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a7cf6aed5f
Author | SHA1 | Date |
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Christopher Arndt | a7cf6aed5f | |
Christopher Arndt | d44a6725b8 |
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# personclass1.py
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class Person:
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name = ""
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age = 0
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gender = None
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person = Person()
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print(person)
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print(person.__class__)
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print(person.name)
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print(person.age)
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print(person.gender)
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person.name = "Joe Doe"
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person.age = 32
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person.gender = "m"
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print(person)
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print(person.__class__)
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print(person.name)
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print(person.age)
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print(person.gender)
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# personclass1.py
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class Person:
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def __init__(self, name="", age=0, gender=None):
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self.name = name
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self.age = age
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self.gender = gender
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person1 = Person()
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print(person1)
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print(person1.__class__)
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print(person1.name)
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print(person1.age)
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print(person1.gender)
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person2 = Person("Joe Doe", 32, gender="m")
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print(person2)
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print(person2.__class__)
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print(person2.name)
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print(person2.age)
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print(person2.gender)
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@ -0,0 +1,142 @@
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# Ausgangslage
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salaries_1 = {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}
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salaries_2 = {'Barbara': 3000, 'Elke': 3300, 'Michael': 2800, 'Johann': 2000}
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# Aufgabe 1:
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salaries_all = salaries_1.copy()
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salaries_all.update(salaries_2)
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print(salaries_all)
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print("-" * 50)
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# Aufgabe 2:
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employee = ""
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max_salary = 0
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sum_salary = 0
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for name, salary in salaries_all.items():
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if salary > max_salary:
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max_salary = salary
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employee = name
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sum_salary += salary
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print("Topverdiener:", employee, max_salary)
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print("-" * 50)
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# Aufgabe 3:
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print("Summe:", sum_salary)
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# Aufgabe 3 alternativ:
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print("Summe:", sum(salaries_all.values()))
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print("-" * 50)
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# Aufgabe 4:
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salaries_new = {}
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for key in salaries_all:
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salaries_new[key] = salaries_all[key] * 1.1
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# Aufgabe 5:
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heading = "Name Gehalt"
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print(heading)
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print("-" * len(heading))
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for name, salary in salaries_new.items():
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print(name.ljust(8), int(salary))
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print("-" * 50)
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# Aufgabe 6:
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heading = "Name Gehalt"
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print(heading)
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print("-" * len(heading))
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for name, salary in salaries_all.items():
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print(name.ljust(8), int(salary))
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print("-" * 50)
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# Aufgabe 5 & 6 mit Funktion
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def print_salary_list(data, col_width=8):
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heading = "Name".ljust(col_width) + "Gehalt"
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print(heading)
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print("-" * len(heading))
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for name, salary in data.items():
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print(name.ljust(col_width), int(salary))
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print_salary_list(salaries_new, 10)
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print("-" * 50)
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print_salary_list(salaries_all, 10)
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print("-" * 50)
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# Bonusaufgabe 1:
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def print_ordered_salary_list(data, col_width=8):
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heading = "Name".ljust(col_width) + "Gehalt"
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print(heading)
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print("-" * len(heading))
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salary_first = []
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for name, salary in data.items():
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salary_first.append((salary, name))
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for salary, name in sorted(salary_first, reverse=True):
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print(name.ljust(col_width), int(salary))
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print_ordered_salary_list(salaries_new)
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print("-" * 50)
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# Bonusaufgabe 2:
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def print_enumerated_salary_list(data, col_width=8):
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heading = "Nr. " + "Name".ljust(col_width) + "Gehalt"
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print(heading)
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print("-" * len(heading))
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salary_first = []
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for name, salary in data.items():
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salary_first.append((salary, name))
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for i, (salary, name) in enumerate(sorted(salary_first, reverse=True)):
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print(str(i+1).rjust(3), name.ljust(col_width), int(salary))
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print_enumerated_salary_list(salaries_new)
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print("-" * 50)
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# Bonusaufgabe 3:
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print("Gehaltsdurchschnitt:", sum(salaries_new.values()) / len(salaries_new))
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print("-" * 50)
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# Bonusaufgabe 4:
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def print_extended_salary_list(data, col_width=8):
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mean_salary = sum(data.values()) / len(data)
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heading = "Nr. " + "Name".ljust(col_width + 1) + "Gehalt".ljust(col_width +1) + "% d. ø"
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print(heading)
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print("-" * len(heading))
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salary_first = []
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for name, salary in data.items():
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salary_first.append((salary, name))
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for i, (salary, name) in enumerate(sorted(salary_first, reverse=True)):
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print(str(i+1).rjust(3), name.ljust(col_width), str(int(salary)).ljust(col_width), round(salary / (mean_salary / 100), 2))
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print_extended_salary_list(salaries_new)
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@ -19,9 +19,12 @@ Schreiben Sie ein Programm, das die beiden Dictionaries in einem neuen
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Dictionary (`salaries_all`) zusammenführt und die Namen aller Mitarbeiter
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Dictionary (`salaries_all`) zusammenführt und die Namen aller Mitarbeiter
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als Liste ausgibt.
