Python2按值合并列表列表 [英] Python2 merge list of lists by value

问题描述

我有4个元素的列表，它看起来像:

I have a list of 4 elements and it looks like:

L = [['grape','green',2, 100], ['grape','purple',3,3], ['apple','red',2, 15], ['apple','greed',3, 10],
     ['apple','red',4, 4], ['banana','yellow',2, 3]]

我希望有一个函数可以通过第一个元素合并该列表，将第二个和第三个元素连接起来，然后对最后一个元素求和以得到输出，例如:

I would like to have a function to merge that list by the first element, concatenate the second and third element, then sum up the last element to get the output like:

[['grape',['green,purple'],[2,3],103],['banana','yellow',2, 3],['apple',['red','green','red'],[2,3,4],29]

我尝试了以下代码:

[(x,sum(map(itemgetter(3),y))) for x,y in itertools.groupby(L, itemgetter(0))], which can successfully sum up the last element, but

[(x,(map(itemgetter(1),y)),(map(itemgetter(2),y)),sum(map(itemgetter(3),y))) for x,y in itertools.groupby(L, itemgetter(0))]

没有按我预期的那样工作.

did not work as I expected.

有什么办法吗?

推荐答案

使用itertools.groupby()您需要两个单独的迭代:

Using itertools.groupby() you need two separate iteration:

In [20]: [[i[0], j, k, sum(z)] for i, j, k, z in [zip(*g) for _, g in groupby(L, itemgetter(0))]]
Out[20]: 
[['grape', ('green', 'purple'), (2, 3), 103],
 ['apple', ('red', 'greed', 'red'), (2, 3, 4), 29],
 ['banana', ('yellow',), (2,), 3]]

这是使用collections.defaultdict的另一种方法，但不是这种优化方法:

Here is another way using collections.defaultdict, but not that optimize :

In [34]: d = defaultdict(list)

In [35]: for i, j, k, z in L:
            d[i].append((j, k, z))
   ....:     

In [36]: [[i, j, k, sum(z)] for i, j, k, z in [[i, *zip(*values)] for i, values in d.items()]]
Out[36]: 
[['banana', ('yellow',), (2,), 3],
 ['apple', ('red', 'greed', 'red'), (2, 3, 4), 29],
 ['grape', ('green', 'purple'), (2, 3), 103]]

请注意，列表中的*zip(...)称为就地解包，仅在python 3.5+中可用.

Note that *zip(...) within a list is called in-place unpacking and is only available in python 3.5+.

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