解析python asciitree输出并打印'标记括号表示法'

Question

Google SyntaxNet提供了类似的输出.

saw VBD ROOT
 +-- Alice NNP nsubj
 |   +-- , , punct
 |   +-- reading VBG rcmod
 |       +-- who WP nsubj
 |       +-- had VBD aux
 |       +-- been VBN aux
 |       +-- about IN prep
 |           +-- SyntaxNet NNP pobj
 +-- , , punct
 +-- Bob NNP dobj
 +-- in IN prep
 |   +-- hallway NN pobj
 |       +-- the DT det
 +-- yesterday NN tmod
 +-- . . punct

我想使用python来读取和解析此输出(String数据).并打印出'标记括号表示法',如 - >

[saw [Alice [,] [reading [who][had][been][about [SyntaxNet]]]][,][Bob][in [hallway [the]]][yesterday][.]]

你能帮助我吗？

Answer 1

您可以使SyntaxNet以conll格式输出所有内容,而不是解析树,这样更易​​于解析.你的句子的conll格式如下所示:

1       Alice   _       NOUN    NNP     _       10      nsubj   _       _
2       ,       _       .       ,       _       1       punct   _       _
3       who     _       PRON    WP      _       6       nsubj   _       _
4       had     _       VERB    VBD     _       6       aux     _       _
5       been    _       VERB    VBN     _       6       aux     _       _
6       reading _       VERB    VBG     _       1       rcmod   _       _
7       about   _       ADP     IN      _       6       prep    _       _
8       SyntaxNet       _       NOUN    NNP     _       7       pobj    _       _
9       ,       _       .       ,       _       10      punct   _       _
10      saw     _       VERB    VBD     _       0       ROOT    _       _
11      Bob     _       NOUN    NNP     _       10      dobj    _       _
12      in      _       ADP     IN      _       10      prep    _       _
13      the     _       DET     DT      _       14      det     _       _
14      hallway _       NOUN    NN      _       12      pobj    _       _
15      yesterday       _       NOUN    NN      _       10      tmod    _       _
16      .       _       .       .       _       10      punct   _       _

可以在此处找到每列的含义.我们现在关注的唯一列是第一个(单词的ID),第二个(单词本身)和第7个(头部,换句话说,父级).根节点的父节点为0.

为了得到conll格式,我们只需要注释掉demo.sh的最后几行(我假设你用来获取输出):

$PARSER_EVAL \
  --input=$INPUT_FORMAT \
  --output=stdout-conll \
  --hidden_layer_sizes=64 \
  --arg_prefix=brain_tagger \
  --graph_builder=structured \
  --task_context=$MODEL_DIR/context.pbtxt \
  --model_path=$MODEL_DIR/tagger-params \
  --slim_model \
  --batch_size=1024 \
  --alsologtostderr \
   | \
  $PARSER_EVAL \
  --input=stdin-conll \
  --output=stdout-conll \
  --hidden_layer_sizes=512,512 \
  --arg_prefix=brain_parser \
  --graph_builder=structured \
  --task_context=$MODEL_DIR/context.pbtxt \
  --model_path=$MODEL_DIR/parser-params \
  --slim_model \
  --batch_size=1024 \
  --alsologtostderr #\
#  | \
#  bazel-bin/syntaxnet/conll2tree \
#  --task_context=$MODEL_DIR/context.pbtxt \
#  --alsologtostderr

(别忘了注释掉前一行的反斜杠)

(我从这里获得了这个技巧,请参阅评论)

当我自己运行demo.sh时,我得到了许多我不需要的信息.你怎么摆脱我留给你弄清楚(让我知道:)).我现在将相关部分写入一个文件,所以我可以把它管道到我要编写的python程序中.如果您可以删除信息,您应该能够将demo.sh直接传递到python程序中.

注意:我对python很新,所以随意改进我的代码.

首先,我们只想从输入中读取conll文件.我喜欢把每个单词都放在一个很好的课堂上.

#!/usr/bin/env python

import sys

# Conll data format:
# http://ilk.uvt.nl/conll/#dataformat
#
# The only parts we need:
# 1: ID
# 2: FORM (The original word)
# 7: HEAD (The ID of its parent)

class Word:
    "A class containing the information of a single line from a conll file."

    def __init__(self, columns):
        self.id = int(columns[0])
        self.form = columns[1]
        self.head = int(columns[6])
        self.children = []

# Read the conll input and put it in a list of words.
words = []
for line in sys.stdin:
    # Remove newline character, split on spaces and remove empty columns.
    line = filter(None, line.rstrip().split(" "))

    words.append(Word(line))

不错,但它还不是树形结构.我们还要做更多的工作.

我可以多次预先查看整个列表,查找每个孩子的每个单词,但这样效率会很低.我通过他们的父母对它们进行排序,然后它应该是一个快速查找,以获得给定父母的每个孩子.

# Sort the words by their head (parent).
lookup = [[] for _ in range(len(words) + 1)]
for word in words:
    lookup[word.head].append(word)

创建树结构:

# Build a tree
def buildTree(head):
    "Find the children for the given head in the lookup, recursively"

    # Get all the children of this parent.
    children = lookup[head]

    # Get the children of the children.
    for child in children:
        child.children = buildTree(child.id)

    return children

# Get the root's child. There should only be one child. The function returns an
# array of children so just get the first one.
tree = buildTree(0)[0] # Start with head = 0 (which is the ROOT node)

为了能够以新格式打印树,您可以向Word类添加一些方法重载:

def __str__(self):
    if len(self.children) == 0:
        return "[" + self.form + "]"
    else:
        return "[" + self.form + " " + "".join(str(child) for child in self.children) + "]"

def __repr__(self):
    return self.__str__()

现在你可以这样做:

print tree

管道就像这样:

cat input.conll | ./my_parser.py

或直接来自syntaxnet:

 echo "Alice, who had been reading about SyntaxNet, saw Bob in the hallway yesterday." | syntaxnet/demo.sh | ./my_parser.py