我打印了几个列表,但值没有排序.
for f, h, u, ue, b, be, p, pe, m, me in zip(filename, human_rating, rating_unigram, percentage_error_unigram, rating_bigram, percentage_error_bigram, rating_pos, percentage_error_pos, machine_rating, percentage_error_machine_rating):
print "{:>6s}{:>5.1f}{:>7.2f}{:>8.2f} {:>7.2f} {:>7.2f} {:>7.2f} {:>8.2f} {:>7.2f} {:>8.2f}".format(f,h,u,ue,b,be,p,pe,m,me)
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根据'filename'中的值对所有这些列表进行排序的最佳方法是什么?
因此,如果:
filename = ['f3','f1','f2']
human_rating = ['1','2','3']
etc.
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然后排序将返回:
filename = ['f1','f2','f3']
human_rating = ['2','3','1']
etc.
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Dav*_*ver 13
我会拉链然后排序:
zipped = zip(filename, human_rating, …)
zipped.sort()
for row in zipped:
print "{:>6s}{:>5.1f}…".format(*row)
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如果你真的想要获得个人列表,我会按照上面的方式对它们进行排序,然后解压缩它们:
filename, human_rating, … = zip(*zipped)
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这个怎么样:zip
进入一个元组列表,对元组列表进行排序,然后“解压缩”?
l = zip(filename, human_rating, ...)
l.sort()
# 'unzip'
filename, human_rating ... = zip(*l)
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或者在一行中:
filename, human_rating, ... = zip(*sorted(zip(filename, human_rating, ...)))
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示例运行:
foo = ["c", "b", "a"]
bar = [1, 2, 3]
foo, bar = zip(*sorted(zip(foo, bar)))
print foo, "|", bar # prints ('a', 'b', 'c') | (3, 2, 1)
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