1

我有这段文字：

2,3,5,1,13,7,17​​,11,89,1,233,29,61,47,1597,19,37,41,421,199,28657,23,3001,521,53,281,514229,31,557,2207, 19801,3571,141961,107,73,9349,135721,2161,2789,211,433494437,43,109441,139,2971215073,1103,97,101,6376021,90481,9553,5779,791,453,5779,659,31, 2521,4513,3010349,35239681,1087,14736206161,9901,269,67,137,71,6673,103681,9375829,54018521,230686501,29134601,988681,79,157,1601,2269,370248451,99194853094755497,83,9521,6709， 173,263,1069,181,741469,4969,4531100550901,6643838879,761,769,193,599786069,197,401,743519377,919,519121,103,8288823481,119218851371,1247833,11128427,827728777,331,1459000305513721,10745088481,677,229,1381,347， 29717,709,159512939815855788121,

这是从我的生成器程序生成的数字，现在问题有源代码限制，所以我不能在我的解决方案中使用上述文本，所以我想压缩它并将其放入 python 中的数据结构中，以便我可以打印它们通过如下索引：

F = [`compressed data`]

并且F[0]会给出2 F[5]这样7的……请给我推荐一种合适的压缩技术。

PS：我是python的新手，所以请解释一下你的方法。

4

3 回答 3

8

当然你可以这样做：

import base64
import zlib
compressed = 'eJwdktkNgDAMQxfqR+5j/8V4QUJQUttx3Nrzl0+f+uunPPpm+Tf3Z/tKX1DM5bXP+wUFA777bCob4HMRfUk14QwfDYPrrA5gcuQB49lQQxdZpdr+1oN2bEA3pW5Nf8NGOFsR19NBszyX7G2raQpkVUEBdbTLuwSRlcDCYiW7GeBaRYJrgImrM3lmI/WsIxFXNd+aszXoRXuZ1PnZRdwKJeqYYYKq6y1++PXOYdgM0TlZcymCOdKqR7HYmYPiRslDr2Sn6C0Wgw+a6MakM2VnBk6HwU6uWqDRz+p6wtKTCg2WsfdKJwfJlHNaFT4+Q7PGfR9hyWK3p3464nhFwpOd7kdvjmz1jpWcxmbG/FJUXdMZgrpzs+jxC11twrBo3TaNgvsf8oqIYwT4r9XkPnNC1XcP7qD5cW7UHSJZ3my5qba+ozncl5kz8gGEEYOQ'
data = zlib.decompress(base64.b64decode(compressed))

请注意，这仅短了 139 个字符。但它有效：

>>> data
'2,3,5,1,13,7,17,11,89,1,233,29,61,47,1597,19,37,41,421,199,28657,23,3001,521,53,281,514229,31,557,2207,19801,3571,141961,107,73,9349,135721,2161,2789,211,433494437,43,109441,139,2971215073,1103,97,101,6376021,90481,953,5779,661,14503,797,59,353,2521,4513,3010349,35239681,1087,14736206161,9901,269,67,137,71,6673,103681,9375829,54018521,230686501,29134601,988681,79,157,1601,2269,370248451,99194853094755497,83,9521,6709,173,263,1069,181,741469,4969,4531100550901,6643838879,761,769,193,599786069,197,401,743519377,919,519121,103,8288823481,119218851371,1247833,11128427,827728777,331,1459000305513721,10745088481,677,229,1381,347,29717,709,159512939815855788121,'

如果您的代码限制真的很短，也许您应该计算这些数据或其他什么？它是什么？

于 2011-01-30T20:31:16.870 回答
3

如果您确实想要压缩， zlib会完成工作。如果你不想压缩，那我的读心能力恐怕会下降。

于 2011-01-30T19:20:25.623 回答
1

在 Python 2.4-2.7 上，pypy、jython：

>>> enc = sdata.encode('zlib').encode('base64')
>>> print enc
eJwdktkNgDAMQxfqR+5j/8V4QUJQUttx3Nrzl0+f+uunPPpm+Tf3Z/tKX1DM5bXP+wUFA777bCob
4HMRfUk14QwfDYPrrA5gcuQB49lQQxdZpdr+1oN2bEA3pW5Nf8NGOFsR19NBszyX7G2raQpkVUEB
dbTLuwSRlcDCYiW7GeBaRYJrgImrM3lmI/WsIxFXNd+aszXoRXuZ1PnZRdwKJeqYYYKq6y1++PXO
YdgM0TlZcymCOdKqR7HYmYPiRslDr2Sn6C0Wgw+a6MakM2VnBk6HwU6uWqDRz+p6wtKTCg2WsfdK
JwfJlHNaFT4+Q7PGfR9hyWK3p3464nhFwpOd7kdvjmz1jpWcxmbG/FJUXdMZgrpzs+jxC11twrBo
3TaNgvsf8oqIYwT4r9XkPnNC1XcP7qD5cW7UHSJZ3my5qba+ozncl5kz8gGEEYOQ
>>> print enc.decode('base64').decode('zlib')[:79]
2,3,5,1,13,7,17,11,89,1,233,29,61,47,1597,19,37,41,421,199,28657,23,3001,521,53
>>> sdata == enc.decode('base64').decode('zlib')
True
>>> F = [int(s) for s in sdata.split(',') if s.strip()]
>>> F[0], F[5]
(2, 7)
于 2011-01-31T10:02:37.227 回答