python打印格式方式未知列表长度

Question

如果我有这样的6长度列表:

l = ["AA","BB","CC","DD"]

我可以打印出来:

print "%-2s %-2s %-2s %-2s" % tuple(l)

输出将是:

AA BB CC DD

但如果列表l可以是任何长度怎么办？有没有办法以相同的格式打印具有未知数量元素的列表？

Answer 1

生成单独的片段并加入它们:

print ' '.join(['%-2s' % (i,) for i in l])

或者你可以使用字符串乘法:

print ('%-2s ' * len(l))[:-1] % tuple(l)

最后[:-1]去除了无关的空间; 你也可以用.rstrip().

演示:

>>> print ' '.join(['%-2s' % (i,) for i in l])
AA BB CC DD
>>> print ' '.join(['%-2s' % (i,) for i in (l + l)])
AA BB CC DD AA BB CC DD
>>> print ('%-2s ' * len(l))[:-1] % tuple(l)
AA BB CC DD
>>> print ('%-2s ' * len(l))[:-1] % tuple(l + l)
AA BB CC DD AA BB CC DD

时间统计:

>>> def joined_snippets(l):
...     ' '.join(['%-2s' % (i,) for i in l])
... 
>>> def joined_template(l):
...     ' '.join(['%-2s' for i in l])%tuple(l)
... 
>>> def multiplied_template(l):
...     ('%-2s ' * len(l))[:-1] % tuple(l)
... 
>>> from timeit import timeit
>>> l = ["AA","BB","CC","DD"]
>>> timeit('f(l)', 'from __main__ import l, joined_snippets as f')
1.3180170059204102
>>> timeit('f(l)', 'from __main__ import l, joined_template as f')
1.080280065536499
>>> timeit('f(l)', 'from __main__ import l, multiplied_template as f')
0.7333378791809082
>>> l *= 10
>>> timeit('f(l)', 'from __main__ import l, joined_snippets as f')
10.041708946228027
>>> timeit('f(l)', 'from __main__ import l, joined_template as f')
5.52706503868103
>>> timeit('f(l)', 'from __main__ import l, multiplied_template as f')
2.8013129234313965

乘法模板选项将其他选项留在灰尘中.