Find element's index in pandas Series

Question

I know this is a very basic question but for some reason I can't find an answer. How can I get the index of certain element of a Series in python pandas? (first occurrence would suffice)

I.e., I'd like something like:

import pandas as pd
myseries = pd.Series([1,4,0,7,5], index=[0,1,2,3,4])
print myseries.find(7) # should output 3

Certainly, it is possible to define such a method with a loop:

def find(s, el):
    for i in s.index:
        if s[i] == el: 
            return i
    return None

print find(myseries, 7)

but I assume there should be a better way. Is there?

Answer 1

>>> 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.

Answer 2

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

Answer 3

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.

Answer 4

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']

Answer 5

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

Answer 6

If you use numpy, you can get an array of the indecies that your value is found:

import numpy as np
import pandas as pd
myseries = pd.Series([1,4,0,7,5], index=[0,1,2,3,4])
np.where(myseries == 7)

This returns a one element tuple containing an array of the indecies where 7 is the value in myseries:

(array([3], dtype=int64),)

Answer 7

you can use Series.idxmax()

>>> import pandas as pd
>>> myseries = pd.Series([1,4,0,7,5], index=[0,1,2,3,4])
>>> myseries.idxmax()
3
>>> 

Answer 8

This is the most native and scalable approach I could find:

>>> myindex = pd.Series(myseries.index, index=myseries)

>>> myindex[7]
3

>>> myindex[[7, 5, 7]]
7    3
5    4
7    3
dtype: int64

Answer 9

Another way to do it that hasn't been mentioned yet is the tolist method:

myseries.tolist().index(7)

should return the correct index, assuming the value exists in the Series.

Answer 10

Often your value occurs at multiple indices:

>>> myseries = pd.Series([0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1])
>>> myseries.index[myseries == 1]
Int64Index([3, 4, 5, 6, 10, 11], dtype='int64')

Answer 11

The Pandas has builtin class Index with a function called get_loc. This function will either return

index (element index)
slice (if the specified number is in sequence)
array (bool array if the number is at multiple indexes)

Example:

import pandas as pd

>>> mySer = pd.Series([1, 3, 8, 10, 13])
>>> pd.Index(mySer).get_loc(10)  # Returns index
3  # Index of 10 in series

>>> mySer = pd.Series([1, 3, 8, 10, 10, 10, 13])
>>> pd.Index(mySer).get_loc(10)  # Returns slice
slice(3, 6, None)  # 10 occurs at index 3 (included) to 6 (not included)


# If the data is not in sequence then it would return an array of bool's.
>>> mySer = pd.Series([1, 10, 3, 8, 10, 10, 10, 13, 10])
>>> pd.Index(mySer).get_loc(10)
array([False, True, False, False, True, True, False, True])

There are many other options too but I found it very simple for me.