Reputation: 465
I have this code that shuffles a list. I first split it into two lists because I have a interleave function that interleaves 2 lists:
def shuffle(xs, n=1):
il=list()
if len(xs)%2==0:
stop=int(len(xs)//2)
a=xs[:stop]
b=xs[stop:]
print(a)
print(b)
else:
stop=int(len(xs)//2)
a=xs[:stop]
b=xs[stop:]
print(a)
print(b)
if n>0:
for i in range(n):
shuffle=interleave(a,b)
else:
return
return shuffle
and when I test it:
>>> shuffle([1,2,3,4,5,6,7],1)
[1, 2, 3]
[4, 5, 6, 7]
1
[7]
[7, 4]
[1, 4, 2, 5, 3, 6, 7, 4]
What is 4 in the list twice and why is it printing 1, [7], 7,4]??
EDIT:
def interleave(xs,ys):
a=xs
b=ys
minlength=[len(a),len(b)]
extralist= list()
interleave= list()
for i in range((minval(minlength))):
pair=a[i],b[i]
interleave.append(pair)
flat=flatten(interleave)
c=a+b
if len(b)>len(a):
remainder=len(b)-len(a)
for j in range(remainder,-1,-1):
extra=b[-j]
extralist.append(extra)
if len(a)>len(b):
remainder=len(a)-len(b)
for j in range(remainder,-1,-1):
extra=a[-j]
extralist.append(extra)
del extralist[-1]
final=flat+extralist
return final
Upvotes: 1
Views: 913
Reputation: 295
Assuming you are wanting to interleave the lists, you can write a simple recursive function to do it n number of times. The one thing with interleaving the list is that the first and last characters will always be the same, I believe.
def shuffle(lst, num):
'''
lst - is a list
num - is the amount of times to shuffle
'''
def interleave(lst1,lst2):
'''
lst1 and lst2 - are lists to be interleaved together
'''
if not lst1:
return lst2
elif not lst2:
return lst1
return lst1[0:1] + interleave(lst2, lst1[1:])
while num > 0:
lst = interleave(lst[:len(lst)/2], lst[len(lst)/2:])
print lst
num -= 1
return lst
Upvotes: 0
Reputation: 2170
Why not just use the standard library?
>>> from random import shuffle
>>> l = list(range(1,20))
>>> l
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> shuffle(l)
>>> l
[17, 15, 9, 13, 19, 7, 10, 18, 5, 1, 12, 3, 2, 16, 4, 14, 8, 6, 11]
>>>
Upvotes: 9