在多列上筛选熊猫数据框的最快方法

Question

I have a pandas dataframe with several columns that labels data in a final column, for example,

df = pd.DataFrame( {'1_label' : ['a1','b1','c1','d1'],
                    '2_label' : ['a2','b2','c2','d2'],
                    '3_label' : ['a3','b3','c3','d3'],
                    'data'    : [1,2,3,4]})

df =      1_label 2_label 3_label  data
     0      a1      a2      a3     1
     1      b1      b2      b3     2
     2      c1      c2      c3     3
     3      d1      d2      d3     4

and a list of tuples,

list_t = [('a1','a2','a3'), ('d1','d2','d3')]

I want to filter this dataframe and return a new dataframe containing only the rows that correspond to the tuples in my list.

result =        1_label 2_label 3_label  data
            0      a1      a2      a3     1
            1      d1      d2      d3     4

My naive (and C++ inspired) solution was to use append (like vector::push_back)

for l1, l2, l3 in list_t:
    if df[(df['1_label'] == l1) & 
          (df['2_label'] == l2) & 
          (df['3_label'] == l3)].empty is False:
        result = result.append(df[(df['1_label'] == l1) & 
                              (df['2_label'] == l2) &
                              (df['3_label'] == l3)]

While my solution works I suspect it is horrendously slow for large dataframes and large list of tuples as I think pandas creates a new dataframe upon each call to append. Could anyone suggest a faster/cleaner way to do this? Thanks!

Answer 1

If I understood correctly, merge should do the job:

pd.DataFrame(list_t, columns=['1_label', '2_label', '3_label']).merge(df)
Out[73]: 
  1_label 2_label 3_label  data
0      a1      a2      a3     1
1      d1      d2      d3     4

Answer 2

Assuming no duplicates, you could create index out of the columns you want to "filter" on:

In [10]: df
Out[10]: 
  1_label 2_label 3_label  data
0      a1      a2      a3     1
1      b1      b2      b3     2
2      c1      c2      c3     3
3      d1      d2      d3     4

In [11]: df.set_index(['1_label', '2_label', '3_label'])\
    .loc[[('a1','a2','a3'), ('d1','d2','d3')]]\
    .reset_index()
Out[11]: 
  1_label 2_label 3_label  data
0      a1      a2      a3     1
1      d1      d2      d3     4