Question

我正在处理2个列表，如下所示：

list_a = [x,y,z,.....]  
list_b = [xa,xb,xc,xd,xe,ya,yb,yc,yd,za,zb,zc,zd,ze,zf]

我想要实现的是，在安排如下数据时制作更多列表：

list_x = [x,xa,xb,xc,xd,xe]  
list_y = [y,ya,yb,yc,yd]  
list_z = [z,za,zb,zc,zd,ze,zf]

现在，如果我使用如下的循环：

final_list=[]
for item in list_a:
    for value in list_b:
    if value[0] == item:
        print item, value

它会过滤数据，但无法达到所需的格式你们能否对此发表一些有价值的评论谢谢

Answer 1

list_a = ['x','y','z']  
list_b = ['xa','xb','xc','xd','xe','ya','yb','yc','yd','za','zb','zc','zd','ze','zf']  

print [[x for x in list_b if x.startswith(y)] for y in list_a]

输出：

[['xa', 'xb', 'xc', 'xd', 'xe'], ['ya', 'yb', 'yc', 'yd'], ['za', 'zb', 'zc', 'zd', 'ze', 'zf']]

或者更确切地说：

print [(y,[x for x in list_b if x.startswith(y)]) for y in list_a]

输出：

[('x', ['xa', 'xb', 'xc', 'xd', 'xe']), ('y', ['ya', 'yb', 'yc', 'yd']), ('z', ['za', 'zb', 'zc', 'zd', 'ze', 'zf'])]

Answer 2

您可以使用此代码：

list_a = ['x','y','z']
list_b = ['xa','xb','xc','xd','xe','ya','yb','yc','yd','za','zb','zc','zd','ze','zf']
print [(key, [_ for _ in list_b if key == _[0]]) for key in list_a]

它为您提供了一个元组列表，第一个条目是单个字母，第二个条目是列表。

或者你这样做没有这样的元组：

print [[key] + [_ for _ in list_b if key == _[0]] for key in list_a]

Answer 3

不是100％关于此格式，但您可以使用列表列表。

list_a = ["x","y","z"]  
list_b = ["xa","xb","xc","xd","xe","ya","yb","yc","yd","za","zb","zc","zd","ze","zf"]  
final_list = []
for item in list_a:
    item_list = [item]
    for value in list_b:
        if value[0] == item:
            item_list.append(value)
    final_list.append(item_list)
print final_list

它返回 [[＆＃39; x＆＃39;，＆＃39; xa＆＃39;，＆＃39; xb＆＃39;，＆＃39; xc＆＃39;，＆＃39; xd＆＃39;，＆＃39 ; xe＆＃39;]，[＆＃39; y＆＃39;，＆＃39; ya＆＃39;，＆＃39; yb＆＃39;，＆＃39; yc＆＃39;，＆＃39; yd＆＃ 39;]，[＆＃39; z＆＃39;，＆＃39; za＆＃39;，＆＃39; zb＆＃39;，＆＃39; zc＆＃39;，＆＃39; zd＆＃39;，＆＃39; ze＆＃39;，＆＃39; zf＆＃39;]]

Answer 4

您可以使用itertools.groupby()和operator.itemgetter()：

来遵循功能方法

In [46]: from itertools import groupby

In [47]: from operator import itemgetter

In [48]: list_b = ['xa','xb','xc','xd','xe','ya','yb','yc','yd','za','zb','zc','zd','ze','zf']

In [49]: [[prefix] + list(group) for prefix, group in groupby(list_b, key=itemgetter(0))]
Out[49]: 
[['x', 'xa', 'xb', 'xc', 'xd', 'xe'],
 ['y', 'ya', 'yb', 'yc', 'yd'],
 ['z', 'za', 'zb', 'zc', 'zd', 'ze', 'zf']]

请务必注意，如果未订购list_b，则传递给groupby()的参数应为sorted(list_b)。

作为旁注，您只需将可选参数operator更改为key=itemgetter(0)即可删除模块key=lambda s: s[0]。

对两个列表进行分类并相应地安排在python中

4 个答案: