[英]Append new Pandas DataFrame to an old one without column names sorted
[英]How to get position for records grouped by one column sorted and sorted by another pandas DataFrame
我有一個非常大的 DataFrame,大約 100M 行,如下所示:
query score1 score2 key
0 query0 97.149704 1.317513 key1
1 query1 86.344880 1.337784 key2
2 query2 85.192480 1.312714 key3
3 query1 86.240326 1.317513 key4
4 query2 85.192480 1.312714 key5
...
我想通過"query"
對數據框進行分組,然后獲取按"score1"
和"score2"
排序的每一行的位置(越高越好),所以輸出應該如下所示 -
query score1 score2 key pos1 pos2
0 query0 97.149704 1.317513 key1 0 0
1 query1 86.344880 1.237784 key2 0 1
2 query2 85.192480 1.312714 key3 1 0
3 query1 86.240326 1.317513 key4 1 0
4 query2 85.492410 1.212714 key5 0 1
目前,我有一個看起來像這樣的函數:
def func(query, df, score1=True):
mini_df = df[df["query"] == query]
mini_df.reset_index(drop=True, inplace=True)
col_name = "pos_score2"
if score1:
col_name = "pos_score1"
mini_df[col_name] = mini_df.index
return mini_df
我從main()
調用:
p = Pool(cpu_count())
df_list = list(p.starmap(func, zip(queries, repeat(df))))
df = pd.concat(df_list, ignore_index=True)
但這需要很長時間。 我在具有 96 個 CPU、Intel Xeon 和 512G 內存的機器上運行它,它仍然需要超過 24 小時。 實現這一目標的更快方法是什么?
使用groupby
和rank
:
df[['pos1', 'pos2']] = (df.groupby('query')[['score1', 'score2']]
.rank(method='max', ascending=False)
.sub(1).astype(int))
print(df)
# Output
query score1 score2 key pos1 pos2
0 query0 97.149704 1.317513 key1 0 0
1 query1 86.344880 1.237784 key2 0 1
2 query2 85.192480 1.312714 key3 1 0
3 query1 86.240326 1.317513 key4 1 0
4 query2 85.492410 1.212714 key5 0 1
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