我有这个代码可以随机播放一个列表。我首先将它分成两个列表,因为我有一个交错函数,交错2个列表:
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
当我测试它时:
>>> 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]
列表中的4是两次,为什么打印1,[7],7,4] ??
编辑:
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
答案 0 :(得分:7)
为什么不使用标准库?
>>> 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]
>>>
答案 1 :(得分:0)
假设您想要交错列表,您可以编写一个简单的递归函数来执行多次。交换列表的一件事是,我相信第一个和最后一个字符将始终相同。
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