基于特定元素合并值不为null/none的数组

Question

我一直试图用一个公共字段连接多个数组，但没有成功。我想从这样的情况出发：

mainArray = [['dog', None, None, 'bark', None], 
             ['dog', 'brown', None, None, None],
             ['dog', None, 'happy', None, None],
             ['cat', 'black', None, None, None],
             ['cat', None, None, 'soft', None],
             ['cat', None, None, None, 'purr']]

对这样的事情：

mainArray = [['dog', 'brown', 'happy', 'bark', None],
             ['cat', 'black', None, 'soft', 'purr']]

我知道这应该是相当简单，但我还没有找到一个正确的实现这一点。请把我引向正确的方向。你知道吗

Answer 1

假设键是每个子列表的第一项，并且不匹配的字段不会重叠，那么使用^{}和^{}有一种方法：

from itertools import groupby, dropwhile

r = [[next(dropwhile(lambda x: x is None, i), None) for i in zip(*g)] 
                       for _, g in groupby(mainArray, lambda x: x[0])]
print(r)
# [['dog', 'brown', 'happy', 'bark', None], 
#  ['cat', 'black', None, 'soft', 'purr']]

Answer 2

您可以这样做，假设每个子数组的每个位置都有一个以上的非None值（如示例数据所示）：

import collections
from pprint import pprint

mainArray = [['dog', None, None, 'bark', None],
             ['dog', 'brown', None, None, None],
             ['dog', None, 'happy', None, None],
             ['cat', 'black', None, None, None],
             ['cat', None, None, 'soft', None],
             ['cat', None, None, None, 'purr']]

temp_dict = collections.OrderedDict()
for row in mainArray:
    key = row[0]
    if key not in temp_dict:
        temp_dict[key] = [row[1:]]
    else:
        temp_dict[key].append(row[1:])

mainArray = []
for key,rows in temp_dict.items():
    merged = [None] * len(rows[0])
    for row in rows:
        for i in range(len(row)):
            if row[i] is not None:
                merged[i] = row[i]
    mainArray.append([key] + merged)

pprint(mainArray)

输出（这是正确的，尽管与您显示的略有不同）：

[['dog', 'brown', 'happy', 'bark', None],
 ['cat', 'black', None, 'soft', 'purr']]