Find element's index in pandas Series

>>> myseries[myseries == 7]
3    7
dtype: int64
>>> myseries[myseries == 7].index[0]
3

Though I admit that there should be a better way to do that, but this at least avoids iterating and looping through the object and moves it to the C level.

Converting to an Index, you can use get_loc

In [1]: myseries = pd.Series([1,4,0,7,5], index=[0,1,2,3,4])

In [3]: Index(myseries).get_loc(7)
Out[3]: 3

In [4]: Index(myseries).get_loc(10)
KeyError: 10

Duplicate handling

In [5]: Index([1,1,2,2,3,4]).get_loc(2)
Out[5]: slice(2, 4, None)

Will return a boolean array if non-contiguous returns

In [6]: Index([1,1,2,1,3,2,4]).get_loc(2)
Out[6]: array([False, False,  True, False, False,  True, False], dtype=bool)

Uses a hashtable internally, so fast

In [7]: s = Series(randint(0,10,10000))

In [9]: %timeit s[s == 5]
1000 loops, best of 3: 203 µs per loop

In [12]: i = Index(s)

In [13]: %timeit i.get_loc(5)
1000 loops, best of 3: 226 µs per loop

As Viktor points out, there is a one-time creation overhead to creating an index (its incurred when you actually DO something with the index, e.g. the is_unique)

In [2]: s = Series(randint(0,10,10000))

In [3]: %timeit Index(s)
100000 loops, best of 3: 9.6 µs per loop

In [4]: %timeit Index(s).is_unique
10000 loops, best of 3: 140 µs per loop

In [92]: (myseries==7).argmax()
Out[92]: 3

This works if you know 7 is there in advance. You can check this with (myseries==7).any()

Another approach (very similar to the first answer) that also accounts for multiple 7's (or none) is

In [122]: myseries = pd.Series([1,7,0,7,5], index=['a','b','c','d','e'])
In [123]: list(myseries[myseries==7].index)
Out[123]: ['b', 'd']

Another way to do this, although equally unsatisfying is:

s = pd.Series([1,3,0,7,5],index=[0,1,2,3,4])

list(s).index(7)

returns: 3

On time tests using a current dataset I'm working with (consider it random):

[64]:    %timeit pd.Index(article_reference_df.asset_id).get_loc('100000003003614')
10000 loops, best of 3: 60.1 µs per loop

In [66]: %timeit article_reference_df.asset_id[article_reference_df.asset_id == '100000003003614'].index[0]
1000 loops, best of 3: 255 µs per loop


In [65]: %timeit list(article_reference_df.asset_id).index('100000003003614')
100000 loops, best of 3: 14.5 µs per loop