Reto #8 - Python

Retos/Reto #10 - LA API [Media]/python/Paula2409.py

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+"""/*
+ * Llamar a una API es una de las tareas más comunes en programación.
+ *
+ * Implementa una llamada HTTP a una API (la que tú quieras) y muestra su
+ * resultado a través de la terminal. Por ejemplo: Pokémon, Marvel...
+ *
+ * Aquí tienes un listado de posibles APIs: 
+ * https://github.com/public-apis/public-apis
+ */
+"""
+import requests
+
+def request_http():
+    request = requests.get('https://developer.oxforddictionaries.com/')
+    print(request.status_code) # prints 200 (ok)
+    print(request.headers) # print(headers of API)
+    print(request.json) # <bound method Response.json of <Response [200]>>
+    print(request.text) # text HTML and CSS
+
+request_http()

Retos/Reto #6 - PIEDRA, PAPEL, TIJERA, LAGARTO, SPOCK [Media]/python/Paula2409.py

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+"""
+/*
+ * Crea un programa que calcule quien gana más partidas al piedra,
+ * papel, tijera, lagarto, spock.
+ * - El resultado puede ser: "Player 1", "Player 2", "Tie" (empate)
+ * - La función recibe un listado que contiene pares, representando cada jugada.
+ * - El par puede contener combinaciones de "🗿" (piedra), "📄" (papel),
+ *   "✂️" (tijera), "🦎" (lagarto) o "🖖" (spock).
+ * - Ejemplo. Entrada: [("🗿","✂️"), ("✂️","🗿"), ("📄","✂️")]. Resultado: "Player 2".
+ * - Debes buscar información sobre cómo se juega con estas 5 posibilidades.
+ */
+"""
+game_rules = {
+    '🗿': ['🦎','✂️'], # rock crushes lizard and scissors
+    '📄': ['🗿','🖖'], # paper covers rock and invalidates spock
+    '✂️': ['📄','🦎'], # scissors cuts paper and beheads lizard
+    '🦎': ['🖖', '📄'], # lizard poison spock and eats paper 
+    '🖖': ['✂️', '🗿'] # spock breaks scissors and crushes rock
+    }
+
+def rock_paper_scissors(play):
+    """
+    Determines the winner of a game based on the rules of 
+    'rock, paper, scissors, lizard, spock'.
+
+    Args:
+        play(list): a list with the option of each player
+
+    Returns:
+        str: returns which player wins
+    """
+    points_player1, points_player2 = 0,0
+    index_play = 0
+    while index_play < len(play):
+        if play[index_play][0] in game_rules[play[index_play][1]]:
+            points_player2 += 1
+            print(f'Points player 2: {points_player2}')
+        else:
+            points_player1 += 1
+            print(f'Points player 1: {points_player1}')
+
+        index_play += 1
+    if points_player1 > points_player2:
+        return 'Player 1'
+    elif points_player2 > points_player1:
+        return 'Player 2'
+    else:
+        return 'Tie'
+
+print(rock_paper_scissors([("🗿","✂️"), ("✂️","🗿"), ("📄","✂️")]))
+
+

Retos/Reto #8 - EL GENERADOR PSEUDOALEATORIO [Media]/python/Paula2409.py

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+import datetime
+
+now = datetime.datetime.now().microsecond
+print(now)
+print(now%101)

Retos/Reto #9 - HETEROGRAMA, ISOGRAMA Y PANGRAMA [Fácil]/python/Paula2409.py

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+"""
+/*
+ * Crea 3 funciones, cada una encargada de detectar si una cadena de
+ * texto es un heterograma, un isograma o un pangrama.
+ * - Debes buscar la definición de cada uno de estos términos.
+ */
+"""
+import string 
+
+def evaluate_text():
+    """
+    The function evaluates whether a given text is a heterogram, isogram, or pangram based on the
+    occurrence of letters in the text.
+
+    Args:
+        text (str): any text given by user
+
+    Returns:
+        str: returns if the text is a heterogram, an isogram and a pangram.
+    """    
+    # Input a word or phrase   
+    text = input("Ingrese un texto o palabra para evaluar: ")
+    # Process the input
+    new_text = text.lower().replace('á', 'a').replace('é', 'e').replace('í', 'i').replace('ó', 'o').replace('ú', 'u').replace(' ', '')
+    simbols = string.punctuation
+
+    letters_present = {}
+
+    for char in new_text:
+        if char in simbols:
+            new_text = new_text.replace(char, "")
+        else:
+            if char in letters_present:
+                letters_present[char] += 1
+            else:
+                letters_present[char] = 1
+
+    values = list(letters_present.values())
+
+    def is_heterogram():
+        # Heterogram: is a word, phrase or sentence in which no letter of the alphabet occurs more than once.
+        if len(set(values)) == 1:
+            return f"The text '{text}' is a heterogram"
+        else:
+            return f"The text '{text}' is not a heterogram"
+
+
+    def is_isogram():
+        # Isogram: word or phrase in which each letter occurs the same number of times.
+        for value in values:
+            if value != values[0]:
+                return f"The text '{text}' is an Isogram"
+        return f"The text '{text}' is not an Isogram"
+
+    def is_pangrama():
+        # Pangrama: word, phrase or sentence in which all letters of the alphabet are present.
+        letters = list(string.ascii_lowercase) + ['ñ']
+        for letter in letters:
+            if letter in text:
+                letters.remove(letter)
+        if len(letters) == 0:
+            return f"The text '{text}' is a pangram"
+        else:
+            return f"The text '{text}' is not a pangram"
+
+    print(is_heterogram())
+    print(is_isogram())
+    print(is_pangrama())
+
+evaluate_text()