在递归调用中交换列表元素 python
swap elements of list in recursive call python
我想用简单的函数交换列表中的随机元素。
但它在递归调用中不起作用。
在第一次递归调用中,元素交换工作,
但嵌套递归调用(或第一次递归调用中的嵌套递归调用)不起作用。
我不知道为什么只有第一个递归调用中的交换有效。
以下是结果。
谢谢大家。
def change(lst):
if len(lst)>4:
a, b = np.random.randint(0, len(lst)), np.random.randint(0, len(lst))
print(lst)
lst[a], lst[b] = lst[b], lst[a]
print(lst)
mid = int(len(lst)/2)
change(lst[:mid])
change(lst[mid:])
k = list(range(0, 20))
change(k)
print(k)
`
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9]
[3, 19, 2, 0, 4, 5, 6, 7, 8, 9]
[3, 19, 2, 0, 4]
[3, 0, 2, 19, 4]
[5, 6, 7, 8, 9]
[5, 6, 8, 7, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[10, 11, 12, 13, 14]
[10, 14, 12, 13, 11]
[15, 16, 17, 18, 1]
[15, 16, 17, 18, 1]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1] <= result.
那是因为您通过 lst[:mid]
、lst[mid:]
创建了原始列表的 副本 。一种解决方案是将相同的列表和(单独)要处理的范围传递给 change()
。
问题是在你的递归调用中:
change(lst<b>[:mid]</b>)
change(lst<b>[mid:]</b>)
您使用 切片运算符。切片运算符构造一个 新列表,因此您的 更改是在新列表 上进行的,不会反映在原始列表中(因为它是复制).
您可以使用索引代替:
def change(lst<b>,frm=0,to=None</b>):
<b>if to is None: # set the default to the end of the list
to = len(lst)</b>
if <b>to-frm</b> > 4:
a, b = np.random.randint(<b>frm,to</b>), np.random.randint(<b>frm,to</b>)
print(lst)
lst[a], lst[b] = lst[b], lst[a]
print(lst)
mid = <b>(frm+to)//2</b>
change(lst<b>,frm,mid</b>)
change(lst<b>,mid,to</b>)
然后我们得到:
>>> k = list(range(0, 20))
>>> change(k)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
>>> print(k)
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
我想用简单的函数交换列表中的随机元素。 但它在递归调用中不起作用。
在第一次递归调用中,元素交换工作, 但嵌套递归调用(或第一次递归调用中的嵌套递归调用)不起作用。
我不知道为什么只有第一个递归调用中的交换有效。
以下是结果。
谢谢大家。
def change(lst):
if len(lst)>4:
a, b = np.random.randint(0, len(lst)), np.random.randint(0, len(lst))
print(lst)
lst[a], lst[b] = lst[b], lst[a]
print(lst)
mid = int(len(lst)/2)
change(lst[:mid])
change(lst[mid:])
k = list(range(0, 20))
change(k)
print(k)
`
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9]
[3, 19, 2, 0, 4, 5, 6, 7, 8, 9]
[3, 19, 2, 0, 4]
[3, 0, 2, 19, 4]
[5, 6, 7, 8, 9]
[5, 6, 8, 7, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 1]
[10, 11, 12, 13, 14]
[10, 14, 12, 13, 11]
[15, 16, 17, 18, 1]
[15, 16, 17, 18, 1]
[0, 19, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1] <= result.
那是因为您通过 lst[:mid]
、lst[mid:]
创建了原始列表的 副本 。一种解决方案是将相同的列表和(单独)要处理的范围传递给 change()
。
问题是在你的递归调用中:
change(lst<b>[:mid]</b>)
change(lst<b>[mid:]</b>)
您使用 切片运算符。切片运算符构造一个 新列表,因此您的 更改是在新列表 上进行的,不会反映在原始列表中(因为它是复制).
您可以使用索引代替:
def change(lst<b>,frm=0,to=None</b>):
<b>if to is None: # set the default to the end of the list
to = len(lst)</b>
if <b>to-frm</b> > 4:
a, b = np.random.randint(<b>frm,to</b>), np.random.randint(<b>frm,to</b>)
print(lst)
lst[a], lst[b] = lst[b], lst[a]
print(lst)
mid = <b>(frm+to)//2</b>
change(lst<b>,frm,mid</b>)
change(lst<b>,mid,to</b>)
然后我们得到:
>>> k = list(range(0, 20))
>>> change(k)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 2, 3, 4, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 12, 6, 7, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 10, 11, 5, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]
>>> print(k)
[0, 1, 4, 3, 2, 7, 6, 12, 8, 9, 5, 11, 10, 13, 14, 15, 16, 17, 18, 19]