返回给定字典键的下一个键，python 3.6+

Question

I am trying to find a way to get the next key of a Python 3.6+ (which are ordered)

For example:

dict = {'one':'value 1','two':'value 2','three':'value 3'}

What I am trying to achieve is a function to return the next key. something like:

next_key(dict, current_key='two')   # -> should return 'three' 

This is what I have so far:

def next_key(dict,key):
    key_iter = iter(dict)  # create iterator with keys
    while k := next(key_iter):    #(not sure if this is a valid way to iterate over an iterator)
        if k == key:   
            #key found! return next key
            try:    #added this to handle when key is the last key of the list
                return(next(key_iter))
            except:
                return False
    return False

well, that is the basic idea, I think I am close, but this code gives a StopIteration error. Please help.

Thank you!

Answer 1

An iterator way...

def next_key(dict, key):
    keys = iter(dict)
    key in keys
    return next(keys, False)

Demo:

>>> next_key(dict, 'two')
'three'
>>> next_key(dict, 'three')
False
>>> next_key(dict, 'four')
False

Answer 2

Looping while k:= next(key_iter) doesn't stop correctly. Iterating manually with iter is done either by catching StopIteration :

iterator = iter(some_iterable)

while True:
    try:
        value = next(iterator)
    except StopIteration:
        # no more items

or by passing a default value to next and letting it catch StopIteration for you, then checking for that default value (but you need to pick a default value that won't appear in your iterable:):

iterator = iter(some_iterable)

while (value := next(iterator, None)) is not None:
    # …

# no more items

but iterators are, themselves, iterable, so you can skip all that and use a plain ol' for loop:

iterator = iter(some_iterable)

for value in iterator:
    # …

# no more items

which translates into your example as:

def next_key(d, key):
    key_iter = iter(d)

    for k in key_iter:
        if k == key:
            return next(key_iter, None)

    return None

Answer 3

You can get the keys of the dictionary as list and use index() to get the next key. You can also check for IndexError with try/except block:

my_dict = {'one':'value 1','two':'value 2','three':'value 3'}

def next_key(d, key):
  dict_keys = list(d.keys())
  try:
    return dict_keys[dict_keys.index(key) + 1]
  except IndexError:
    print('Item index does not exist')
    return -1

nk = next_key(my_dict, key="two")
print(nk)

And you better not use dict , list etc as variable names.

Answer 4

# Python3 code to demonstrate working of 
# Getting next key in dictionary Using list() + index()

# initializing dictionary 
test_dict = {'one':'value 1','two':'value 2','three':'value 3'}

def get_next_key(dic, current_key):
    """ get the next key of a dictionary.

    Parameters
    ----------
    dic: dict
    current_key: string

    Return
    ------
    next_key: string, represent the next key in dictionary.
    or
    False If the value passed in current_key can not be found in the dictionary keys,
    or it is last key in the dictionary
    """

    l=list(dic) # convert the dict keys to a list

    try:
        next_key=l[l.index(current_key) + 1] # using index method to get next key
    except (ValueError, IndexError):
        return False
    return next_key

get_next_key(test_dict, 'two')

'three'

get_next_key(test_dict, 'three')

False

get_next_key(test_dict, 'one')

'two'

get_next_key(test_dict, 'NOT EXISTS')

False