将Python列表拆分为5组，然后打印

Question

我有一个清单项目，看起来如下：

dblist=['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve']

它可能有超过12个项目(可能是无限的)。--我想把这个列表分成五个组打印出来，并希望我的输出如下所示：

env1: one two three four five
env2: six seven eight nine ten
env3: eleven twelve

我怎样才能做到这一点？

我发现了一些很好的技巧，似乎能够完成这项工作，但我似乎无法使它与我的代码一起工作。例如，以下内容似乎可以做我想做的事情：

def n_split(iterable, n, fillvalue=None):
    num_extra = len(iterable) % n
    zipped = zip(*[iter(iterable)] * n)
    return zipped if not num_extra else zipped + [iterable[-num_extra:], ]

for ints in n_split(range(1,32), 5):
    print(([str(i) for i in ints]))

具有以下产出：

['1', '2', '3', '4', '5']
['6', '7', '8', '9', '10']
['11', '12', '13', '14', '15']
['16', '17', '18', '19', '20']
['21', '22', '23', '24', '25']
['26', '27', '28', '29', '30']
['31', '32', '33']

如开头所示，如何使用此技术以五个组的形式打印dblist？

Answer 1

使用str.join()

for group in n_split(dblist, 5):
    print ' '.join(group)

您可以添加enumerate()以添加计数器：

for count, group in enumerate(n_split(dblist, 5), 1):
    print 'env{}: {}'.format(count, ' '.join(group))

演示：

>>> def n_split(iterable, n, fillvalue=None):
...     num_extra = len(iterable) % n
...     zipped = zip(*[iter(iterable)] * n)
...     return zipped if not num_extra else zipped + [iterable[-num_extra:], ]
... 
>>> dblist=['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve']
>>> for group in n_split(dblist, 5):
...     print ' '.join(group)
... 
one two three four five
six seven eight nine ten
eleven twelve
>>> for count, group in enumerate(n_split(dblist, 5), 1):
...     print 'env{}: {}'.format(count, ' '.join(group))
... 
env1: one two three four five
env2: six seven eight nine ten
env3: eleven twelve

Answer 2

这是你想要的吗？

dblist=['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve']
for i in range(0,len(dblist),5):
    print(dblist[i:i+5])

['one', 'two', 'three', 'four', 'five']
['six', 'seven', 'eight', 'nine', 'ten']
['eleven', 'twelve']