Question

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)

Answer 1

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......

Answer 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']]

Answer 3

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)]

Python根据模式将列表拆分为子集

3 个答案: