Pandas: Find an element's Index in Series [7 Ways]

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# Pandas: Find an element's Index in Series

To find an element's index in a Series in Pandas:

Use bracket notation to check for equality.
The expression will return a new Series containing the element's index in the first position.
Use the index attribute to get the first element in the Series.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# 2    5
# dtype: int64
print(series[series == 5])

print('-' * 50)

index_of_5 = series[series == 5].index[0]
print(index_of_5)  # 👉️ 2

pandas find element index in series

The code for this article is available on GitHub

We first used bracket notation to get a new Series object that contains the element's position at index 0.

main.py

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# 2    5
# dtype: int64
print(series[series == 5])

The next step is to use the Series.index property to get the first element.

main.py

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index_of_5 = series[series == 5].index[0]
print(index_of_5)  # 👉️ 2

This approach also works when there are multiple occurrences of the same value in the Series.

main.py

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import pandas as pd

series = pd.Series([5, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# 0    5
# 2    5
# dtype: int64
print(series[series == 5])

print('-' * 50)

print(series[series == 5].index[0])  # 👉️ 0
print(series[series == 5].index[1])  # 👉️ 2

The code for this article is available on GitHub

If you need to convert the Series of indices to a list, use the list class.

main.py

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import pandas as pd

series = pd.Series([5, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# 👇️ [0, 2]
print(list(series[series == 5].index))

# Making sure the specified element exists in the Series with `try/except`

When using this approach, you have to make sure the specified element exists in the Series, otherwise, you'd get an IndexError.

You can use a try/except statement if you need to handle the potential error.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# Series([], dtype: int64)
print(series[series == 100])

print('-' * 50)

try:
    index_of_100 = series[series == 100].index[0]
    print(index_of_100)
except IndexError:
    # 👇️ this runs
    print('The given element does NOT exist in the Series')

making sure the specified element exists in the series with try except

The code for this article is available on GitHub

The Series in the example doesn't contain a 100 value, so the expression returns a new, empty Series object.

Trying to access the empty Series object at index 0 causes an error, so we wrapped the code in a try/except block.

If the given element doesn't exist, the except block runs.

# Pandas: Find an element's Index in Series using `Index()` class

You can also find an element's index in a Series by converting the Series to an Index object and using the get_loc() method.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])


index_of_5 = pd.Index(series).get_loc(5)
print(index_of_5)  # 👉️ 2

print(series[index_of_5])  # 👉️ 5

find element index in series using index class

The code for this article is available on GitHub

We used the Index() constructor to convert the Series to an Index object.

Index objects are immutable sequences that are used for indexing and alignment.

Index objects have a get_loc method that returns the index of the supplied value.

If the given value doesn't exist in the Series, you would get a KeyError exception.

You can use a try/except block to handle the potential error.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])


try:
    index_of_100 = pd.Index(series).get_loc(100)
    print(index_of_100)
except KeyError:
    # 👇️ This runs
    print('The given value does NOT exist in the Series')

handle key error exception if value not in series

If there are multiple occurrences of the same value in the Series, the get_loc method returns a boolean array.

main.py

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import pandas as pd

series = pd.Series([5, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

index_of_5 = pd.Index(series).get_loc(5)

# 👇️ [ True False  True False False False]
print(index_of_5)

print(series[index_of_5].index[0])  # 👉️ 0
print(series[index_of_5].index[1])  # 👉️ 2

find element index in series for multiple occurrences

The array contains True values for the matching elements.

# Pandas: Find an element's Index in a Series using `where()`

You can also use the Series.where() method to find an element's index in a Series.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

index_of_5 = series.where(series == 5).first_valid_index()
print(index_of_5)  # 👉️ 2

find element index in series using where

The code for this article is available on GitHub

The Series.where method replaces the values where the given condition is False.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# 0    NaN
# 1    NaN
# 2    5.0
# 3    NaN
# 4    NaN
# 5    NaN
# dtype: float64
print(series.where(series == 5))

The last step is to use the first_valid_index method to get the index of the first non-NA value.

If the given value is not in the Series, the first_valid_index() method will return None.

You can use an if statement if you need to check if the variable is or is not None.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])


index_of_100 = series.where(series == 100).first_valid_index()
print(index_of_100)  # 👉️ None

if index_of_100 is not None:
    print(index_of_100)
else:
    # 👇️ this runs
    print('The supplied value is NOT in the Series')

The supplied value doesn't exist in the Series, so None is returned.

# Pandas: Find an element's Index in a Series using `argmax()`

You can also use the Series.argmax() method to find an element's index in a Series.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

index_of_5 = (series == 5).argmax()
print(index_of_5)  # 👉️ 2

find element index in series using argmax

The code for this article is available on GitHub

The Series.argmax() method returns the index of the largest value in the Series.

We first used a condition to get a boolean Series that contains a True value for the item that meets the condition.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

# 0    False
# 1    False
# 2     True
# 3    False
# 4    False
# 5    False
# dtype: bool
print(series == 5)

True values get converted to 1 and False values get converted to 0, so the argmax() method returns the index of the True value.

# Pandas: Find an element's Index in a Series using `list()`

You can also convert the Series to a list and use the list.index() method to find an element's index in a Series.

main.py

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import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

index_of_5 = list(series).index(5)
print(index_of_5)  # 👉️ 2

find element index in series using conversion to list

The code for this article is available on GitHub

We first used the list() class to convert the Series to a list.

This enables us to use the list.index() method.

The list.index() method returns the first index of the supplied value.

If the value doesn't exist in the list, a ValueError is raised.

You can use a try/except block if you need to handle the potential error.

main.py

Copied!
import pandas as pd

series = pd.Series([1, 3, 5, 7, 9, 11], index=[0, 1, 2, 3, 4, 5])

try:
    index_of_100 = list(series).index(100)
    print(index_of_100)
except ValueError:
    # 👇️ this runs
    print('The value does NOT exist in the Series')

handle potential value error if value not in series

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Pandas: Find an element's Index in Series [7 Ways]

# Table of Contents