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# Python – Extract Dictionary Items with Summation Greater than K

• Last Updated : 29 Aug, 2020

Given dictionary with value lists, extract all those items with values summing over K.

Input : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “geeks” : [1, 2, 3, 4]}, K = 10
Output : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “geeks” : [1, 2, 3, 4]}
Explanation : All elements are greater than 10.

Input : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “geeks” : [1, 2, 3, 4]}, K = 50
Output : {}
Explanation : No elements are greater than 50.

Method #1 :  Using dictionary comprehension + sum()

This is one of the ways in which this problem can be solved. In this one-liner, we iterate through the keys and append the dictionary item only if its value’s total crosses K computed using sum().

## Python3

 `# Python3 code to demonstrate working of ``# Extract Dictionary Items with Summation Greater than K``# Using dictionary comprehension + sum()`` ` `# initializing dictionary``test_dict ``=` `{``"Gfg"` `: [``6``, ``3``, ``4``], ``             ``"is"` `:  [``8``, ``10``, ``12``], ``             ``"Best"` `: [``10``, ``16``, ``14``],  ``             ``"for"` `: [``7``, ``4``, ``3``], ``             ``"geeks"` `: [``1``, ``2``, ``3``, ``4``]}`` ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))`` ` `# initializing K ``K ``=` `15`` ` `# summation using sum(), values extracted using items()``res ``=` `{key: val ``for` `key, val ``in` `test_dict.items() ``if` `sum``(val) > K}`` ` `# printing result ``print``(``"The computed dictionary : "` `+` `str``(res)) `
Output
```The original dictionary is : {'Gfg': [6, 3, 4], 'is': [8, 10, 12], 'Best': [10, 16, 14], 'for': [7, 4, 3], 'geeks': [1, 2, 3, 4]}
The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}
```

Method #2 : Using filter() + lambda() + sum() + dict()

This is another way in which this task can be performed. In this, computation part is done using filter() and lambda, summation using sum() and result converted to dictionary using dict().

## Python3

 `# Python3 code to demonstrate working of ``# Extract Dictionary Items with Summation Greater than K``# Using filter() + lambda() + sum() + dict()`` ` `# initializing dictionary``test_dict ``=` `{``"Gfg"` `: [``6``, ``3``, ``4``], ``             ``"is"` `:  [``8``, ``10``, ``12``], ``             ``"Best"` `: [``10``, ``16``, ``14``],  ``             ``"for"` `: [``7``, ``4``, ``3``], ``             ``"geeks"` `: [``1``, ``2``, ``3``, ``4``]}`` ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))`` ` `# initializing K ``K ``=` `15`` ` `# summation using sum(), values extracted using items()``# filter() + lambda used for computation``res ``=` `dict``(``filter``(``lambda` `ele: ``sum``(ele[``1``]) > K, test_dict.items()))`` ` `# printing result ``print``(``"The computed dictionary : "` `+` `str``(res)) `
Output
```The original dictionary is : {'Gfg': [6, 3, 4], 'is': [8, 10, 12], 'Best': [10, 16, 14], 'for': [7, 4, 3], 'geeks': [1, 2, 3, 4]}
The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}
```

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