Iterate over a list in Python

• Difficulty Level : Medium
• Last Updated : 30 Jan, 2023

List is equivalent to arrays in other languages, with the extra benefit of being dynamic in size. In Python, the list is a type of container in Data Structures, which is used to store multiple data at the same time. Unlike Sets, lists in Python are ordered and have a definite count.

There are multiple ways to iterate over a list in Python. Let’s see all the different ways to iterate over a list in Python, and performance comparison between them.

Method #1: Using For loop

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]`` ` `# Using for loop``for` `i ``in` `list``:``    ``print``(i)`

Output:

```1
3
5
7
9```

Method #2: For loop and range()
In case we want to use the traditional for loop which iterates from number x to number y.

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]`` ` `# getting length of list``length ``=` `len``(``list``)`` ` `# Iterating the index``# same as 'for i in range(len(list))'``for` `i ``in` `range``(length):``    ``print``(``list``[i])`

Output:

```1
3
5
7
9```

Iterating using the index is not recommended if we can iterate over the elements (as done in Method #1).

Method #3: Using while loop

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]`` ` `# Getting length of list``length ``=` `len``(``list``)``i ``=` `0`` ` `# Iterating using while loop``while` `i < length:``    ``print``(``list``[i])``    ``i ``+``=` `1`

Output:

```1
3
5
7
9```

Method #4: Using list comprehension (Possibly the most concrete way).

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]`` ` `# Using list comprehension``[``print``(i) ``for` `i ``in` `list``]`

Output:

```1
3
5
7
9```

Method #5: Using enumerate()
If we want to convert the list into an iterable list of tuples (or get the index based on a condition check, for example in linear search you might need to save the index of minimum element), you can use the enumerate() function.

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]`` ` `# Using enumerate()``for` `i, val ``in` `enumerate``(``list``):``    ``print` `(i, ``","``,val)`

Output:

```0 , 1
1 , 3
2 , 5
3 , 7
4 , 9```

Note: Even method #2 can be used to find the index, but method #1 can’t (Unless an extra variable is incremented every iteration) and method #5 gives a concise representation of this indexing.

Method #6: Using numpy
For very large n-dimensional lists (for example an image array), it is sometimes better to use an external library such as numpy.

Python3

 `# Python program for``# iterating over array``import` `numpy as geek`` ` `# creating an array using ``# arrange method``a ``=` `geek.arange(``9``)`` ` `# shape array with 3 rows ``# and 4 columns``a ``=` `a.reshape(``3``, ``3``)`` ` `# iterating an array``for` `x ``in` `geek.nditer(a):``    ``print``(x)`

Output:

```0
1
2
3
4
5
6
7
8```

We can use np.ndenumerate() to mimic the behavior of enumerating. The extra power of NumPy comes from the fact that we can even control the way to visit the elements (Fortran order rather than C order, say :)) but the one caveat is that the np.nditer treats the array as read-only by default, so one must pass extra flags such as op_flags=[‘readwrite’] for it to be able to modify elements.

Method #7:  Using the iter function and the next function

Here is an additional approach using the iter function and the next function:

Python3

 `# Python3 code to iterate over a list``list` `=` `[``1``, ``3``, ``5``, ``7``, ``9``]` `# Create an iterator object using the iter function``iterator ``=` `iter``(``list``)` `# Use the next function to retrieve the elements of the iterator``try``:``    ``while` `True``:``        ``element ``=` `next``(iterator)``        ``print``(element)``except` `StopIteration:``    ``pass``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```1
3
5
7
9```

This approach creates an iterator object using the iter function and then uses the next function to retrieve the elements of the iterator one at a time. The try-except block is used to catch the StopIteration exception that is raised by the next function when the iterator is exhausted.

This approach has a time complexity of O(n), where n is the length of the list, because it iterates through the elements of the list once and performs a constant number of operations on each element. The space complexity is O(1), because it does not create any additional data structures to store intermediate results or the final result.

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