Dada una lista, la tarea es encontrar el promedio de cada segmento consecutivo de n longitud donde cada segmento contiene n elementos.
Ejemplo:
Input : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 13, 14, 15, 16, 17, 18, 19, 20]
Output: [3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 13, 14, 15, 16, 17, 18]
Explanation:
Segment 1 - [1, 2, 3, 4, 5] => 15/5 = 3
Segment 2 - [2, 3, 4, 5, 6] => 20/5 = 4
Segment 3 - [3, 4, 5, 6, 7] => 25/5 = 5
and so on.....
Método #1: Usar la comprensión de listas
# Python code to find average of each consecutive segment # List initialisation Input = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20] # Defining Splits splits = 5 # Finding average of each consecutive segment Output = [sum(Input[i:i + splits])/splits for i in range(len(Input) - splits + 1)] # printing output print(Output)
Producción:
[3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0]
Método #2: Usando mean la función
# Python code to find average of each consecutive segment # Importing from statistics import mean from itertools import islice # List initialisation Input = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20] # Finding average of each consecutive segment zip_list = zip(*(islice(Input, i, None) for i in range(5))) Output = list(map(mean, zip_list)) # printing output print(Output)
Producción:
[3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
Publicación traducida automáticamente
Artículo escrito por everythingispossible y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA