用一根柱子或另一根柱子将一只熊猫的数据显示出来。

This is my dataframe :

这是我的dataframe:

df = pd.DataFrame({'name' : ['name1', 'name2', 'name1', 'name3'],
                    'rate' : [1,2,2,3],
                    'id' : range(4)})

   id   name  rate
0   0  name1     1
1   1  name2     2
2   2  name1     2
3   3  name3     3

I want to group the rows of a pandas dataframe if they have the same values in column name OR in column rate.

如果在列名或列速率中具有相同的值，我想对熊猫数据存储器的行进行分组。

          id            name       rate
0  [0, 1, 2]  [name1, name2]  [1, 2, 2]
1        [3]           name3        [3]

I have a huge dataframe so I don't want to iterate over each row (unless that the only solution). What should I do ?

我有一个很大的dataframe，所以我不想遍历每一行(除非这是唯一的解决方案)。我该怎么办?

(I can use Numpy arrays instead of Pandas dataframe)

(我可以使用Numpy数组而不是熊猫dataframe)

1 个解决方案

#1

Your conditions are unboundedly transitive. Say in rows 2i, 2i + 1 the name is shared and in rows 2i + 1, 2i + 2 the rate is shared, you need to keep linking rows.

你的条件是无限的过渡性的。在第2i行，2i + 1行名称是共享的，在第2i + 1行，2i + 2行速率是共享的，你需要保持链接行。

One way to solve this is using the graph theory's connected components algorithm.

解决这个问题的一种方法是使用图论的连通分量算法。

For this you can use networkx. In code, it could be as follows:

为此，您可以使用networkx。在代码中，可以如下所示:

import networkx as nx
import itertools

G = nx.Graph()
G.add_nodes_from(df.id)
G.add_edges_from(
    [(r1[1]['id'], r2[1]['id']) for (r1, r2) in itertools.product(df.iterrows(), df.iterrows()) if r1[1].id < r2[1].id and (r1[1]['rate'] == r2[1]['rate'] or r1[1]['name'] == r2[1]['name'])]
)

Let's create a group column, indicating, for each row, its group:

让我们创建一个组列，为每一行表示它的组:

df['group'] = df['id'].map(
    dict(itertools.chain.from_iterable([[(ee, i) for ee in e] for (i, e) in enumerate(nx.connected_components(G))])))
>>> df.group
0    0
1    0
2    0
3    1

Now you just need to groupby the group column, and apply a list.

现在只需要将group列分组，并应用一个列表。

#1