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如何在python中显示类似msdos tree命令的树?

tree python

BPL · 技术社区 · 6 年前

import sys


dct = {
    -1: [0, 60000],
    0: [100, 20],
    100: [30],
    30: [400, 500],
    60000: [70, 80]
}


def ptree(parent, tree, indent=''):
    print(parent)
    if parent not in tree:
        return

    for child in tree[parent][:-1]:
        print(indent + '|' + '-' * 4, end='')
        ptree(child, tree, indent + '|' + ' ' * 4)
    child = tree[parent][-1]
    print(indent + '`' + '-' * 4, end='')
    ptree(child, tree, indent + '  ' * 4)


ptree(-1, dct)

这个小代码有两个问题:

在本文中,Item-1被认为是“不可见的根项”,因此我不想打印它
这个命令的输出非常难看,而且对齐也不好,我想得到一个与 tree

为了解决第一个要点,我考虑在代码中引入这些丑陋的条件/黑客:

def ptree(parent, tree, indent=''):
    if parent != -1:
        print(parent)
    if parent not in tree:
        return

    for child in tree[parent][:-1]:
        if parent != -1:
            print(indent + '|' + '-' * 4, end='')
            ptree(child, tree, indent + '|' + ' ' * 4)
        else:
            ptree(child, tree, indent)
    child = tree[parent][-1]
    if parent != -1:
        print(indent + '`' + '-' * 4, end='')
        ptree(child, tree, indent + '  ' * 4)
    else:
        ptree(child, tree, indent)

问题

2 回复 | 直到 6 年前

Darkonaut 6 年前

经过反复尝试,我终于找到了一个解决方案,似乎能准确地说出原始树产生的结果:

def ptree(start, tree, indent_width=4):

    def _ptree(start, parent, tree, grandpa=None, indent=""):
        if parent != start:
            if grandpa is None:  # Ask grandpa kids!
                print(parent, end="")
            else:
                print(parent)
        if parent not in tree:
            return
        for child in tree[parent][:-1]:
            print(indent + "â" + "â" * indent_width, end="")
            _ptree(start, child, tree, parent, indent + "â" + " " * 4)
        child = tree[parent][-1]
        print(indent + "â" + "â" * indent_width, end="")
        _ptree(start, child, tree, parent, indent + " " * 5)  # 4 -> 5

    parent = start
    _ptree(start, parent, tree)


dct = {
    -1: [0, 60000],
    0: [100, 20, 10],
    100: [30],
    30: [400, 500],
    60000: [70, 80, 600],
    500: [495, 496, 497]
}

除了使用正确的连接器外,检查grander并将上次ptree调用的缩进从4增加到5是关键。

ptree(-1, dct)
# Out
âââââ0
â    âââââ100
â    â    âââââ30
â    â         âââââ400
â    â         âââââ500
â    â              âââââ495
â    â              âââââ496
â    â              âââââ497
â    âââââ20
â    âââââ10
âââââ60000
     âââââ70
     âââââ80
     âââââ600

Prune 6 年前

缩进量是根据节点值的打印图像进行移位的问题,而不是4个空格的常量。这个 math log10 和 ceil 做这项工作的方法。

import sys
import math


dct = {
    -1: [0, 60000],
    0: [100, 20, 7],
    100: [30],
    30: [400, 500],
    60000: [70, 80],
    7: [9, 11, 13],
}


def ptree(parent, tree, indent=''):

    if parent != -1:
        print(parent)

    if parent not in tree:
        return

    shift = math.ceil(math.log10(parent)) \
            if parent >= 10 else 1
    indent += ' ' * shift

    for child in tree[parent][:-1]:
        print(indent + '|' + '-' * 4, end='')
        ptree(child, tree, indent + '|' + ' ' * 4)

    child = tree[parent][-1]
    print(indent + '`' + '-' * 4, end='')
    ptree(child, tree, indent + ' ' * 4)


ptree(-1, dct)

输出:

 |----0
 |     |----100
 |     |      `----30
 |     |            |----400
 |     |            `----500
 |     |----20
 |     `----7
 |          |----9
 |          |----11
 |          `----13
 `----60000
          |----70
          `----80