如何按值对列表进行分组？

Question

假设我有这样一个列表：

[['John', 1], ['Fred', 1], ['Carolyn', 1], ['Kenneth', 3], ['Ronald', 3], ['Dorothy', 2], ['Joyce', 2], ['Julia', 3], ['Deborah', 1], ['Jonathan', 2], ['Aaron', 2], ['Marie', 1], ['Adam', 2], ['Kevin', 2], ['Alice', 2], ['Jerry', 1], ['Kimberly', 1], ['Lawrence', 1], ['Louis', 2], ['Anthony', 1], ['Carolyn', 3], ['Edward', 2], ['Samuel', 3], ['Rachel', 1], ['Kathleen', 1], ['Fred', 3], ['Fred', 3], ['Gerald', 2], ['Donna', 2], ['Keith', 3], ['Matthew', 3], ['Stephanie', 1]]

如何获得此输出：

John Dorothy Kenneth
Fred Joyce Ronald
Carolyn Jonathan Julia
Deborah Aaron Carolyn
Marie Adam Samuel
Jerry Kevin Fred
Kimberly Alice Fred
Lawrence Louis Keith
Anthony Edward Matthew

我要做的是把三个元素分组。如果一个组不是由3个名称组成的，则不会显示该组。你知道吗

另一个例子：

输入：

[['Heather', 2], ['Evelyn', 1], ['Norma', 1], ['Evelyn', 3], ['Harry', 3], ['Sean', 1], ['Anna', 1], ['Jerry', 3], ['Anna', 3], ['Julia', 1], ['Dorothy', 2]]

预期产量：

Evelyn Heather Evelyn
Norma Dorothy Harry

到目前为止，我只根据每个数字（1、2、3）将他们的名字分组到子列表中。你知道吗

r = [[] for i in range(3)]
for i in l:
    if i[1] == 1:
        r[0].append(i[0])
    elif i[1] == 2:
        r[1].append(i[0])
    elif i[1] == 3:
        r[2].append(i[0])
print r

r = [['John', 'Fred', 'Carolyn', 'Deborah', 'Marie', 'Jerry', 'Kimberly', 'Lawrence', 'Anthony', 'Rachel', 'Kathleen', 'Stephanie'], ['Dorothy', 'Joyce', 'Jonathan', 'Aaron', 'Adam', 'Kevin', 'Alice', 'Louis', 'Edward', 'Gerald', 'Donna'], ['Kenneth', 'Ronald', 'Julia', 'Carolyn', 'Samuel', 'Fred', 'Fred', 'Keith', 'Matthew']]

Answer 1

下面是一个类似于this问题的方法：

l1 = [['John', 1], ['Fred', 1], ['Carolyn', 1], ['Kenneth', 3], ['Ronald', 3], ['Dorothy', 2], ['Joyce', 2], ['Julia', 3], ['Deborah', 1], ['Jonathan', 2], ['Aaron', 2], ['Marie', 1], ['Adam', 2], ['Kevin', 2], ['Alice', 2], ['Jerry', 1], ['Kimberly', 1], ['Lawrence', 1], ['Louis', 2], ['Anthony', 1], ['Carolyn', 3], ['Edward', 2], ['Samuel', 3], ['Rachel', 1], ['Kathleen', 1], ['Fred', 3], ['Fred', 3], ['Gerald', 2], ['Donna', 2], ['Keith', 3], ['Matthew', 3], ['Stephanie', 1]]
l2 = [['Heather', 2], ['Evelyn', 1], ['Norma', 1], ['Evelyn', 3], ['Harry', 3], ['Sean', 1], ['Anna', 1], ['Jerry', 3], ['Anna', 3], ['Julia', 1], ['Dorothy', 2]]

def get_exp(l):
    v = set(map(lambda x:x[1], l))
    nl = [[y[0] for y in l if y[1]==x] for x in v]
    return '\n'.join(list(map(' '.join, zip(*nl))))
output_l1 = get_exp(l1)
output_l2 = get_exp(l2)

输出\u l1：

John Dorothy Kenneth
Fred Joyce Ronald
Carolyn Jonathan Julia
Deborah Aaron Carolyn
Marie Adam Samuel
Jerry Kevin Fred
Kimberly Alice Fred
Lawrence Louis Keith
Anthony Edward Matthew

输出l2：

Evelyn Heather Evelyn
Norma Dorothy Harry

Answer 2

怎么样：

l = [['Heather', 2], ['Evelyn', 1], ['Norma', 1], ['Evelyn', 3], ['Harry', 3], ['Sean', 1], ['Anna', 1], ['Jerry', 3], ['Anna', 3], ['Julia', 1], ['Dorothy', 2]]
from itertools import groupby
def keyfunc(arr) :
   return arr[1]

l = sorted(l, key=keyfunc)
s =[[*x,] for  i,x in groupby(data , keyfunc)]
combinations = [*zip(*s),]

然后你就可以通过以下操作打印出元素：

for l in combinations :
   print(' '.join([x[0] for x in l]))

打印输出：

John Dorothy Kenneth
Fred Joyce Ronald
Carolyn Jonathan Julia
Deborah Aaron Carolyn
Marie Adam Samuel
Jerry Kevin Fred
Kimberly Alice Fred
Lawrence Louis Keith
Anthony Edward Matthew

Answer 3

您可以使用zip和itertools.groupby的组合非常简洁地实现这一点。首先，按数字对列表排序，然后分组和压缩。如果需要字符串，则可以加入：

from operator import itemgetter
from itertools import groupby

l = [['John', 1], ['Fred', 1], ['Carolyn', 1], ['Kenneth', 3], ['Ronald', 3], ['Dorothy', 2], ['Joyce', 2], ['Julia', 3], ['Deborah', 1], ['Jonathan', 2], ['Aaron', 2], ['Marie', 1], ['Adam', 2], ['Kevin', 2], ['Alice', 2], ['Jerry', 1], ['Kimberly', 1], ['Lawrence', 1], ['Louis', 2], ['Anthony', 1], ['Carolyn', 3], ['Edward', 2], ['Samuel', 3], ['Rachel', 1], ['Kathleen', 1], ['Fred', 3], ['Fred', 3], ['Gerald', 2], ['Donna', 2], ['Keith', 3], ['Matthew', 3], ['Stephanie', 1]]

l.sort(key = itemgetter(1))
groups = zip(*([name for name, g in n] for k, n in groupby(l, itemgetter(1))))

[" ".join(names) for names in groups]

输出：

['John Dorothy Kenneth',
 'Fred Joyce Ronald',
 'Carolyn Jonathan Julia',
 'Deborah Aaron Carolyn',
 'Marie Adam Samuel',
 'Jerry Kevin Fred',
 'Kimberly Alice Fred',
 'Lawrence Louis Keith',
 'Anthony Edward Matthew']