Python – Get next key in Dictionary

Sometimes, while working with Python dictionaries, we can have a problem in which we need to extract the next key in order of dictionary. This can have application as from Python 3.6 onwards the dictionaries are ordered. Lets discuss certain ways in which this task can be performed. 

Method #1 : Using index() + loop (O (n))

The combination of above functions can be used to perform this task. In this, we perform the conversion of dictionary elements to list. And then index() is used to check for index and append the index count to get the next item.

Python3

# Python3 code to demonstrate working of
# Get next key in Dictionary
# Using index() + loop

# initializing dictionary
test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# initializing key
test_key = 'is'

# Get next key in Dictionary
# Using index() + loop
temp = list(test_dict)
try:
    res = temp[temp.index(test_key) + 1]
except (ValueError, IndexError):
    res = None

# printing result
print("The next key is : " + str(res))

Output : 

The original dictionary is : {'gfg': 1, 'best': 3, 'is': 2}
The next key is : best

Method #2 : Using iter() + next() (O (n))

This is yet another way in which this task can be performed. In this, we convert the dictionary to iterator using iter() and then extract the next key using next(). 

Python3

# Python3 code to demonstrate working of
# Get next key in Dictionary
# Using iter() + next()

# initializing dictionary
test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# initializing key
test_key = 'is'

# Get next key in Dictionary
# Using iter() + next()
res = None
temp = iter(test_dict)
for key in temp:
    if key == test_key:
        res = next(temp, None)

# printing result
print("The next key is : " + str(res))

Output : 

The original dictionary is : {'gfg': 1, 'best': 3, 'is': 2}
The next key is : best

Method #3 :  If you have lots of queries, they should be fast -> (O (1))

Create two additional dictionaries “index_of_key” and “key_of_index” in O (n). Then it is easy finding an index of any key of original dictionary and choose any other plus or minus this, check if it is in dictionary “key_of_index” and if, then get it.

Python3

#!/usr/bin/python3

# Python3 code to demonstrate working of
# Get next key in Dictionary
# Using two additional dictionaries
# "index_of_key" and "key_of_index"

# initializing dictionary
test_dict = {'gfg': 1, 'is': 2, 'best': 3}

# prepare additional dictionaries
ki = dict()
ik = dict()
for i, k in enumerate(test_dict):
    ki[k] = i   # dictionary index_of_key
    ik[i] = k     # dictionary key_of_index

# printing original dictionary
print("The original dictionary is:", test_dict)

# initializing key and offset
test_key = 'is'
offset = 1  # (1 for next key, but can be any existing distance)

# Get next key in Dictionary
index_of_test_key = ki['is']
index_of_next_key = index_of_test_key + offset
res = ik[index_of_next_key] if index_of_next_key in ik else None

# printing result
print("The next key is:", res)

Output : 

The original dictionary is : {'gfg': 1, 'best': 3, 'is': 2}
The next key is : best