Parsing a lisp file with Python

Question

I have the following lisp file, which is from the UCI machine learning database. I would like to convert it into a flat text file using python. A typical line looks like th

Answer 1

Since the data is already in Lisp, use lisp itself:

(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)))

Answer 2

As shown in this answer, pyparsing appears to be the right tool for that:

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']]]]

For the completeness' sake, this is how to format the results (using 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       |
+---------+--------+----+-------+-----+---------+

To convert that data back to the lisp notation:

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

d = lisp(d[0])

Answer 3

Separate it into pairs with a regular expression:

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']]

Then make it into a dictionary:

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

The result:

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

Printing:

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]

Proper formatting is left as an exercise for the reader...

Answer 4

If you know that the data is correct and the format uniform (seems so at a first sight), and if you need just this data and don't need to solve the general problem... then why not just replacing every non-numeric with a space and then going with split?

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 ...