Question

我正在从我的控件之外的源解析一些文本，这不是一种非常方便的格式。我有这样的台词：

问题类别：人类的努力问题子类别：太空探索问题类型：无法启动软件版本：9.8.77.omni.3问题详情：信号障碍室问题。

我想用这样的键分割线：

Problem_Category = "Human Endeavors"
Problem_Subcategory = "Space Exploration"
Problem_Type = "Failure to Launch"
Software_Version = "9.8.77.omni.3"
Problem_Details = "Issue with signal barrier chamber."

键将始终采用相同的顺序，并且后面跟着一个分号，但在值和下一个键之间不一定有空格或换行符。我不确定什么可以用作分隔符来解析它，因为冒号和空格也可以出现在值中。我该如何解析这个文本？

Answer 1

如果你的文本块是这个字符串：

text = 'Problem Category: Human Endeavors Problem Subcategory: Space ExplorationProblem Type: Failure to LaunchSoftware Version: 9.8.77.omni.3Problem Details: Issue with signal barrier chamber.'

然后

import re
names = ['Problem Category', 'Problem Subcategory', 'Problem Type', 'Software Version', 'Problem Details']

text = 'Problem Category: Human Endeavors Problem Subcategory: Space ExplorationProblem Type: Failure to LaunchSoftware Version: 9.8.77.omni.3Problem Details: Issue with signal barrier chamber.'

pat = r'({}):'.format('|'.join(names))
data = dict(zip(*[iter(re.split(pat, text, re.MULTILINE)[1:])]*2))
print(data)

产生字典

{'Problem Category': ' Human Endeavors ',
 'Problem Details': ' Issue with signal barrier chamber.',
 'Problem Subcategory': ' Space Exploration',
 'Problem Type': ' Failure to Launch',
 'Software Version': ' 9.8.77.omni.3'}

所以你可以分配

text = df_dict['NOTE_DETAILS'][0]
...
df_dict['NOTE_DETAILS'][0] = data

然后您可以使用dict索引访问子类别：

df_dict['NOTE_DETAILS'][0]['Problem_Category']

但是要小心。深层嵌套的dicts列表/ DataFrames通常是一个糟糕的设计。正如Zen of Python所说， Flat优于嵌套。

Answer 2

鉴于您提前知道关键字，请将文本分区为“当前关键字”，“剩余文本”，然后继续使用下一个关键字对剩余文本进行分区。

# get input from somewhere
raw = 'Problem Category: Human Endeavors Problem Subcategory: Space ExplorationProblem Type: Failure to LaunchSoftware Version: 9.8.77.omni.3Problem Details: Issue with signal barrier chamber.'

# these are the keys, in order, without the colon, that will be captured
keys = ['Problem Category', 'Problem Subcategory', 'Problem Type', 'Software Version', 'Problem Details']
prev_key = None
remaining = raw
out = {}

for key in keys:
    # get the value from before the key and after the key
    prev_value, _, remaining = remaining.partition(key + ':')

    # start storing values after the first iteration, since we need to partition the second key to get the first value
    if prev_key is not None:
        out[prev_key] = prev_value.strip()

    # what key to store next iteration
    prev_key = key

# capture the final value (since it lags behind the parse loop)
out[prev_key] = remaining.strip()

# out now contains the parsed values, print it out nicely
for key in keys:
    print('{}: {}'.format(key, out[key]))

打印：

Problem Category: Human Endeavors
Problem Subcategory: Space Exploration
Problem Type: Failure to Launch
Software Version: 9.8.77.omni.3
Problem Details: Issue with signal barrier chamber.

Answer 3

我讨厌并担心正则表达式，所以这里只是使用内置方法的解决方案。

#splits a string using multiple delimiters.
def multi_split(s, delims):
    strings = [s]
    for delim in delims:
        strings = [x for s in strings for x in s.split(delim) if x]
    return strings

s = "Problem Category: Human Endeavors Problem Subcategory: Space ExplorationProblem Type: Failure to LaunchSoftware Version: 9.8.77.omni.3Problem Details: Issue with signal barrier chamber."
categories = ["Problem Category", "Problem Subcategory", "Problem Type", "Software Version", "Problem Details"]
headers = [category + ": " for category in categories]

details = multi_split(s, headers)
print details

details_dict = dict(zip(categories, details))
print details_dict

结果（为了便于阅读，我添加了换行符）：

[
    'Human Endeavors ', 
    'Space Exploration', 
    'Failure to Launch', 
    '9.8.77.omni.3', 
    'Issue with signal barrier chamber.'
]

{
    'Problem Subcategory': 'Space Exploration', 
    'Problem Details': 'Issue with signal barrier chamber.', 
    'Problem Category': 'Human Endeavors ', 
    'Software Version': '9.8.77.omni.3', 
    'Problem Type': 'Failure to Launch'
}

Answer 4

这只是一般BNF解析的工作，可以很好地处理歧义。我使用了perl和Marpa，一般的BNF解析器。希望这会有所帮助。

use 5.010;
use strict;
use warnings;

use Marpa::R2;

my $g = Marpa::R2::Scanless::G->new( { source => \(<<'END_OF_SOURCE'),

    :default ::= action => [ name, values ]

    pairs ::= pair+

    pair ::= name (' ') value

    name ::= 'Problem Category:'
    name ::= 'Problem Subcategory:'
    name ::= 'Problem Type:'
    name ::= 'Software Version:'
    name ::= 'Problem Details:'

    value ::= [\s\S]+

    :discard ~ whitespace
    whitespace ~ [\s]+

END_OF_SOURCE
} );

my $input = <<EOI;
Problem Category: Human Endeavors Problem Subcategory: Space ExplorationProblem Type: Failure to LaunchSoftware Version: 9.8.77.omni.3Problem Details: Issue with signal barrier chamber.
EOI

my $ast = ${ $g->parse( \$input ) };

my @pairs;

ast_traverse($ast);

for my $pair (@pairs){
    my ($name, $value) = @$pair;
    say "$name = $value";
}

sub ast_traverse{
    my $ast = shift;
    if (ref $ast){
        my ($id, @children) = @$ast;
        if ($id eq 'pair'){

            my ($name, $value) = @children;

            chop $name->[1];

            shift @$value;
            $value = join('', @$value);
            chomp $value;

            push @pairs, [ $name->[1], '"' . $value . '"' ];
        }
        else {
            ast_traverse($_) for @children;
        }
    }
}

打印：

Problem Category = "Human Endeavors "
Problem Subcategory = "Space Exploration"
Problem Type = "Failure to Launch"
Software Version = "9.8.77.omni.3"
Problem Details = "Issue with signal barrier chamber."

根据特定键从文本块中解析值

4 个答案: