Anonymous
public text v1 · immutablea=[0,1,2,3,4,5,6,7,8,9,0]
n=3
b=[[a[i:i+3]] for i in range(0,len(a)-len(a)%n,n)]
b.append(a[len(a)-len(a)%n:])a=[0,1,2,3,4,5,6,7,8,9,0]
n=3
b=[[a[i:i+3]] for i in range(0,len(a)-len(a)%n,n)]
b.append(a[len(a)-len(a)%n:])