial de un número guardado en una variable, sólo si la variable contiene un número entero mayor a 0"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"7) Crear un ciclo for dentro de un ciclo while"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 70,
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"metadata": {},
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"outputs": [],
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"source": [
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"n=0\n",
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"while (n > 3):\n",
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" n += 1\n",
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" for i in range (1,n):\n",
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" print (i,' ciclo For')\n",
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" print ('Ciclo while' , n)\n",
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" \n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = 0\n",
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"while(n < 5):\n",
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" n += 1\n",
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" for i in range(1,n):\n",
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" print('Ciclo while nro ' + str(n))\n",
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" print('Ciclo for nro ' + str(i))"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"8) Crear un ciclo while dentro de un ciclo for"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 57,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = 5\n",
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"for i in range (1,n):\n",
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" while (n < 5):\n",
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" n -= 1\n",
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" print('Ciclo ', n)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 56,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = 5\n",
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"for i in range(1, n):\n",
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" while(n < 5):\n",
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" n -= 1\n",
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" print('Ciclo while nro ' + str(n))\n",
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" print('Ciclo for nro ' + str(i))"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"9) Imprimir los números primos existentes entre 0 y 30"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"10) ¿Se puede mejorar el proceso del punto 9? Utilizar las sentencias break y/ó continue para tal fin"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.1"
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},
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "89a5003fa4d04edff07a126a2c6f2ee59ee1adb31e9fcd60b29f2b77c18232ea"
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}
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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