I’m impressed with all the answers here. This is not a new answer, just an attempt to summarize the timings of all these methods. I considered the case of a series with 25 elements and assumed the general case where the index could contain any values and you want the index value corresponding to the search value which is towards the end of the series.

Here are the speed tests on a 2012 Mac Mini in Python 3.9.10 with Pandas version 1.4.0.

In [1]: import pandas as pd

In [2]: import numpy as np

In [3]: data = [406400, 203200, 101600, 76100, 50800, 25400, 19050, 12700, 950
   ...: 0, 6700, 4750, 3350, 2360, 1700, 1180, 850, 600, 425, 300, 212, 150, 1
   ...: 06, 75, 53, 38]

In [4]: myseries = pd.Series(data, index=range(1,26))

In [5]: assert(myseries[21] == 150)

In [6]: %timeit myseries[myseries == 150].index[0]
179 µs ± 891 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [7]: %timeit myseries[myseries == 150].first_valid_index()
205 µs ± 3.67 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [8]: %timeit myseries.where(myseries == 150).first_valid_index()
597 µs ± 4.03 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [9]: %timeit myseries.index[np.where(myseries == 150)[0][0]]
110 µs ± 872 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [10]: %timeit pd.Series(myseries.index, index=myseries)[150]
125 µs ± 2.56 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [11]: %timeit myseries.index[pd.Index(myseries).get_loc(150)]
49.5 µs ± 814 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [12]: %timeit myseries.index[list(myseries).index(150)]
7.75 µs ± 36.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [13]: %timeit myseries.index[myseries.tolist().index(150)]
2.55 µs ± 27.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [14]: %timeit dict(zip(myseries.values, myseries.index))[150]
9.89 µs ± 79.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [15]: %timeit {v: k for k, v in myseries.items()}[150]
9.99 µs ± 67 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

@Jeff’s answer seems to be the fastest – although it doesn’t handle duplicates.

Correction: Sorry, I missed one, @Alex Spangher’s solution using the list index method is by far the fastest.

Update: Added @EliadL’s answer.

Hope this helps.

Amazing that such a simple operation requires such convoluted solutions and many are so slow. Over half a millisecond in some cases to find a value in a series of 25.

2022-02-18 Update

Updated all the timings with the latest Pandas version and Python 3.9. Even on an older computer, all the timings have significantly reduced (10 to 70%) compared to the previous tests (version 0.25.3).

Plus: Added two more methods utilizing dictionaries.