Iterating over dictionary keys in order?

Question

I am trying to iterate over a dictionary (inside another dictionary specifically, but I don't think that part matters), and I'm unable to get the for loop to iterate over the values in the order that they were placed in. I would like to be able to take the first value of each dictionary. I thought that after python 3.6, dictionaries kept their order (here), but these wont stay in order.

This is what my dictionary looks like:

count_dict = {'dessert': Counter({'cake': 1}), 'vegetable': Counter({'carrots': 1, 'beet': 1}), 'grain': Counter({'wheat': 3, 'rice': 2, 'zantham': 1}), 'meat': Counter({'chicken': 2, 'pork': 1})}

This is my code that tried:

for it in count_dict:
        for food in count_dict[it]:
            out_dict[it] = food
return(out_dict)

I get:

{'dessert': 'cake', 'vegetable': 'beet', 'grain': 'rice', 'meat': 'chicken'}

but need to get the first value from each dictionary:

{'dessert': 'cake', 'vegetable': 'carrots', 'grain': 'wheat', 'meat': 'chicken'}

I looked for ever for an answer on Stack but couldn't find an answer. (sorry if the answers obvious, I'm kinda new to Python)

Notice that your actual output and expected output have an identical ordering by keys. That is what is meant by the ordering being retained: the keys keep the same order. Clearly, in this case, the dictionaries that counter uses do not keep their order. — Kraigolas, May 26 '21 at 04:27
`beet` is not a first value, but you said you want first value from. — nobleknight, May 26 '21 at 04:48

score 1 · Accepted Answer · edited May 26 '21 at 04:51

Dictionaries (after python 3.6) maintain their insertion order (order in which keys and values are inserted), not the sorted order.

I did get an answer.

from collections import Counter

count_dict = {'dessert': Counter({'cake': 1}), 'vegetable': Counter({'carrots': 1, 'beet': 1}), 'grain': Counter({'wheat': 3, 'rice': 2, 'zantham': 1}), 'meat': Counter({'chicken': 2, 'pork': 1})}

new_dict = {}

for a, b in count_dict.items():
    new_dict[a] = sorted(b, key=lambda x: (-b[x], x))[0]

print(new_dict)

This will sort inner dictionary first by the values then alphabetically by keys.

sorted(b, key=lambda x: (-b[x], x))

Sample Run 1 (sorting alphabetically, because values are same.)

b = {'carrots': 1, 'beet': 1}
sorted(b, key=lambda x: (-b[x], x))

Output

['beet', 'carrots']

Sample Run 2 (sorting by value an the alphabetically.)

b = {'wheat': 3, 'rice': 2, 'zantham': 1}
sorted(b, key=lambda x: (-b[x], x))

Output

['wheat', 'rice', 'zantham']

thanks, I had this in the back of my mind, but was not sure how to do it. Sorry to ask for more, but how does the (-b[x]), x) section work in the lambda function? — magic.mrr, May 26 '21 at 05:06
`b[x]` gives value for key `x`. `sorted` will sort it by ascending order of values. When we change it to `-b[x]`, maximum value will be minimum. So we will get it reverse order by values. You can say that why don't we used `reverse` parameter, but we also want it to sorted by alphabetically by keys. If values for two keys are same, then `sorted` will check key `x` and sort it according to key `x`. — nobleknight, May 26 '21 at 05:15

score 0 · Answer 2 · answered May 26 '21 at 06:52

If you strictly want to return the first key value only, regardless of alphabetical key order or numerical value order, you could simply do the following:

from collections import Counter

count_dict = {'dessert': Counter({'cake': 1}), 'vegetable': Counter({'carrots': 1, 'beet': 1}), 'grain': Counter({'wheat': 3, 'rice': 2, 'zantham': 1}), 'meat': Counter({'chicken': 2, 'pork': 1})}
out_dict = {}

for it in count_dict:
        for food in count_dict[it]:
            out_dict[it] = food
            break  # Break out the loop so that you only iterate over the first key in dictionary
print(out_dict)

Iterating over dictionary keys in order?

2 Answers2