5

我有一本使用 Python 创建的字典。

d = {'a': ['Adam', 'Book', 4], 'b': ['Bill', 'TV', 6, 'Jill', 'Sports', 1, 'Bill', 'Computer', 5], 'c': ['Bill', 'Sports', 3], 'd': ['Quin', 'Computer', 3, 'Adam', 'Computer', 3], 'e': ['Quin', 'TV', 2, 'Quin', 'Book', 5], 'f': ['Adam', 'Computer', 7]}

我想以横向树格式而不是在控制台上打印出来。我尝试过漂亮的打印，但是当字典变长时，它变得难以阅读。

例如，使用这个字典，它会返回：

a -> Book -> Adam -> 4
b -> TV -> Bill -> 6
  -> Sports -> Jill -> 1
  -> Computer -> Bill -> 5
c -> Sports -> Bill -> 3
d -> Computer -> Quin -> 3
              -> Adam -> 3
e -> TV -> Quin -> 2
    Book -> Quin -> 5
f -> Computer -> Adam -> 7

本质上，漂亮的打印是按活动组织的，或者是列表中第二个位置的项目，然后是名称，然后是数字。

上面的示例输出只是一个示例。我尝试使用Pretty print 一棵树，但无法弄清楚如何将其转换为横向格式。

4

3 回答 3

8

您可以查看ETE 工具包的代码。_asciiArt函数可以生成很好的树表示，即使有内部节点标签

from ete2 import Tree
t = Tree("(((A,B), C), D);")
print t

#               /-A
#          /---|
#     /---|     \-B
#    |    |
#----|     \-C
#    |
#     \-D
于 2013-06-23T11:23:47.293 回答
3

这就是我将如何做到的。由于树只有两层深——不管你想要的输出格式似乎暗示了什么——没有必要使用递归来遍历它的内容，因为迭代工作得很好。可能这与您引用的#​​f 代码完全不同，因为我不懂该语言，但它更短且更具可读性——至少对我而言。

from itertools import izip

def print_tree(tree):
    for key in sorted(tree.iterkeys()):
        data = tree[key]
        previous = data[0], data[1], data[2]
        first = True
        for name, activity, value in izip(*[iter(data)]*3):  # groups of three
            activity = activity if first or activity != previous[1] else ' '*len(activity)
            print '{} ->'.format(key) if first else '    ',
            print '{} -> {} -> {}'.format(activity, name, value)
            previous = name, activity, value
            first = False

d = {'a': ['Adam', 'Book', 4],
     'b': ['Bill', 'TV', 6, 'Jill', 'Sports', 1, 'Bill', 'Computer', 5],
     'c': ['Bill', 'Sports', 3],
     'd': ['Quin', 'Computer', 3, 'Adam', 'Computer', 3],
     'e': ['Quin', 'TV', 2, 'Quin', 'Book', 5],
     'f': ['Adam', 'Computer', 7]}

print_tree(d)

输出：

a -> Book -> Adam -> 4
b -> TV -> Bill -> 6
     Sports -> Jill -> 1
     Computer -> Bill -> 5
c -> Sports -> Bill -> 3
d -> Computer -> Quin -> 3
              -> Adam -> 3
e -> TV -> Quin -> 2
     Book -> Quin -> 5
f -> Computer -> Adam -> 7

更新

要按名称而不是活动组织输出，您需要更改三行，如下所示：

from itertools import izip

def print_tree(tree):
    for key in sorted(tree.iterkeys()):
        data = tree[key]
        previous = data[0], data[1], data[2]
        first = True
        for name, activity, value in sorted(izip(*[iter(data)]*3)):  # changed
            name = name if first or name != previous[0] else ' '*len(name) # changed
            print '{} ->'.format(key) if first else '    ',
            print '{} -> {} -> {}'.format(name, activity, value) # changed
            previous = name, activity, value
            first = False

修改后的输出：

a -> Adam -> Book -> 4
b -> Bill -> Computer -> 5
          -> TV -> 6
     Jill -> Sports -> 1
c -> Bill -> Sports -> 3
d -> Adam -> Computer -> 3
     Quin -> Computer -> 3
e -> Quin -> Book -> 5
          -> TV -> 2
f -> Adam -> Computer -> 7
于 2012-09-04T09:39:12.573 回答
0 
def treePrint(tree):
    for key in tree:
        print key, # comma prevents a newline character
        treeElem = tree[key] # multiple lookups is expensive, even amortized O(1)!
        for subElem in treeElem:
            print " -> ", subElem,
            if type(subElem) != str: # OP wants indenting after digits
                print "\n " # newline and a space to match indenting
        print "" # forces a newline
于 2012-09-04T01:56:23.860 回答