Question

我有一个构建huffman树的方法，如下所示：

def buildTree(tuples) :
    while len(tuples) > 1 :
        leastTwo = tuple(tuples[0:2])                  # get the 2 to combine
        theRest  = tuples[2:]                          # all the others
        combFreq = leastTwo[0][0] + leastTwo[1][0]     #enter code here the branch points freq
        tuples   = theRest + [(combFreq,leastTwo)]     # add branch point to the end
        tuples.sort()                                  # sort it into place
    return tuples[0]            # Return the single tree inside the list

但是我用以下参数提供函数：

[(1, 'b'), (1, 'd'), (1, 'g'), (2, 'c'), (2, 'f'), (3, 'a'), (5, 'e')]

我收到错误

  File "<stdin>", line 7, in buildTree
    tuples.sort()
TypeError: '<' not supported between instances of 'tuple' and 'str'

调试时我发现错误在tuples.sort()。

Answer 1

因为您正在以$(window).scroll(function(){ var a = 112; var pos = $(window).scrollTop(); if(pos > a) { $("#compareHead").css({ position: 'fixed', top: '51px' }); } });形式创建内部节点而引发错误。对于相同的优先级，Python然后尝试将来自叶节点的符号（因此(priority, (node, node))节点元组中的第二个元素）与来自内部节点的(priority, symbol)元组进行比较：

(node, node)

为了构建一个huffman树，如果你想对你的节点数组进行排序，你只需要对优先级进行排序，忽略其余的元组（符号或子节点）：

>>> inner = (combFreq, leastTwo)
>>> inner
(2, ((1, 'b'), (1, 'd')))
>>> theRest[1]
(2, 'c')
>>> theRest[1] < inner
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: '<' not supported between instances of 'str' and 'tuple'

通过该更正，您的tuples.sort(key=lambda t: t[0])函数会生成一个树：

buildTree()

就个人而言，我会使用优先级队列，每次都避免排序。见How to implement Priority Queues in Python?

TypeError：＆＃39;＆lt;＆＃39; ＆＃39; tuple＆＃39;的实例之间不支持和＆＃39; str＆＃39;

1 个答案: