我这样做但感觉这可以用更少的代码来实现。毕竟是Python。从列表开始,我根据字符串前缀将该列表拆分为子集。
# Splitting a list into subsets
# expected outcome:
# [['sub_0_a', 'sub_0_b'], ['sub_1_a', 'sub_1_b']]
mylist = ['sub_0_a', 'sub_0_b', 'sub_1_a', 'sub_1_b']
def func(l, newlist=[], index=0):
newlist.append([i for i in l if i.startswith('sub_%s' % index)])
# create a new list without the items in newlist
l = [i for i in l if i not in newlist[index]]
if len(l):
index += 1
func(l, newlist, index)
func(mylist)
答案 0 :(得分:16)
您可以使用itertools.groupby
:
>>> import itertools
>>> mylist = ['sub_0_a', 'sub_0_b', 'sub_1_a', 'sub_1_b']
>>> for k,v in itertools.groupby(mylist,key=lambda x:x[:5]):
... print k, list(v)
...
sub_0 ['sub_0_a', 'sub_0_b']
sub_1 ['sub_1_a', 'sub_1_b']
或完全按照您的指定:
>>> [list(v) for k,v in itertools.groupby(mylist,key=lambda x:x[:5])]
[['sub_0_a', 'sub_0_b'], ['sub_1_a', 'sub_1_b']]
当然,常见的警告适用(确保您的列表使用您用于分组的相同键进行排序),并且您可能需要稍微复杂的关键函数来实现真实世界数据......
答案 1 :(得分:2)
In [28]: mylist = ['sub_0_a', 'sub_0_b', 'sub_1_a', 'sub_1_b']
In [29]: lis=[]
In [30]: for x in mylist:
i=x.split("_")[1]
try:
lis[int(i)].append(x)
except:
lis.append([])
lis[-1].append(x)
....:
In [31]: lis
Out[31]: [['sub_0_a', 'sub_0_b'], ['sub_1_a', 'sub_1_b']]
答案 2 :(得分:2)
使用itertools
'groupby
:
def get_field_sub(x): return x.split('_')[1]
mylist = sorted(mylist, key=get_field_sub)
[ (x, list(y)) for x, y in groupby(mylist, get_field_sub)]