如何按值分组列表？

时间：2019-06-15 18:32:10

标签： python python-2.7 list grouping element

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]]

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

我想做的是将3个元素组成一组。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

直到现在，我只设法根据每个数字（1、2、3）将其名称分组为子列表。

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']]

7 个答案:

答案 0 :(得分：1)

这是一种类似于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)

output_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

output_l2 ：

Evelyn Heather Evelyn
Norma Dorothy Harry

答案 1 :(得分：1)

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),]

然后您可以通过执行以下操作打印出元素：

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

打印出：

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

答案 2 :(得分：1)

结合使用zip和itertools.groupby可以非常简洁地完成此操作。首先，按编号对列表进行排序，然后按分组和邮政编码。如果您想要字符串，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']

答案 3 :(得分：0)

我认为最简单的答案是使用字典根据您拥有的号码（将是字典的键）来收集结果。然后您可以按长度筛选保存在字典中的结果：

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']

答案 4 :(得分：0)

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

答案 5 :(得分：0)

通过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

答案 6 :(得分：0)

使用numpy是另一种方式：

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']
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