英文:
pandas: group dataframe rows into different clusters
问题
我有这个数据框：
df = pd.DataFrame({
'forms_a_cluster': [False, False, True, True, True, False, False, False,
True, True, False, True, True, True, False],
'cluster_number':[False, False, 1, 1, 1, False, False, False,
2, 2, False, 3, 3, 3, False]})
想法是，当满足某些条件的行时，将这些情况选择为True，当连续的行满足条件时，它们就形成了一个簇。我想能够为每个簇添加标签，如cluster_1，cluster_2，cluster_3等。我已经提供了希望得到的输出示例，使用cluster_number列。但是，由于在实际数据中，我需要在不同数据集上多次执行此操作，这些数据集的行数不同，每次簇的大小也不同，我不知道如何做到这一点。提前感谢您的帮助！
英文:
I have this dataframe:
df = pd.DataFrame({
'forms_a_cluster': [False, False, True, True, True, False, False, False,
True, True, False, True, True, True, False],
'cluster_number':[False, False, 1, 1, 1, False, False, False,
2, 2, False, 3, 3, 3, False]})
The idea is that I have some criteria which, when certain rows have met it, selects those cases as True, and when consecutive rows meet the criteria, they then form a cluster. I want to be able to label each cluster as cluster_1, cluster_2, cluster_3 etc. I've given an example of the hoped for output with the column cluster_number. But I have no idea how to do this, given that in the real data, I have to do it many times on different datasets which have a different number of rows and the cluster sizes will be different every time. Do you have any idea how to go about this? Thanks in advance!
答案1
得分: 2
你可以使用过滤后的 groupby.ngroup，然后 reindex 来添加 False：
df['cluster_number'] = (df[df['forms_a_cluster']]
.groupby('id').ngroup().add(1)
.reindex(df.index, fill_value=False)
)
输出：
id forms_a_cluster cluster_number
0 1 False False
1 1 False False
2 2 True 1
3 2 True 1
4 2 True 1
5 3 False False
6 3 False False
7 3 False False
8 4 True 2
9 4 True 2
10 5 False False
11 6 True 3
12 6 True 3
13 6 True 3
14 7 False False
请注意，这是代码的翻译和输出的示例。如有其他问题或需要进一步的帮助，请告诉我。
英文:
You can use a filtered groupby.ngroup then reindex to add the False:
df['cluster_number'] = (df[df['forms_a_cluster']]
.groupby('id').ngroup().add(1)
.reindex(df.index, fill_value=False)
)
Output:
id forms_a_cluster cluster_number
0 1 False False
1 1 False False
2 2 True 1
3 2 True 1
4 2 True 1
5 3 False False
6 3 False False
7 3 False False
8 4 True 2
9 4 True 2
10 5 False False
11 6 True 3
12 6 True 3
13 6 True 3
14 7 False False
答案2
得分: 0
你可以使用ngroup来完成这个操作：
df["cluster_number"] = False
df.loc[df["forms_a_cluster"], "cluster_number"] = (
df.loc[df["forms_a_cluster"]].groupby(["id", "forms_a_cluster"]).ngroup() + 1
)
print(df)
id forms_a_cluster cluster_number
0 1 False False
1 1 False False
2 2 True 1
3 2 True 1
4 2 True 1
5 3 False False
6 3 False False
7 3 False False
8 4 True 2
9 4 True 2
10 5 False False
11 6 True 3
12 6 True 3
13 6 True 3
14 7 False False
英文:
You can use ngroup to do this:
df["cluster_number"] = False
df.loc[df["forms_a_cluster"], "cluster_number"] = (
df.loc[df["forms_a_cluster"]].groupby(["id", "forms_a_cluster"]).ngroup() + 1
)
print(df)
id forms_a_cluster cluster_number
0 1 False False
1 1 False False
2 2 True 1
3 2 True 1
4 2 True 1
5 3 False False
6 3 False False
7 3 False False
8 4 True 2
9 4 True 2
10 5 False False
11 6 True 3
12 6 True 3
13 6 True 3
14 7 False False
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