Сократить словарь операторов if

Вы можете использовать цикл for и использовать функцию Python, которая for может быть объединена с else, где часть else выполняется, если не было выполнено ни одного оператора break.

for i in students:
    # Check credentials if it's correct on the right email and password.
    if (email == students[i]["email"] and passwords == students[i]["password"]):
        print("Welcome", students[i]["name"], students[i]["lastname"], "you are successfully logged in!")
        break
    elif count > 5:
        tijd = time.localtime()
        clock = time.strftime("%I:%M:%S %p", tijd)
        print("On", clock, "You have logged incorrectly 5 times. You are blocked for 15 minutes")
        # time.sleep(900) #Sleep 15 minutes.
        break
else:
    print("Incorrect E-mail or password!")
    print("")

Если вы можете изменить структуру данных, вы можете использовать адрес электронной почты в качестве ключа словаря вместо числа.

students = {  # Make a static dictenoary called students
    "[email protected]": {"name": "Daniel", "lastname": "maker", "email": "[email protected]", "password": "123456"},
    "[email protected]": {"name": "Sandy", "lastname": "Mango", "email": "[email protected]", "password": "sdasdas!@"},
    "[email protected]": {"name": "Kreeft", "lastname": "Garnaal", "email": "[email protected]", "password": "Mekalekkea!@"},
    "[email protected]": {"name": "Alfred", "lastname": "Skylake", "email": "[email protected]", "password": "suiker!@"}
}

[...]
    try:
        student = students[email]
        if passwords == student["password"]:
            print("Welcome", student["name"], student["lastname"], "you are successfully logged in!")
            break
        elif count > 5:
            tijd = time.localtime()
            clock = time.strftime("%I:%M:%S %p", tijd)
            print("On", clock, "You have logged incorrectly 5 times. You are blocked for 15 minutes")
            # time.sleep(900) #Sleep 15 minutes.
            break
        else:
            print("Incorrect E-mail or password!")
            print("")
    except KeyError:
        print("Incorrect E-mail or password!")
        print("")

Вы, конечно, не хотите хранить пароли в виде простого текста и вместо этого использовать хэш.

Используйте понимание списка

selected_students = [s for s in students if s["email"] == email and s["password"] == password]

Если len(seleced_students) == 0, то вы ничего не нашли.

Это не сломается и всегда будет проходить через полный dict, но он более питонический, чем for, а также находит дубликаты.

Вы можете заменить свои операторы if-else простым циклом for:

import time 

students = { # Make a static dictenoary called students
    1: {"name": "Daniel","lastname": "maker","email" : "[email protected]","password" : "123456"},
    2: {"name": "Sandy","lastname": "Mango","email" : "[email protected]","password" : "sdasdas!@"},
    3: {"name": "Kreeft","lastname": "Garnaal","email" : "[email protected]","password" : "Mekalekkea!@"},
    4: {"name": "Alfred","lastname": "Skylake","email" : "[email protected]","password" :"suiker!@"}
}


count = 0
while True:
    email = input("\nEnter your E-mail account: ")
    passwords = input("Enter your password: ")
    count += 1 #count the number of failed login attempts
    login = False
    print ("")
    
    for student in students:
        if students[student]["email"] == email:
            if students[student]["password"] == passwords:
                print (f"Welcome, {students[student]["name"]},{students[student]["lastname"]}, you are successfully logged in!")
                count = 0
                login = True
                break
    if not login:
        print ("Incorrect E-mail or password!\n")
    if count > 5:
        current_time = time.localtime()
        clock = time.strftime("%I:%M:%S %p", current_time)
        print("On", clock, "You have logged incorrectly 5 times. You are blocked for 15 minutes")
        #time.sleep(900) #Sleep 15 minutes.
        #count = 0 
        break