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als Liste ausgibt.
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Tipp: Machen Sie zunächst eine Kopie des ersten Dictionaries (`dict.copy()`)
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Tipp: Machen Sie zunächst eine Kopie des ersten Dictionaries ([dict.copy]`()`)
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und fügen Sie dann die Einträge des zweiten Dictionaries der Kopie hinzu
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und fügen Sie dann die Einträge des zweiten Dictionaries der Kopie hinzu
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(`dict.update()`).
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([dict.update]`()`).
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[dict.copy]: https://docs.python.org/3/library/stdtypes.html#dict.copy
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[dict.update]: https://docs.python.org/3/library/stdtypes.html#dict.update
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## Aufgabe 2:
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## Aufgabe 2:
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@ -40,8 +43,8 @@ ermittelt und diese ausgibt.
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## Aufgabe 4:
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## Aufgabe 4:
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Schreiben Sie ein Programm, das das Gehalt aller Mitarbeiter um 10% erhöht und
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Schreiben Sie ein Programm, das das Gehalt aller Mitarbeiter um 10% erhöht und
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diese Daten in einem *neuen* Dictionary mit der selben Struktur speichert. Das
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diese Daten in einem *neuen* Dictionary (z.B. `salaries_new`) mit der selben
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ursprüngliche Dictionary soll nicht verändert werden.
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Struktur speichert. Das ursprüngliche Dictionary soll *nicht* verändert werden.
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## Aufgabe 5:
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## Aufgabe 5:
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@ -56,7 +59,10 @@ Mark ???
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...
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...
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```
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```
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Tipp: verwenden Sie die Escape-Sequenz für Tabulatoren `\t`.
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Tipp: verwenden Sie die String-Methode [ljust]`()` für die Ausgabe der ersten
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Spalte, um die zweite Spalte auszurichten.
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[ljust]: https://docs.python.org/3/library/stdtypes.html#str.ljust
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## Aufgabe 6:
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## Aufgabe 6:
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@ -70,7 +76,7 @@ um zu zeigen, dass das ursprüngliche Dictionary nicht verändert wurde.
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1. Sortieren Sie die Liste nach Höhe des Gehalts (absteigend).
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1. Sortieren Sie die Liste nach Höhe des Gehalts (absteigend).
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Tipp: verwenden Sie die Funktion `sorted()` zum Sortieren.
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Tipp: verwenden Sie die Funktion [sorted]`()` zum Sortieren.
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2. Nummerieren Sie die Listeneinträge in der Ausgabe, startend mit 1.
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2. Nummerieren Sie die Listeneinträge in der Ausgabe, startend mit 1.
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@ -85,6 +91,9 @@ um zu zeigen, dass das ursprüngliche Dictionary nicht verändert wurde.
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Wert im Dictionary ablegen.
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Wert im Dictionary ablegen.
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[sorted]: https://docs.python.org/3/library/functions.html#sorted
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## Ressourcen
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## Ressourcen
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Hilfe zu Dictionaries:
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Hilfe zu Dictionaries:
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