如何获取按一列分组的记录的位置，该列由另一个熊猫数据框排序和排序

Question

I have a very large DataFrame with ~100M rows that looks like this:

    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
...

I want to group the dataframe by "query" and then get the position of each row sorted by "score1" and "score2" (higher is better) so the output should look like this -

    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

Currently, I have a function that looks something like this:

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

which I call from main() :

p = Pool(cpu_count())
df_list = list(p.starmap(func, zip(queries, repeat(df))))
df = pd.concat(df_list, ignore_index=True)

but it takes a long time. I am running this on machine with 96 CPUs Intel Xeon with 512G memory and it still takes more than 24 hrs. What would be a much faster way to achieve this?

Answer 1

Use groupby and 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