Vectorized lookup on a pandas dataframe
I have two DataFrames . . .
df1 is a table I need to pull values from using index, column pairs retrieved from multiple columns in df2.
I see there is a function get_value which works perfectly when given an index and column value, but when trying to vectorize this function to create a new column I am failing...
df1 = pd.DataFrame(np.arange(20).reshape((4, 5)))
df1.columns = list('abcde')
df1.index = ['cat', 'dog', 'fish', 'bird']
a b c d e
cat 0 1 2 3 4
dog 5 6 7 8 9
fish 10 11 12 13 14
bird 15 16 17 18 19
df1.get_value('bird, 'c')
17
Now what I need to do is to create an entire new column on df2 -- when indexing df1 based on index, column pairs from the animal, letter columns specified in df2 effectively vectorizing the pd.get_value function above.
df2 = pd.DataFrame(np.arange(20).reshape((4, 5)))
df2['animal'] = ['cat', 'dog', 'fish', 'bird']
df2['letter'] = list('abcd')
0 1 2 3 4 animal letter
0 0 1 2 3 4 cat a
1 5 6 7 8 9 dog b
2 10 11 12 13 14 fish c
3 15 16 17 18 19 bird d
resulting in . . .
0 1 2 3 4 animal letter looked_up
0 0 1 2 3 4 cat a 0
1 5 6 7 8 9 dog b 6
2 10 11 12 13 14 fish c 12
3 15 16 17 18 19 bird d 18
Deprecation Notice:
lookupwas deprecated in v1.2.0
There's a function aptly named lookup that does exactly this.
df2['looked_up'] = df1.lookup(df2.animal, df2.letter)
df2
0 1 2 3 4 animal letter looked_up
0 0 1 2 3 4 cat a 0
1 5 6 7 8 9 dog b 6
2 10 11 12 13 14 fish c 12
3 15 16 17 18 19 bird d 18
If looking for a bit faster approach then zip will help in case of small dataframe i.e
k = list(zip(df2['animal'].values,df2['letter'].values))
df2['looked_up'] = [df1.get_value(*i) for i in k]
Output:
0 1 2 3 4 animal letter looked_up 0 0 1 2 3 4 cat a 0 1 5 6 7 8 9 dog b 6 2 10 11 12 13 14 fish c 12 3 15 16 17 18 19 bird d 18
As John suggested you can simplify the code which will be much faster.
df2['looked_up'] = [df1.get_value(r, c) for r, c in zip(df2.animal, df2.letter)]
In case of missing data use if else i.e
df2['looked_up'] = [df1.get_value(r, c) if not pd.isnull(c) | pd.isnull(r) else pd.np.nan for r, c in zip(df2.animal, df2.letter) ]
For small dataframes
%%timeit
df2['looked_up'] = df1.lookup(df2.animal, df2.letter)
1000 loops, best of 3: 801 µs per loop
k = list(zip(df2['animal'].values,df2['letter'].values))
df2['looked_up'] = [df1.get_value(*i) for i in k]
1000 loops, best of 3: 399 µs per loop
[df1.get_value(r, c) for r, c in zip(df2.animal, df2.letter)]
10000 loops, best of 3: 87.5 µs per loop
For large dataframe
df3 = pd.concat([df2]*10000)
%%timeit
k = list(zip(df3['animal'].values,df3['letter'].values))
df2['looked_up'] = [df1.get_value(*i) for i in k]
1 loop, best of 3: 185 ms per loop
df2['looked_up'] = [df1.get_value(r, c) for r, c in zip(df3.animal, df3.letter)]
1 loop, best of 3: 165 ms per loop
df2['looked_up'] = df1.lookup(df3.animal, df3.letter)
100 loops, best of 3: 8.82 ms per loop
lookup and get_value are great answers if your values exist in lookup dataframe.
However, if you've (row, column) pairs not present in the lookup dataframe, and want the lookup value be NaN -- merge and stack is one way to do it
In [206]: df2.merge(df1.stack().reset_index().rename(columns={0: 'looked_up'}),
left_on=['animal', 'letter'], right_on=['level_0', 'level_1'],
how='left').drop(['level_0', 'level_1'], 1)
Out[206]:
0 1 2 3 4 animal letter looked_up
0 0 1 2 3 4 cat a 0
1 5 6 7 8 9 dog b 6
2 10 11 12 13 14 fish c 12
3 15 16 17 18 19 bird d 18
Test with adding non-existing (animal, letter) pair
In [207]: df22
Out[207]:
0 1 2 3 4 animal letter
0 0.0 1.0 2.0 3.0 4.0 cat a
1 5.0 6.0 7.0 8.0 9.0 dog b
2 10.0 11.0 12.0 13.0 14.0 fish c
3 15.0 16.0 17.0 18.0 19.0 bird d
4 NaN NaN NaN NaN NaN dummy NaN
In [208]: df22.merge(df1.stack().reset_index().rename(columns={0: 'looked_up'}),
left_on=['animal', 'letter'], right_on=['level_0', 'level_1'],
how='left').drop(['level_0', 'level_1'], 1)
Out[208]:
0 1 2 3 4 animal letter looked_up
0 0.0 1.0 2.0 3.0 4.0 cat a 0.0
1 5.0 6.0 7.0 8.0 9.0 dog b 6.0
2 10.0 11.0 12.0 13.0 14.0 fish c 12.0
3 15.0 16.0 17.0 18.0 19.0 bird d 18.0
4 NaN NaN NaN NaN NaN dummy NaN NaN