Question

我有这样的文字：

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 ，2207,19801,3571,141961,107,73,9349,135721,2161,2789,211,433494437,43,109441,139,2971215073,1103,97,101,6376021,90481,953,5779,661,14503,797 ，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，

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，以便我可以通过索引来打印它们：

F = [`compressed data`]

和F[0]会2 F[5]给7这样的... {建议我使用合适的压缩技术。

PS：我是python的新手，所以请解释一下你的方法。

Answer 1

当然可以这样做：

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139个字符。但它的确有效：

>>> 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那么短，也许您应该计算这些数据或其他什么？它是什么？

Answer 2

如果你确实想要压缩，

zlib会完成工作。如果你不想要压缩，那么我担心我的思维阅读能力正在下降。

Answer 3

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)

python中的TEXT压缩

3 个答案: