这是一个非常有趣的问题！
我认为这取决于以下几个方面：
按索引访问单行（ ）应具有运行时O(m)，其中m << n_rows
按索引访问单行（ ）应该具有运行时O(n_rows)
通过索引访问单行（ ）应该有运行O(m)在那里m < n_rows）
通过布尔索引访问行（独立于索引）应具有运行时 O(n_rows)
演示：
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
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
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
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