Question

我有以下lisp文件，来自UCI machine learning database。我想使用python将其转换为平面文本文件。典型的线条如下所示：

(1 ((st 8) (pitch 67) (dur 4) (keysig 1) (timesig 12) (fermata 0))((st 12) (pitch 67) (dur 8) (keysig 1) (timesig 12) (fermata 0)))

我想将其解析为文本文件，如：

time pitch duration keysig timesig fermata
8    67    4        1      12      0
12   67    8        1      12      0

是否有一个python模块可以智能地解析这个？这是我第一次看到lisp。

Answer 1

如this answer所示，pyparsing似乎是正确的工具：

inputdata = '(1 ((st 8) (pitch 67) (dur 4) (keysig 1) (timesig 12) (fermata 0))((st 12) (pitch 67) (dur 8) (keysig 1) (timesig 12) (fermata 0)))'

from pyparsing import OneOrMore, nestedExpr

data = OneOrMore(nestedExpr()).parseString(inputdata)
print data

# [['1', [['st', '8'], ['pitch', '67'], ['dur', '4'], ['keysig', '1'], ['timesig', '12'], ['fermata', '0']], [['st', '12'], ['pitch', '67'], ['dur', '8'], ['keysig', '1'], ['timesig', '12'], ['fermata', '0']]]]

为了完整性，这是如何格式化结果（使用texttable）：

from texttable import Texttable

tab = Texttable()
for row in data.asList()[0][1:]:
    row = dict(row)
    tab.header(row.keys())
    tab.add_row(row.values())
print tab.draw()

+---------+--------+----+-------+-----+---------+
| timesig | keysig | st | pitch | dur | fermata |
+=========+========+====+=======+=====+=========+
| 12      | 1      | 8  | 67    | 4   | 0       |
+---------+--------+----+-------+-----+---------+
| 12      | 1      | 12 | 67    | 8   | 0       |
+---------+--------+----+-------+-----+---------+

将该数据转换回lisp表示法：

def lisp(x):
    return '(%s)' % ' '.join(lisp(y) for y in x) if isinstance(x, list) else x

d = lisp(d[0])

Answer 2

如果您知道数据是正确的并且格式统一（初看起来似乎如此），并且如果您只需要这些数据并且不需要解决一般问题...那么为什么不只是更换每个非数字的空格，然后分裂？

import re
data = open("chorales.lisp").read().split("\n")
data = [re.sub("[^-0-9]+", " ", x) for x in data]
for L in data:
    L = map(int, L.split())
    i = 1  # first element is chorale number
    while i < len(L):
        st, pitch, dur, keysig, timesig, fermata = L[i:i+6]
        i += 6
        ... your processing goes here ...

Answer 3

用正则表达式将它分成几对：

In [1]: import re

In [2]: txt = '(((st 8) (pitch 67) (dur 4) (keysig 1) (timesig 12) (fermata 0))((st 12) (pitch 67) (dur 8) (keysig 1) (timesig 12) (fermata 0)))'

In [3]: [p.split() for p in re.findall('\w+\s+\d+', txt)]
Out[3]: [['st', '8'], ['pitch', '67'], ['dur', '4'], ['keysig', '1'], ['timesig', '12'], ['fermata', '0'], ['st', '12'], ['pitch', '67'], ['dur', '8'], ['keysig', '1'], ['timesig', '12'], ['fermata', '0']]

然后把它变成字典：

dct = {}
for p in data:
    if not p[0] in dct.keys():
        dct[p[0]] = [p[1]]
    else:
        dct[p[0]].append(p[1])

结果：

In [10]: dct
Out[10]: {'timesig': ['12', '12'], 'keysig': ['1', '1'], 'st': ['8', '12'], 'pitch': ['67', '67'], 'dur': ['4', '8'], 'fermata': ['0', '0']}

印刷：

print 'time pitch duration keysig timesig fermata'
for t in range(len(dct['st'])):
    print dct['st'][t], dct['pitch'][t], dct['dur'][t], 
    print dct['keysig'][t], dct['timesig'][t], dct['fermata'][t]

正确的格式留作读者的练习......

Answer 4

由于数据已经在Lisp中，请使用lisp本身：

(let ((input '(1 ((ST 8) (PITCH 67) (DUR 4) (KEYSIG 1) (TIMESIG 12) (FERMATA 0))
            ((ST 12) (PITCH 67) (DUR 8) (KEYSIG 1) (TIMESIG 12) (FERMATA 0)))))

       (let ((row-headers (mapcar 'car (second input)))
          (row-data (mapcar (lambda (row) (mapcar 'second row)) (cdr input))))

     (format t "~{~A~^ ~}~%" row-headers)
     (format t "~{~{~A~^ ~}~^ ~%~}" row-data)))

使用Python解析lisp文件

4 个答案: