使用__str __（）在python中漂亮地打印节点的链接列表

Question

I have a node class in python which is something like this

class Node:
  def __init__(self, value, next, prev, rand):
    self.value = value
    self.next = next
    self.prev = prev
    self.rand = rand

I create a list out of this by creating nodes and setting the appropriate pointers. Now, i want to pretty print this list : like [1] --> [2] --> [3] and so on. If there is a random pointer along with next it will be [1] --> [2] --> [3,4] -->[4] for example. here 2 next is 3 and 2 rand is 4 while 3 next is 4. I am trying to do this with the built-in str () method in the node class it self. I have it as below as of now

#pretty print a node
def __str__(self):
  if self.next is None:
    return '[%d]' % (self.value)
  else:
    return '[%d]-->%s' % (self.value, str(self.next))

This pretty prints it without the random pointer. But I am not able to incorporate the random pointer printing in this. I tried couple of approaches, but it's messing up the brackets. How would i go about this ?

Thanks

Answer 1

Try factoring the printing of the pointers out, that way it may be clearer:

#pretty print a node
def __str__(self):
  if self.next is None:
    return self.stringify_pointers()
  else:
    return '%s-->%s' % (self.stringify_pointers(), str(self.next))

def stringify_pointers(self):
    s = '['
    if self.next:
        s += str(self.next) + (',' if self.rand else '')
    if self.rand:
        # <> to distinguish rand from next
        s += '<%s>' % str(self.rand)
    return s + ']'

Answer 2

One way to do it is using array slices on the string and injecting the random value into the next node string like this

def __str__(self):
        if self.next is None:
            return '[%d]' % (self.value)
        else:
            nn = str(self.next)
            if self.rand != None:
                nn = '%s,%d%s' % (nn[:2], self.rand, nn[2:])                             
            return '[%d]-->%s' % (self.value, nn)