In time series data, it's fairly common to need to compute the last known value "as of" a particular date. However, missing data is the norm, so it's a touch more complicated than doing a simple binary search. Here is an implementation using array operations that takes these things into account:

def asof_locs(stamps, where, mask):
"""
Parameters
----------
stamps : array of timestamps
where : array of timestamps
Values to determine the "as of" for
mask : array of booleans where data is not NA

Returns
-------
locs : array of ints
Locations of each "as of" value, -1 for NA
"""
locs = np.where(locs > 0, locs - 1, -1)

result[(locs == 0) & (where < stamps[first])] = -1

return result


Some algorithmic notes. First, let's run this through line_profiler on a large time series to see where time is being taken up:

In [9]: rng = date_range('1/1/2000', '6/1/2000', freq='10s')

In [10]: s = Series(np.random.randn(len(rng)), index=rng)

In [11]: rng
Out[11]:
<class 'pandas.tseries.index.DatetimeIndex'>
[2000-01-01 00:00:00, ..., 2000-06-01 00:00:00]
Length: 1313281, Freq: 10S, Timezone: None

In [12]: s = s.resample('s')

In [13]: len(s)
Out[13]: 13132801

In [25]: lprun -f asof_locs asof_locs(s.index, s.index[5::5], mask)
Timer unit: 1e-06 s

File: foo.py
Function: asof_locs at line 3
Total time: 0.793543 s

Line #   % Time  Line Contents
==============================================================
4               """
5               Parameters
6               ----------
7               stamps : array of timestamps
8               where : array of timestamps
9                   Values to determine the "as of" for
10               mask : array of booleans where data is NA
11
12               Returns
13               -------
14               locs : array of ints
15                   Locations of each "as of" value, -1 for NA
16               """
17    63.3       locs = stamps[mask].searchsorted(where, side='right')
18     8.3       locs = np.where(locs > 0, locs - 1, -1)
19
21
23     3.6       result[(locs == 0) & (where < stamps[first])] = -1
24
25     0.0       return result


The main trickiness here is this step:

result = np.arange(len(stamps))[mask].take(locs)


Since the indices returned by searchsorted are relative to the filtered values, you need to remap to what would have been the indices in the original array, and this does exactly that. Lastly, you might be interested in breaking up stamps[mask].searchsorted(where, side='right') to see how much time is spend in stamps[mask] versus searchsorted:

In [28]: timeit stamps[mask]
10 loops, best of 3: 127 ms per loop