python – Change the name of a key in dictionary

The Question :

403 people think this question is useful

I want to change the key of an entry in a Python dictionary.

Is there a straightforward way to do this?

The Question Comments :

The Answer 1

789 people think this answer is useful

Easily done in 2 steps:

dictionary[new_key] = dictionary[old_key]
del dictionary[old_key]

Or in 1 step:

dictionary[new_key] = dictionary.pop(old_key)

which will raise KeyError if dictionary[old_key] is undefined. Note that this will delete dictionary[old_key].

>>> dictionary = { 1: 'one', 2:'two', 3:'three' }
>>> dictionary['ONE'] = dictionary.pop(1)
>>> dictionary
{2: 'two', 3: 'three', 'ONE': 'one'}
>>> dictionary['ONE'] = dictionary.pop(1)
Traceback (most recent call last):
  File "<input>", line 1, in <module>
KeyError: 1

The Answer 2

69 people think this answer is useful

if you want to change all the keys:

d = {'x':1, 'y':2, 'z':3}
d1 = {'x':'a', 'y':'b', 'z':'c'}

In [10]: dict((d1[key], value) for (key, value) in d.items())
Out[10]: {'a': 1, 'b': 2, 'c': 3}

if you want to change single key: You can go with any of the above suggestion.

The Answer 3

39 people think this answer is useful

pop’n’fresh

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

The Answer 4

30 people think this answer is useful

In python 2.7 and higher, you can use dictionary comprehension: This is an example I encountered while reading a CSV using a DictReader. The user had suffixed all the column names with ‘:’

ori_dict = {'key1:' : 1, 'key2:' : 2, 'key3:' : 3}

to get rid of the trailing ‘:’ in the keys:

corrected_dict = { k.replace(':', ''): v for k, v in ori_dict.items() }

The Answer 5

11 people think this answer is useful

Since keys are what dictionaries use to lookup values, you can’t really change them. The closest thing you can do is to save the value associated with the old key, delete it, then add a new entry with the replacement key and the saved value. Several of the other answers illustrate different ways this can be accomplished.

The Answer 6

7 people think this answer is useful

If you have a complex dict, it means there is a dict or list within the dict:

myDict = {1:"one",2:{3:"three",4:"four"}}
myDict[2][5] = myDict[2].pop(4)
print myDict

Output
{1: 'one', 2: {3: 'three', 5: 'four'}}

The Answer 7

5 people think this answer is useful

No direct way to do this, but you can delete-then-assign

d = {1:2,3:4}

d[newKey] = d[1]
del d[1]

or do mass key changes:

d = dict((changeKey(k), v) for k, v in d.items())

The Answer 8

5 people think this answer is useful
d = {1:2,3:4}

suppose that we want to change the keys to the list elements p=[‘a’ , ‘b’]. the following code will do:

d=dict(zip(p,list(d.values()))) 

and we get

{'a': 2, 'b': 4}

The Answer 9

4 people think this answer is useful

To convert all the keys in the dictionary

Suppose this is your dictionary:

>>> sample = {'person-id': '3', 'person-name': 'Bob'}

To convert all the dashes to underscores in the sample dictionary key:

>>> sample = {key.replace('-', '_'): sample.pop(key) for key in sample.keys()}
>>> sample
>>> {'person_id': '3', 'person_name': 'Bob'}

The Answer 10

4 people think this answer is useful

this function gets a dict, and another dict specifying how to rename keys; it returns a new dict, with renamed keys:

def rekey(inp_dict, keys_replace):
    return {keys_replace.get(k, k): v for k, v in inp_dict.items()}

test:

def test_rekey():
    assert rekey({'a': 1, "b": 2, "c": 3}, {"b": "beta"}) == {'a': 1, "beta": 2, "c": 3}

The Answer 11

2 people think this answer is useful

In case of changing all the keys at once. Here I am stemming the keys.

a = {'making' : 1, 'jumping' : 2, 'climbing' : 1, 'running' : 2}
b = {ps.stem(w) : a[w] for w in a.keys()}
print(b)
>>> {'climb': 1, 'jump': 2, 'make': 1, 'run': 2} #output

The Answer 12

2 people think this answer is useful

This will lowercase all your dict keys. Even if you have nested dict or lists. You can do something similar to apply other transformations.

def lowercase_keys(obj):
  if isinstance(obj, dict):
    obj = {key.lower(): value for key, value in obj.items()}
    for key, value in obj.items():         
      if isinstance(value, list):
        for idx, item in enumerate(value):
          value[idx] = lowercase_keys(item)
      obj[key] = lowercase_keys(value)
  return obj 

json_str = {"FOO": "BAR", "BAR": 123, "EMB_LIST": [{"FOO": "bar", "Bar": 123}, {"FOO": "bar", "Bar": 123}], "EMB_DICT": {"FOO": "BAR", "BAR": 123, "EMB_LIST": [{"FOO": "bar", "Bar": 123}, {"FOO": "bar", "Bar": 123}]}}

lowercase_keys(json_str)


Out[0]: {'foo': 'BAR',
 'bar': 123,
 'emb_list': [{'foo': 'bar', 'bar': 123}, {'foo': 'bar', 'bar': 123}],
 'emb_dict': {'foo': 'BAR',
  'bar': 123,
  'emb_list': [{'foo': 'bar', 'bar': 123}, {'foo': 'bar', 'bar': 123}]}}

The Answer 13

0 people think this answer is useful

You can associate the same value with many keys, or just remove a key and re-add a new key with the same value.

For example, if you have keys->values:

red->1
blue->2
green->4

there’s no reason you can’t add purple->2 or remove red->1 and add orange->1

The Answer 14

0 people think this answer is useful

Method if anyone wants to replace all occurrences of the key in a multi-level dictionary.

Function checks if the dictionary has a specific key and then iterates over sub-dictionaries and invokes the function recursively:

def update_keys(old_key,new_key,d):
    if isinstance(d,dict):
        if old_key in d:
            d[new_key] = d[old_key]
            del d[old_key]
        for key in d:
            updateKey(old_key,new_key,d[key])

update_keys('old','new',dictionary)

The Answer 15

0 people think this answer is useful

An example of complete solution

Declare a json file which contains mapping you want

{
  "old_key_name": "new_key_name",
  "old_key_name_2": "new_key_name_2",
}

Load it

with open("<filepath>") as json_file:
    format_dict = json.load(json_file)

Create this function to format a dict with your mapping

def format_output(dict_to_format,format_dict):
  for row in dict_to_format:
    if row in format_dict.keys() and row != format_dict[row]:
      dict_to_format[format_dict[row]] = dict_to_format.pop(row)
  return dict_to_format

The Answer 16

0 people think this answer is useful

Be aware of the position of pop:
Put the key you want to delete after pop()
orig_dict[‘AAAAA’] = orig_dict.pop(‘A’)

orig_dict = {'A': 1, 'B' : 5,  'C' : 10, 'D' : 15}   
# printing initial 
print ("original: ", orig_dict) 

# changing keys of dictionary 
orig_dict['AAAAA'] = orig_dict.pop('A')
  
# printing final result 
print ("Changed: ", str(orig_dict)) 


The Answer 17

0 people think this answer is useful

I wrote this function below where you can change the name of a current key name to a new one.

def change_dictionary_key_name(dict_object, old_name, new_name):
    '''
    [PARAMETERS]: 
        dict_object (dict): The object of the dictionary to perform the change
        old_name (string): The original name of the key to be changed
        new_name (string): The new name of the key
    [RETURNS]:
        final_obj: The dictionary with the updated key names
    Take the dictionary and convert its keys to a list.
    Update the list with the new value and then convert the list of the new keys to 
    a new dictionary
    '''
    keys_list = list(dict_object.keys())
    for i in range(len(keys_list)):
        if (keys_list[i] == old_name):
            keys_list[i] = new_name

    final_obj = dict(zip(keys_list, list(dict_object.values()))) 
    return final_obj

Assuming a JSON you can call it and rename it by the following line:

data = json.load(json_file)
for item in data:
    item = change_dictionary_key_name(item, old_key_name, new_key_name)

Conversion from list to dictionary keys has been found here:
https://www.geeksforgeeks.org/python-ways-to-change-keys-in-dictionary/

The Answer 18

-4 people think this answer is useful

I haven’t seen this exact answer:

dict['key'] = value

You can even do this to object attributes. Make them into a dictionary by doing this:

dict = vars(obj)

Then you can manipulate the object attributes like you would a dictionary:

dict['attribute'] = value

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