# Python – Sort dictionaries list by Key’s Value list index

• Last Updated : 01 Aug, 2020

Given list of dictionaries, sort dictionaries on basis of Key’s index value.

Input : [{“Gfg” : [6, 7, 8], “is” : 9, “best” : 10},
{“Gfg” : [2, 0, 3], “is” : 11, “best” : 19},
{“Gfg” : [4, 6, 9], “is” : 16, “best” : 1}], K = “Gfg”, idx = 0
Output : [{‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}, {‘Gfg’: [6, 7, 8], ‘is’: 9, ‘best’: 10}]
Explanation : 2<4<6, hence dictionary ordered in that way by 0th index of Key.

Input : [{“Gfg” : [6, 7, 8], “is” : 9, “best” : 10},
{“Gfg” : [2, 0, 3], “is” : 11, “best” : 19},
{“Gfg” : [4, 6, 9], “is” : 16, “best” : 1}], K = “Gfg”, idx = 1
Output : [{‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}, {‘Gfg’: [6, 7, 8], ‘is’: 9, ‘best’: 10}]
Explanation : 0<6<7, hence dictionary ordered in that way by 1st index.

Method #1 : Using sorted() + lambda

The combination of above functions can be used to solve this problem. In this, we perform sort using sorted and logic based on list index is provided in lambda function.

## Python3

 `# Python3 code to demonstrate working of ``# Sort dictionaries list by Key's Value list index``# Using sorted() + lambda`` ` `# initializing lists``test_list ``=` `[{``"Gfg"` `: [``6``, ``7``, ``8``], ``"is"` `: ``9``, ``"best"` `: ``10``}, ``             ``{``"Gfg"` `: [``2``, ``0``, ``3``], ``"is"` `: ``11``, ``"best"` `: ``19``},``             ``{``"Gfg"` `: [``4``, ``6``, ``9``], ``"is"` `: ``16``, ``"best"` `: ``1``}]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `"Gfg"`` ` `# initializing idx ``idx ``=` `2`` ` `# using sorted() to perform sort in basis of 1 parameter key and ``# index``res ``=` `sorted``(test_list, key ``=` `lambda` `ele: ele[K][idx])``     ` `# printing result ``print``(``"The required sort order : "` `+` `str``(res))`

Output

The original list : [{‘Gfg’: [6, 7, 8], ‘is’: 9, ‘best’: 10}, {‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}]
The required sort order : [{‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [6, 7, 8], ‘is’: 9, ‘best’: 10}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}]

Method #2 : Using sorted() + lambda (Additional parameter in case of tie)

This is modification in sorting of values, adding another parameter in case of tie of values among list.

## Python3

 `# Python3 code to demonstrate working of ``# Sort dictionaries list by Key's Value list index``# Using sorted() + lambda (Additional parameter in case of tie)`` ` `# initializing lists``test_list ``=` `[{``"Gfg"` `: [``6``, ``7``, ``9``], ``"is"` `: ``9``, ``"best"` `: ``10``}, ``             ``{``"Gfg"` `: [``2``, ``0``, ``3``], ``"is"` `: ``11``, ``"best"` `: ``19``},``             ``{``"Gfg"` `: [``4``, ``6``, ``9``], ``"is"` `: ``16``, ``"best"` `: ``1``}]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `"Gfg"`` ` `# initializing idx ``idx ``=` `2`` ` `# initializing K2 ``K2 ``=` `"best"`` ` `# using sorted() to perform sort in basis of 2 parameter key``# inner is evaluated after the outer key in lambda order``res ``=` `sorted``(``sorted``(test_list, key ``=` `lambda` `ele: ele[K2]), key ``=` `lambda` `ele: ele[K][idx])``     ` `# printing result ``print``(``"The required sort order : "` `+` `str``(res))`

Output

The original list : [{‘Gfg’: [6, 7, 9], ‘is’: 9, ‘best’: 10}, {‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}]
The required sort order : [{‘Gfg’: [2, 0, 3], ‘is’: 11, ‘best’: 19}, {‘Gfg’: [4, 6, 9], ‘is’: 16, ‘best’: 1}, {‘Gfg’: [6, 7, 9], ‘is’: 9, ‘best’: 10}]

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