假设我在 Pandas 数据框中有以下数据：
| Paper ID | Author ID |
|---|---|
| Paper_1 | Author_1 |
| Paper_1 | Author_2 |
| Paper_2 | Author_2 |
| Paper_3 | Author_1 |
| Paper_3 | Author_2 |
| Paper_3 | Author_3 |
| Paper_4 | Author_1 |
| Paper_4 | Author_3 |
我需要找出非零协作的数量。所以，输出应该是：
(Author_1,Author_2) --> 2
(Author_1,Author_3) --> 1
任何帮助或建议将不胜感激。
【问题讨论】：
标签： python pandas dataframe numpy data-cleaning
如果数据相当小，那么在Paper ID 上合并将生成可以折叠/聚合的对：
# assume df has columns Paper ID, Author ID
df_merged = df.merge(df, on="Paper ID")
# keep only one instance of a collaboration
mask = df_merged["Author ID_x"] > df_merged["Author ID_y"]
# aggregate (note the use of the mask to avoid double-
# counting and self-collaborations as noted in the
# comment by Riccardo Bucco)
counts = (
df_merged[mask]
.groupby(["Author ID_x", "Author ID_y"])
.agg(collaboration_count=("Paper ID", "count"))
)
【讨论】：