Dada una lista de elementos, la tarea es imprimir el elemento en un grupo de k hasta n-iteración en rango circular.
Ejemplos:
Input: list = [1, 2, 3, 4, 5, 6, 7], k = 3, n =10 Output: [1, 2, 3] [4, 5, 6] [7, 1, 2] [3, 4, 5] [6, 7, 1] [2, 3, 4] [5, 6, 7] [1, 2, 3] [4, 5, 6] [7, 1, 2] Input: list = [10, 20, 30, 40, 50, 60, 70], k = 4, n = 5 Output: [10, 20, 30, 40] [50, 60, 70, 10] [20, 30, 40, 50] [60, 70, 10, 20] [30, 40, 50, 60]
Podemos usar itertools with zippara hacer esta tarea.
Ejemplo 1:
# Python code to print element in group # of 5 till 9 iteration in circular range. # Importing from itertools import cycle # list initialization List = [90, 99, 192, 0, 43, 55] # Defining no of iterations n = 9 # Defining no of grouping k = 5 for index, *ans in zip(range(n), *[cycle(List)] * k): # printing ans print(ans)
Producción:
[90, 99, 192, 0, 43] [55, 90, 99, 192, 0] [43, 55, 90, 99, 192] [0, 43, 55, 90, 99] [192, 0, 43, 55, 90] [99, 192, 0, 43, 55] [90, 99, 192, 0, 43] [55, 90, 99, 192, 0] [43, 55, 90, 99, 192]
Ejemplo #2:
# Python code to print element in group # of 6 till 4 iteration in circular range. # Importing from itertools import cycle # list initialization List = ['Geeks', 'for', 'geeks', 'is', 'portal'] # Defining no of iterations n = 4 # Defining no of grouping k = 6 for index, *ans in zip(range(n), *[cycle(List)] * k): # printing ans print(ans)
Producción:
['Geeks', 'for', 'geeks', 'is', 'portal', 'Geeks'] ['for', 'geeks', 'is', 'portal', 'Geeks', 'for'] ['geeks', 'is', 'portal', 'Geeks', 'for', 'geeks'] ['is', 'portal', 'Geeks', 'for', 'geeks', 'is']
Publicación traducida automáticamente
Artículo escrito por everythingispossible y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA