CODEWITHSUNDEEP
pythonpandas
scima96
scima96

Reputation: 383

Fastest way to find a Pandas index-column value pair

I have a largish DataFrame with a date index ['Date'] and several columns. One column is a string identifier ['Type'], with related data in the remaining columns. I need to add newData to the DataFrame, but only if the date-type pair (i.e. index-ColumnValue pair) is not already present in the DataFrame. Checking for the existing pairing takes up ~95% of my code computing time, so I really need to find a quicker way to do it.

Options already considered, with timings in increasing order of speed::

existing_pair = len(compiledData[(compiledData['Type'] == newData['Type']) 
& (compiledData.index == newData['Date'])]) > 0 
# average = 114 ms

existing_pair = newData['Date'] in compiledData[compiledData['Type'] == 
                newData['Type']].index
# average = 68 ms

existing_pair = compiledData[compiledData.index == newData['Type']]['Type'].
                isin([newData['Date']]).any()
# average = 44 ms

I am relatively new to Python so I am sure there are better (= faster) ways of checking for an index-colVal pair. Or, it may be that my entire data structure is wrong. Would appreciate any pointers anyone can offer.

Edit: Sample of the compiledData dataframe:

           Type  ColA     ColB  ColC  ColD    ColE  ColF
2021-01-19    B  83.0  -122.15  0.0  11.0   11.000  11.0
2021-01-19    D  83.0 -1495.48  0.0  11.0   11.000  11.0
2021-03-25    D  83.0   432.00  0.0  11.0   11.000  11.0
2021-04-14    D  83.0   646.00  0.0  11.0   11.000  11.0
2021-04-16    A  20.0    11.00  0.0  30.0   11.000  11.0
2021-04-25    D  83.0   -26.82  0.0  11.0   11.000  11.0
2021-04-28    B  83.0  -651.00  0.0  11.0   11.000  11.0

Upvotes: 5

Views: 1157

Answers (2)

anon01
anon01

Reputation: 11171

Index value lookup is faster than column value lookup. I don't know the implementation details (it looks like lookup depends on number of rows). Here is a performance comparison:

def test_value_matches(df, v1, v2):
    # return True if v1, v2 found in df columns, else return False
    if any(df[(df.c1 == v1) & (df.c2 == v2)]):
        return True
    return False

def test_index_matches(df, v1, v2):
    # returns True if (v1, v2) found in (multi) index, else returns False
    if (v1, v2) in df.index:
        return True
    return False

# test dependence of funcs above on num rows in df:
for n in [int(j) for j in [1e4, 1e5, 1e6, 1e7]]:
    df = pd.DataFrame(np.random.random(size=(n, 2)), columns=["c1", "c2"])
    v1, v2 = df.sample(n=1).iloc[0]
    %timeit test_value_matches(df, v1, v2)
    
    # create an index based on column values:
    df2 = df.set_index(["c1", "c2"])
    %timeit test_index_matches(df2, v1, v2)

output

421 µs ± 22.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
10.5 µs ± 175 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

557 µs ± 5.35 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
10.3 µs ± 143 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

3.77 ms ± 166 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.5 µs ± 185 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

22.4 ms ± 2.06 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
28.1 µs ± 10.2 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

Note that this ignores indexing time itself, which may be significant; this approach probably works best with repeated lookups on the same df. For n=1e7, the performance is something like your problem on my machine; the indexed version is ~1000x faster (although apparently growing with n).

Upvotes: 2

tdy
tdy

Reputation: 41327

Try using Index.difference with a MultiIndex:

  1. It looks like compiledData already has a date index, so append Type to the index:

    compiledData = compiledData.set_index('Type', append=True)

  2. And it looks like newData has Date as an independent column, so set its index to ['Date', 'Type']:

    newData = newData.set_index(['Date', 'Type'])

  3. Now that both have a date/type MultiIndex, take their Index.difference to get the unique newData indexes:

    unique = newData.index.difference(compiledData.index)

So newData.loc[unique] rows can be added using append:

compiledData.append(newData.loc[unique]).reset_index(level=1)

Or concat:

pd.concat([compiledData, newData.loc[unique]]).reset_index(level=1)

Upvotes: 1

Related Questions