如何按值对列表进行分组？

Question

假设我有一个这样的列表：

[['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]]

我怎样才能得到这个输出：

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

我想要做的是将 3 个元素分组。如果一个组不是由 3 个名称组成，则不会显示。

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]]

预期输出：

Evelyn Heather Evelyn
Norma Dorothy Harry

到目前为止，我只设法根据每个数字（1、2、3）将他们的名字分组到子列表中。

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

这是类似于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_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

您可以使用zip 和itertools.groupby 的组合非常简洁地做到这一点。首先，按数字对列表进行排序，然后分组和压缩。如果你想要字符串，你可以加入：

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']

Answer 3

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),]

然后您可以通过以下方式打印出元素：

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

打印出来：

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 4

我认为最简单的答案是使用字典按您拥有的数字（这将是字典键）收集结果。然后可以按长度筛选字典中保存的结果：

In [7]: from collections import defaultdict                                     

In [8]: results = defaultdict(list)                                             

In [9]: name_list = [['bob', 1], ['cindy', 1], ['ted', 2]]                      

In [10]: for (value, key) in name_list: 
    ...:     results[key].append(value) 
    ...:                                                                        

In [11]: results                                                                
Out[11]: defaultdict(list, {1: ['bob', 'cindy'], 2: ['ted']})



In [13]: for key in results: 
    ...:     if len(results.get(key)) == 2: 
    ...:         print( 'found a result of length 2: ', results.get(key)) 
    ...:                                                                        
found a result of length 2:  ['bob', 'cindy']

Answer 5

lst1 = [['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]]
lst2 = [['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 my_print(lst):
    d = {v: list(g) for v, g in groupby(sorted(lst, key=lambda k: k[-1]), lambda v: v[-1])}
    while True:
        try:
            i1 = d[1].pop(0)
            i2 = d[2].pop(0)
            i3 = d[3].pop(0)
            print('{} {} {}'.format(i1[0], i2[0], i3[0]))
        except IndexError:
            break

my_print(lst1)
print('*' * 80)
my_print(lst2)

打印：

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
********************************************************************************
Evelyn Heather Evelyn
Norma Dorothy Harry

Answer 6

使用numpy 是另一种方法：

import math
import numpy as np

data = [['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]]

array = np.array([x[0] for x in data])
array = np.resize(array,(3,math.ceil(len(array)/3))).T

[" ".join(x) for x in array]

输出：

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

Answer 7

按第二个元素将名称分组到字典中，然后将它们压缩在一起。

def gum(l):
    g = {}
    for n, k in l:
        g.setdefault(k, []).append(n)
    return zip(*g.values())

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]]

print '\n\n'.join('\n'.join(' '.join(n) for n in l) for l in [gum(l1), gum(l2)])

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

Evelyn Heather Evelyn
Norma Dorothy Harry