Question

我正在寻找有关如何通过解析文件来创建数据结构的建议。这是我文件中的列表。

'01bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',
'01bpar( 3)=  0.00000000E+00',
'02epar( 1)=  0.49998963E+02',
'02epar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',
'02epar( 3)=  0.00000000E+00',
'02epar( 4)=  0.17862340E-01  half_life=  0.3880495E+02  relax_time=  0.5598371E+02',
'02bpar( 1)=  0.49998962E+02',
'02bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',

我需要做的是构建一个如下所示的数据结构：

http://img11.imageshack.us/img11/7645/datastructure.gif

（由于新的用户限制而无法发布）

我已经设法获得所有正则表达式过滤器以获得所需，但我无法构建结构。想法？

Answer 1

理论上可以让pyparsing使用解析操作创建整个结构，但如果你只是按照我的名字命名各个字段，那么构建结构也不算太糟糕。如果你想转换为使用RE，那么这个例子可以让你初步看看事情的样子：

source = """\
'01bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02', 
'01bpar( 3)=  0.00000000E+00', 
'02epar( 1)=  0.49998963E+02', 
'02epar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02', 
'02epar( 3)=  0.00000000E+00', 
'02epar( 4)=  0.17862340E-01  half_life=  0.3880495E+02  relax_time=  0.5598371E+02', 
'02bpar( 1)=  0.49998962E+02', 
'02bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02', """

from pyparsing import Literal, Regex, Word, alphas, nums, oneOf, OneOrMore, quotedString, removeQuotes

EQ = Literal('=').suppress()
scinotationnum = Regex(r'\d\.\d+E[+-]\d+')
dataname = Word(alphas+'_')
key = Word(nums,exact=2) + oneOf("bpar epar")
index = '(' + Word(nums) + ')'

keyedValue = key + EQ + scinotationnum

# define an item in the source - suppress values with keys, just want the unkeyed ones
item = key('key') + index + EQ + OneOrMore(keyedValue.suppress() | scinotationnum)('data')

# initialize summary structure
from collections import defaultdict
results = defaultdict(lambda : {'epar':[], 'bpar':[]})

# extract quoted strings from list
quotedString.setParseAction(removeQuotes)
for raw in quotedString.searchString(source):
    parts = item.parseString(raw[0])
    num,par = parts.key
    results[num][par].extend(parts.data)

# dump out results, or do whatever
from pprint import pprint
pprint(dict(results.iteritems()))

打印：

{'01': {'bpar': ['0.23103878E-01', '0.00000000E+00'], 'epar': []},
 '02': {'bpar': ['0.49998962E+02', '0.23103878E-01'],
        'epar': ['0.49998963E+02',
                 '0.23103878E-01',
                 '0.00000000E+00',
                 '0.17862340E-01']}}

Answer 2

考虑使用dicts的词典。

#!/usr/bin/env python
import re
import pprint
raw = """'01bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',
'01bpar( 3)=  0.00000000E+00',
'02epar( 1)=  0.49998963E+02',
'02epar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',
'02epar( 3)=  0.00000000E+00',
'02epar( 4)=  0.17862340E-01  half_life=  0.3880495E+02  relax_time=  0.5598371E+02',
'02bpar( 1)=  0.49998962E+02',
'02bpar( 2)=  0.23103878E-01  half_life=  0.3000133E+02  relax_time=  0.4328278E+02',"""

datastruct = {}
pattern = re.compile(r"""\D(?P<digits>\d+)(?P<field>[eb]par)[^=]+=\D+(?P<number>\d+\.\d+E[+-]\d+)""")
for line in raw.splitlines():
    result = pattern.search(line)
    parts = result.groupdict()
    if not parts['digits'] in datastruct:
        datastruct[parts['digits']] = {'epar':[], 'bpar':[]}
    datastruct[parts['digits']][parts['field']].append(parts['number'])

pprint.pprint(datastruct, depth=4)

产地：

{'01': {'bpar': ['0.23103878E-01', '0.00000000E+00'], 'epar': []},
 '02': {'bpar': ['0.49998962E+02', '0.23103878E-01'],
        'epar': ['0.49998963E+02',
                 '0.23103878E-01',
                 '0.00000000E+00',
                 '0.17862340E-01']}}

根据评论修改版本：

pattern = re.compile(r"""\D(?P<digits>\d+)(?P<field>[eb]par)[^=]+=\D+(?P<number>\d+\.\d+E[+-]\d+)""")

default = lambda : dict((('epar',[]), ('bpar',[])))
datastruct = defaultdict( default)

for line in raw.splitlines():
    result = pattern.search(line)
    parts = result.groupdict()
    datastruct[parts['digits']][parts['field']].append(parts['number'])

pprint.pprint(datastruct.items())

产生：

[('02',
  {'bpar': ['0.49998962E+02', '0.23103878E-01'],
   'epar': ['0.49998963E+02',
            '0.23103878E-01',
            '0.00000000E+00',
            '0.17862340E-01']}),
 ('01', {'bpar': ['0.23103878E-01', '0.00000000E+00'], 'epar': []})]

Answer 3

您的顶级结构是位置结构，因此它是列表的完美选择。由于列表可以包含任意项，因此named tuple是完美的。元组中的每个项目都可以包含一个包含该元素的列表。

所以，你的代码应该看起来像这个伪代码：

from collections import named tuple
data = []
newTuple = namedtuple('stuff', ['epar','bpar'])
for line in theFile.readlines():
    eparVals = regexToGetThemFromString()
    bparVals = regexToGetThemFromString()
    t = newTuple(eparVals, bparVals)
    data.append(t)

你说你已经可以遍历文件了，并且有各种正则表达式来获取数据，所以我没有打扰添加所有细节。

通过解析输出文件来创建复杂的数据结构

3 个答案: