This is a very interesting question!
I think it depends on the following aspects:
accessing single row by index (index is sorted and unique) should have runtime O(m)
where m << n_rows
accessing single row by index (index is NOT unique and is NOT sorted) should have runtime O(n_rows)
accessing single row by index (index is NOT unique and is sorted) should have runtime O(m)
where m < n_rows
)
accessing row(s) (independently of an index) by boolean indexing should have runtime O(n_rows)
Demo:
index is sorted and unique:
In [49]: df = pd.DataFrame(np.random.rand(10**5,6), columns=list('abcdef'))
In [50]: %timeit df.loc[random.randint(0, 10**4)]
The slowest run took 27.65 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 331 μs per loop
In [51]: %timeit df.iloc[random.randint(0, 10**4)]
1000 loops, best of 3: 275 μs per loop
In [52]: %timeit df.query("a > 0.9")
100 loops, best of 3: 7.84 ms per loop
In [53]: %timeit df.loc[df.a > 0.9]
100 loops, best of 3: 2.96 ms per loop
index is NOT sorted and is NOT unique:
In [54]: df = pd.DataFrame(np.random.rand(10**5,6), columns=list('abcdef'), index=np.random.randint(0, 10000, 10**5))
In [55]: %timeit df.loc[random.randint(0, 10**4)]
100 loops, best of 3: 12.3 ms per loop
In [56]: %timeit df.iloc[random.randint(0, 10**4)]
1000 loops, best of 3: 262 μs per loop
In [57]: %timeit df.query("a > 0.9")
100 loops, best of 3: 7.78 ms per loop
In [58]: %timeit df.loc[df.a > 0.9]
100 loops, best of 3: 2.93 ms per loop
index is NOT unique and is sorted:
In [64]: df = pd.DataFrame(np.random.rand(10**5,6), columns=list('abcdef'), index=np.random.randint(0, 10000, 10**5)).sort_index()
In [65]: df.index.is_monotonic_increasing
Out[65]: True
In [66]: %timeit df.loc[random.randint(0, 10**4)]
The slowest run took 9.70 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 478 μs per loop
In [67]: %timeit df.iloc[random.randint(0, 10**4)]
1000 loops, best of 3: 262 μs per loop
In [68]: %timeit df.query("a > 0.9")
100 loops, best of 3: 7.81 ms per loop
In [69]: %timeit df.loc[df.a > 0.9]
100 loops, best of 3: 2.95 ms per loop