用Python中的itertools/more-itertools对多列列表的项进行分组和组合。

Question

此代码：

from itertools import groupby, count 
L = [38, 98, 110, 111, 112, 120, 121, 898] 
groups = groupby(L, key=lambda item, c=count():item-next(c))
tmp = [list(g) for k, g in groups]

取[38, 98, 110, 111, 112, 120, 121, 898]，将其按连续数字分组，并将它们与最后的输出合并：

['38', '98', '110,112', '120,121', '898']

如何对具有多列的列表进行同样的处理，如下面的列表，您可以根据名称和第二列值的顺序对它们进行分组，然后合并。

换言之，这些数据：

L= [
['Italy','1','3']
['Italy','2','1'],
['Spain','4','2'],
['Spain','5','8'],
['Italy','3','10'],
['Spain','6','4'],
['France','5','3'],
['Spain','20','2']]

应提供以下输出：

[['Italy','1-2-3','3-1-10'],
['France','5','3'],
['Spain','4-5-6','2-8-4'],
['Spain','20','2']]

more-itertools应该更适合这个任务吗？

用Python中的itertools/more-itertools对多列列表的项进行分组和组合。

Answer 1

这基本上是相同的分组技术，但它不是使用itertools.count，而是使用enumerate来生成索引。

首先，我们对数据进行排序，以便将给定国家的所有项目组合在一起，并对数据进行排序。然后，我们使用groupby为每个国家组成一个小组。然后，我们在内部循环中使用groupby对每个国家的连续数据进行分组。最后，我们使用zip & .join将数据重新排列成所需的输出格式。

from itertools import groupby
from operator import itemgetter

lst = [
    ['Italy','1','3'],
    ['Italy','2','1'],
    ['Spain','4','2'],
    ['Spain','5','8'],
    ['Italy','3','10'],
    ['Spain','6','4'],
    ['France','5','3'],
    ['Spain','20','2'],
]

newlst = [[country] + ['-'.join(s) for s in zip(*[v[1][1:] for v in g])]
    for country, u in groupby(sorted(lst), itemgetter(0))
        for _, g in groupby(enumerate(u), lambda t: int(t[1][1]) - t[0])]

for row in newlst:
    print(row)

输出

['France', '5', '3']
['Italy', '1-2-3', '3-1-10']
['Spain', '20', '2']
['Spain', '4-5-6', '2-8-4']

我承认lambda有点神秘；可能最好使用适当的def函数。几分钟后我会在这里加进去。

下面是使用一个更易读的键函数的相同的事情。

def keyfunc(t):
    # Unpack the index and data
    i, data = t
    # Get the 2nd column from the data, as an integer
    val = int(data[1])
    # The difference between val & i is constant in a consecutive group
    return val - i

newlst = [[country] + ['-'.join(s) for s in zip(*[v[1][1:] for v in g])]
    for country, u in groupby(sorted(lst), itemgetter(0))
        for _, g in groupby(enumerate(u), keyfunc)]

Answer 2

您可以在相同的菜谱上构建，并修改lambda函数，以包含来自每一行的第一个条目(Country)。其次，您需要首先根据列表中国家的最后出现情况对列表进行排序。

from itertools import groupby, count


L = [
    ['Italy', '1', '3'],
    ['Italy', '2', '1'],
    ['Spain', '4', '2'],
    ['Spain', '5', '8'],
    ['Italy', '3', '10'],
    ['Spain', '6', '4'],
    ['France', '5', '3'],
    ['Spain', '20', '2']]


indices = {row[0]: i for i, row in enumerate(L)}
sorted_l = sorted(L, key=lambda row: indices[row[0]])
groups = groupby(
    sorted_l,
    lambda item, c=count(): [item[0], int(item[1]) - next(c)]
)
for k, g in groups:
    print [k[0]] + ['-'.join(x) for x in zip(*(x[1:] for x in g))]

输出：

['Italy', '1-2-3', '3-1-10']
['France', '5', '3']
['Spain', '4-5-6', '2-8-4']
['Spain', '20', '2']

Answer 3

而不是使用需要多个排序、检查等的itertools.groupby。下面是一种使用字典进行算法优化的方法：

d = {}
flag = False
for country, i, j in L:
    temp = 1
    try:
        item = int(i)
        for counter, recs in  d[country].items():
            temp += 1
            last = int(recs[-1][0])
            if item in {last - 1, last, last + 1}:
                recs.append([i, j])
                recs.sort(key=lambda x: int(x[0]))
                flag = True
                break
        if flag:
            flag = False
            continue
        else:
            d[country][temp] = [[i, j]]
    except KeyError:
        d[country] = {}
        d[country][1] = [[i, j]]

在一个更复杂的示例中演示：

L = [['Italy', '1', '3'],
 ['Italy', '2', '1'],
 ['Spain', '4', '2'],
 ['Spain', '5', '8'],
 ['Italy', '3', '10'],
 ['Spain', '6', '4'],
 ['France', '5', '3'],
 ['Spain', '20', '2'],
 ['France', '5', '44'],
 ['France', '9', '3'],
 ['Italy', '3', '10'],
 ['Italy', '5', '17'],
 ['Italy', '4', '13'],]

{'France': {1: [['5', '3'], ['5', '44']], 2: [['9', '3']]},
 'Spain': {1: [['4', '2'], ['5', '8'], ['6', '4']], 2: [['20', '2']]},
 'Italy': {1: [['1', '3'], ['2', '1'], ['3', '10'], ['3', '10'], ['4', '13']], 2: [['5', '17']]}}

# You can then produce the results in your intended format as below:
for country, recs in d.items():
    for rec in recs.values():
        i, j = zip(*rec)
        print([country, '-'.join(i), '-'.join(j)])

['France', '5-5', '3-44']
['France', '9', '3']
['Italy', '1-2-3-3-4', '3-1-10-10-13']
['Italy', '5', '17']
['Spain', '4-5-6', '2-8-4']
['Spain', '20', '2']