Python 2D列表到字典

Question

我有一个二维列表，必须从二维列表中获取 2 列，并将每列中的值作为键:值对放置。

示例:

table = [[15, 29, 6, 2],
        [16, 9, 8, 0],
        [7, 27, 16, 0]]

def averages(table, col, by):

    columns = tuple(([table[i][col] for i in range(len(table))]))  #Place col column into tuple so it can be placed into dictionary
    groupby = tuple(([table[i][by] for i in range(len(table))]))   #Place groupby column into tuple so it can be placed into dictionary

    avgdict = {}
    avgdict[groupby] = [columns]
    print(avgdict)

averages(table, 1, 3)

输出是:

{(2, 0, 0): [(29, 9, 27)]}

我试图让输出相等:

{0:36, 2:29}

所以基本上 0 的 2 个键都添加了它们的值

我很难理解如何将每个键与其值分开 如果键相等，则将值相加。

编辑:我只使用 Python 标准库，并没有为这个问题实现 numpy。

Answer 1

您可以创建一个空字典，然后遍历 groupby 的每个元素。如果字典中存在 groupby 中的元素，则将 columns 中对应的元素添加到字典中的值中。否则，将groupby中的元素作为key，将columns中对应的元素作为value。实现如下:

table = [[15, 29, 6, 2],
    [16, 9, 8, 0],
    [7, 27, 16, 0]]

def averages(table, col, by):
    columns = tuple(([table[i][col] for i in range(len(table))]))  #Place col column into tuple so it can be placed into dictionary
    groupby = tuple(([table[i][by] for i in range(len(table))]))   #Place groupby column into tuple so it can be placed into dictionary

    avgdict = {}

    for x in range(len(groupby)):
        key = groupby[x]
        if key in avgdict:
            avgdict[key] += columns[x]
        else:
            avgdict[key] = columns[x]

    print(avgdict)

averages(table, 1, 3)

否则，如果你想保留你的初始平均数，那么你可以将 averages() 函数更改为

def averages(table, col, by):
    columns = tuple(([table[i][col] for i in range(len(table))]))  #Place col column into tuple so it can be placed into dictionary
    groupby = tuple(([table[i][by] for i in range(len(table))]))   #Place groupby column into tuple so it can be placed into dictionary

    avgdict = {}
    avgdict[groupby] = [columns]

    newdict = {}

    for key in avgdict:
        for x in range(len(key)):
            if key[x] in newdict:
                newdict[key[x]] += avgdict[key][0][x]
            else:
                newdict[key[x]] = avgdict[key][0][x]

    print(newdict)