在 Python 中打印树数据结构

Question

我一直在寻找树打印的可能实现，它以用户友好的方式打印树，而不是作为对象的实例。

我在网上遇到了这个解决方案：

资料来源： http ://cbio.ufs.ac.za/live_docs/nbn_tut/trees.html

class node(object):
    def __init__(self, value, children = []):
        self.value = value
        self.children = children

    def __repr__(self, level=0):
        ret = "\t"*level+repr(self.value)+"\n"
        for child in self.children:
            ret += child.__repr__(level+1)
        return ret

此代码按以下方式打印树：

'grandmother'
    'daughter'
        'granddaughter'
        'grandson'
    'son'
        'granddaughter'
        'grandson'

是否可以在不更改__repr__方法的情况下获得相同的结果，因为我将它用于其他目的。

Answer 1

是的，将__repr__代码移动到__str__ ，然后在树上调用str()或将其传递给print语句。 记得在递归调用中也使用__str__ ：

class node(object):
    def __init__(self, value, children = []):
        self.value = value
        self.children = children

    def __str__(self, level=0):
        ret = "\t"*level+repr(self.value)+"\n"
        for child in self.children:
            ret += child.__str__(level+1)
        return ret

    def __repr__(self):
        return '<tree node representation>'

演示：

>>> root = node('grandmother')
>>> root.children = [node('daughter'), node('son')]
>>> root.children[0].children = [node('granddaughter'), node('grandson')]
>>> root.children[1].children = [node('granddaughter'), node('grandson')]
>>> root
<tree node representation>
>>> str(root)
"'grandmother'\n\t'daughter'\n\t\t'granddaughter'\n\t\t'grandson'\n\t'son'\n\t\t'granddaughter'\n\t\t'grandson'\n"
>>> print root
'grandmother'
    'daughter'
        'granddaughter'
        'grandson'
    'son'
        'granddaughter'
        'grandson'

Answer 2

你为什么不将其存储为一个treelib对象 ，并打印出来类似于我们如何绝版的CHAID树在这里与有关你的使用情况更相关节点的描述？

([], {0: 809, 1: 500}, (sex, p=1.47145310169e-81, chi=365.886947811, groups=[['female'], ['male']]))
├── (['female'], {0: 127, 1: 339}, (embarked, p=9.17624191599e-07, chi=24.0936494474, groups=[['C', '<missing>'], ['Q', 'S']]))
│   ├── (['C', '<missing>'], {0: 11, 1: 104}, <Invalid Chaid Split>)
│   └── (['Q', 'S'], {0: 116, 1: 235}, <Invalid Chaid Split>)
└── (['male'], {0: 682, 1: 161}, (embarked, p=5.017855245e-05, chi=16.4413525404, groups=[['C'], ['Q', 'S']]))
    ├── (['C'], {0: 109, 1: 48}, <Invalid Chaid Split>)
    └── (['Q', 'S'], {0: 573, 1: 113}, <Invalid Chaid Split>)

Answer 3

不修改__repr__和__str__解决方案

def other_name(self, level=0):
    print('\t' * level + repr(self.value))
    for child in self.children:
        child.other_name(level + 1)

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_{这个答案是作为对问题Printing a Tree data structure in Python的编辑发布的，由 OP Kristof Pal在 CC BY-SA 3.0 下发布。}