使用Python中的itertools/more-itertools对多列列表的项目进行分组和合并 [英] Group and combine items of multiple-column lists with itertools/more-itertools in Python

问题描述

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]

采用[38, 98, 110, 111, 112, 120, 121, 898]，将其按连续数字分组，并将其与最终输出合并:

Takes [38, 98, 110, 111, 112, 120, 121, 898] , groups it by consecutive numbers and merge them with this final output:

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

如何使用具有多列的列表的列表来完成同样的操作，例如下面的列表，您可以在列表中按名称将其分组，并使用其第二列的值进行合并，然后合并.

How can the same be done with a list of lists with multiple columns, like this list below where you can group them by name and the consecution of its second column value and then merge.

换句话说，该数据:

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']]

更多itertools 是否更适合此任务?

使用Python中的itertools/more-itertools将多列列表的项目分组并合并

Group and combine items of multiple-column lists with itertools/more-itertools in Python

推荐答案

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

This is essentially the same grouping technique, but rather than using itertools.count it uses enumerate to produce the indices.

首先，我们对数据进行排序，以便将给定国家/地区的所有商品归为一组，然后对数据进行排序.然后，我们使用groupby为每个国家/地区分组.然后，在内部循环中使用groupby将每个国家/地区的连续数据分组在一起.最后，我们使用zip& .join将数据重新排列为所需的输出格式.

First, we sort the data so that all items for a given country are grouped together, and the data is sorted. Then we use groupby to make a group for each country. Then we use groupby in the inner loop to group together the consecutive data for each country. Finally, we use zip & .join to re-arrange the data into the desired output format.

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；最好使用适当的def函数.我将在几分钟之内将其添加.

I admit that lambda is a bit cryptic; it'd probably better to use a proper def function instead. I'll add that here in a few minutes.

使用更易读的键功能也是一样.

Here's the same thing using a much more readable key function.

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)]

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