Return next key of a given dictionary key, 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!

Why not just use `iteritems()`? What do you mean by `next`? If you iterate over all keys, use built-in `keys()`, why isn't it enough? — Aaron_ab, Apr 06 '20 at 11:28
You *can* do this, but it’s really awkward (not a fast operation on dicts, dicts preserve order but aren’t ordered per se – less than `OrderedDict`s, anyway). Is there a problem one step removed that might have a better solution you could ask about? — Ry-, Apr 06 '20 at 11:29
Does this answer your question? [How to get the "next" item in an OrderedDict?](https://stackoverflow.com/questions/12328184/how-to-get-the-next-item-in-an-ordereddict) — Mayank Porwal, Apr 06 '20 at 11:30
@F.A If you have already found the answer, I would suggest to take this question down. No point in having multiple same questions. — Mayank Porwal, Apr 06 '20 at 11:37
With `next_key(dict, key='two')` your code returns `thee` for me on Python 3.8.1. Only change I did was replace current_key with key in the function call. Also you shouldn't use variable names that conflict with builtin functions (i.e. names such as `dict, list, . ..`). — DarrylG, Apr 06 '20 at 11:40
@Mayank Porwal after looking at the question you mentioned, it is similar than the one I posted but it refers to a OrderedDict. The one I posted is using a regular dictionary — F. A, Apr 06 '20 at 11:52
@F.A The first line of your question says `(Python 3.6+, ordered dictionary)`. I went with this. And if you are talking about regular dict, then there's no concept of particular next as regular dict is an unordered object. — Mayank Porwal, Apr 06 '20 at 11:57
@MayankPorwal When are people going to stop claiming dicts are still unordered? — Kelly Bundy, Apr 06 '20 at 12:59

score 4 · Answer 1 · answered Apr 06 '20 at 11:36

4

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

answered Apr 06 '20 at 11:36

Kelly Bundy

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1

Wow. That’s terrible, but also beautiful. – Ry- Apr 06 '20 at 11:38
@Ry- Do you mean terrible because of the complexity? (I doubt it can be improved if the dict is all we have.) – Kelly Bundy Apr 06 '20 at 11:41
No, the use of `in` for side effects :D – Ry- Apr 06 '20 at 11:41
1

@Ry- Ah. Heh :-). I like doing that. [Here's](https://stackoverflow.com/a/60458724/12671057) another recent one. – Kelly Bundy Apr 06 '20 at 11:44
@heapoverflow, Amazing very short solution! Although I do not understand how it works. What happens exactly in the line "key in keys"? – F. A Apr 06 '20 at 12:23
1

@F.A It's not explicitly covered [where it should be](https://docs.python.org/3/reference/expressions.html#membership-test-operations), but I believe it falls in the same category as *"user-defined classes which do not define `__contains__()` but do define `__iter__()`"*, so it consumes the iterator until it finds the value (or until the end, if it's not in there). – Kelly Bundy Apr 06 '20 at 12:51

Ry- · Accepted Answer · 2020-04-06T11:43:52.930

4

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

edited Apr 06 '20 at 11:43

answered Apr 06 '20 at 11:38

Ry-

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Thank you very mutch for the answer! I have lerned a lot from reading it. Would you say your proposed solution is better than the one proposed by @heapoverflow ? you mentioned something about side effects of using "in", I am not sure what you mean. I like his solution as it avoids the for loop so it is shorter – F. A Apr 06 '20 at 12:00
1

@F.A: I would say the best solution is probably one that never involves this function at all :] Like I commented, it’s a weird thing to do with a `dict`, so I think you should take a look at the problem you’re trying to solve by doing it (and feel free to ask about that problem here! See also: https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). As for `in`: most people will be surprised to see that, but it’s not too bad when its intent is clear the way it is in a small function. – Ry- Apr 06 '20 at 12:03

score 1 · Answer 3 · answered Apr 06 '20 at 11:35

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.

score 0 · Answer 4 · answered Apr 06 '20 at 12:46

# 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

Return next key of a given dictionary key, python 3.6+

4 Answers4