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# Python | Summation of dictionary list values

• Last Updated : 25 Jan, 2023

Sometimes, while working with Python dictionaries, we can have its values as lists. In this can, we can have a problem in that we just require the count of elements in those lists as a whole. This can be a problem in Data Science in which we need to get total records in observations. Let’s discuss certain ways in which this task can be performed.

Method #1: Using sum() + list comprehension
This task can be performed using sum function which can be used to get the summation and internal list comprehension can provide a mechanism to iterate this logic to all the keys of the dictionary.

## Python3

 `# Python3 code to demonstrate working of``# Summation of dictionary list values``# using sum() + list comprehension` `# initialize dictionary``test_dict ``=` `{``'gfg'``: [``5``, ``6``, ``7``], ``'is'``: [``10``, ``11``], ``'best'``: [``19``, ``31``, ``22``]}` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))` `# Summation of dictionary list values``# using sum() + list comprehension``res ``=` `sum``(``len``(sub) ``for` `sub ``in` `test_dict.values())` `# printing result``print``(``"Summation of dictionary list values are : "` `+` `str``(res))`

Output :

```The original dictionary is : {'best': [19, 31, 22], 'is': [10, 11], 'gfg': [5, 6, 7]}
Summation of dictionary list values are : 8```

Method #2: Using sum() + map()
This task can also be performed using map function in place of list comprehension to extend the logic of finding the length, rest all the functionality remaining same as the above method.

## Python3

 `# Python3 code to demonstrate working of``# Summation of dictionary list values``# using sum() + map()` `# initialize dictionary``test_dict ``=` `{``'gfg'` `: [``5``, ``6``, ``7``], ``'is'` `: [``10``, ``11``], ``'best'` `: [``19``, ``31``, ``22``]}` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))` `# Summation of dictionary list values``# using sum() + map()``res ``=` `sum``(``map``(``len``, test_dict.values()))` `# printing result``print``(``"Summation of dictionary list values are : "` `+` `str``(res))`

Output :

```The original dictionary is : {'best': [19, 31, 22], 'is': [10, 11], 'gfg': [5, 6, 7]}
Summation of dictionary list values are : 8```

Method #3: Using values(), extend() and len() methods

## Python3

 `# Python3 code to demonstrate working of``# Summation of dictionary list values` `# initialize dictionary``test_dict ``=` `{``'gfg'` `: [``5``, ``6``, ``7``], ``'is'` `: [``10``, ``11``], ``'best'` `: [``19``, ``31``, ``22``]}` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))` `# Summation of dictionary list values``x``=``list``(test_dict.values())``a``=``[]``for` `i ``in` `x:``    ``a.extend(i)``res``=``len``(a)``    ` `# printing result``print``(``"Summation of dictionary list values are : "` `+` `str``(res))`

Output

```The original dictionary is : {'gfg': [5, 6, 7], 'is': [10, 11], 'best': [19, 31, 22]}
Summation of dictionary list values are : 8```

Method #4: Using reduce()

Using reduce() and len(): This approach uses the reduce() function from the functools module to iteratively combine the values of the dictionary into a single iterable, and then uses the len() function to get the count of elements in the iterable. Time complexity is O(n) where n is the total number of elements in the lists, and auxiliary space is O(n) where n is the total number of elements in the lists.

## Python3

 `from` `functools ``import` `reduce` `# initialize dictionary``test_dict ``=` `{``'gfg'``: [``5``, ``6``, ``7``], ``'is'``: [``10``, ``11``], ``'best'``: [``19``, ``31``, ``22``]}` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))` `# Summation of dictionary list values``# Using reduce() and len()``# reduce() method applies a given function to all elements in an iterable and returns a single value``# len() method returns the length of the passed iterable``res ``=` `len``(``reduce``(``lambda` `x, y: x ``+` `y, test_dict.values()))` `# printing result``print``(``"Summation of dictionary list values are : "` `+` `str``(res))``# This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original dictionary is : {'gfg': [5, 6, 7], 'is': [10, 11], 'best': [19, 31, 22]}
Summation of dictionary list values are : 8```

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