# Python – Convert list of dictionaries to Dictionary Value list

Given a list of dictionary, convert to dictionary with same key mapped with all values in as value list.

Input : test_list = [{“Gfg” : 6, “is” : 9, “best” : 10},
{“Gfg” : 8, “is” : 11, “best” : 19}]
Output : {‘Gfg’: [6, 8], ‘is’: [9, 11], ‘best’: [10, 19]}
Explanation : 6, 8 of “Gfg” mapped as value list, similarly every other.

Input : test_list = [{“Gfg” : 6, “best” : 10},
{“Gfg” : 8, “best” : 19}]
Output : {‘Gfg’: [6, 8], ‘best’: [10, 19]}
Explanation : Same as above, conversion.

Method #1 : Using loop

This is brute way to solve this problem. In this, we iterate through all the dictionaries, and extract each key and convert to required dictionary in nested loops.

## Python3

 `# Python3 code to demonstrate working of  ` `# Convert list of dictionaries to Dictionary Value list ` `# Using loop ` `from` `collections ``import` `defaultdict ` ` `  `# initializing lists ` `test_list ``=` `[{``"Gfg"` `: ``6``, ``"is"` `: ``9``, ``"best"` `: ``10``},  ` `             ``{``"Gfg"` `: ``8``, ``"is"` `: ``11``, ``"best"` `: ``19``}, ` `             ``{``"Gfg"` `: ``2``, ``"is"` `: ``16``, ``"best"` `: ``10``}, ` `             ``{``"Gfg"` `: ``12``, ``"is"` `: ``1``, ``"best"` `: ``8``}, ` `             ``{``"Gfg"` `: ``22``, ``"is"` `: ``6``, ``"best"` `: ``8``}] ` ` `  `# printing original list ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# using loop to get dictionaries ` `# defaultdict used to make default empty list  ` `# for each key ` `res ``=` `defaultdict(``list``) ` `for` `sub ``in` `test_list: ` `    ``for` `key ``in` `sub: ` `        ``res[key].append(sub[key]) ` `     `  `# printing result  ` `print``(``"The extracted dictionary : "` `+` `str``(``dict``(res))) `

Output

```The original list : [{'Gfg': 6, 'is': 9, 'best': 10}, {'Gfg': 8, 'is': 11, 'best': 19}, {'Gfg': 2, 'is': 16, 'best': 10}, {'Gfg': 12, 'is': 1, 'best': 8}, {'Gfg': 22, 'is': 6, 'best': 8}]
The extracted dictionary : {'Gfg': [6, 8, 2, 12, 22], 'is': [9, 11, 16, 1, 6], 'best': [10, 19, 10, 8, 8]}
```

Method #2 : Using list comprehension + dictionary comprehension

The combination of above functionalities can be used to solve this problem. In this, we use dictionary comprehension to construct dictionary and list comprehension is used to extract values from original list.

## Python3

 `# Python3 code to demonstrate working of  ` `# Convert list of dictionaries to Dictionary Value list ` `# Using list comprehension + dictionary comprehension ` `from` `collections ``import` `defaultdict ` ` `  `# initializing lists ` `test_list ``=` `[{``"Gfg"` `: ``6``, ``"is"` `: ``9``, ``"best"` `: ``10``},  ` `             ``{``"Gfg"` `: ``8``, ``"is"` `: ``11``, ``"best"` `: ``19``}, ` `             ``{``"Gfg"` `: ``2``, ``"is"` `: ``16``, ``"best"` `: ``10``}, ` `             ``{``"Gfg"` `: ``12``, ``"is"` `: ``1``, ``"best"` `: ``8``}, ` `             ``{``"Gfg"` `: ``22``, ``"is"` `: ``6``, ``"best"` `: ``8``}] ` ` `  `# printing original list ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# dictionary and list comprehension  ` `# for shorthand to solution of problem ` `res ``=` `defaultdict(``list``) ` `{res[key].append(sub[key]) ``for` `sub ``in` `test_list ``for` `key ``in` `sub}  ` `     `  `# printing result  ` `print``(``"The extracted dictionary : "` `+` `str``(``dict``(res))) `

Output

```The original list : [{'Gfg': 6, 'is': 9, 'best': 10}, {'Gfg': 8, 'is': 11, 'best': 19}, {'Gfg': 2, 'is': 16, 'best': 10}, {'Gfg': 12, 'is': 1, 'best': 8}, {'Gfg': 22, 'is': 6, 'best': 8}]
The extracted dictionary : {'Gfg': [6, 8, 2, 12, 22], 'is': [9, 11, 16, 1, 6], 'best': [10, 19, 10, 8, 8]}
```

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