Get first value in dict from a list of keys

Question

Let's suppose that I have the following set:

labels = set(["foo", "bar"])

And I have a dict with theses values

d = { "foo": "some value", "asdf": "another value", }

How can I get the first value of the dictionary based on any value of the set labels?

In other words, how can I get the value "some value" from the values of the set?

What is the expected output? A dict is an unordered data structure so I'm not sure why you'd want to do that — turtle
– turtle, Commented Jan 14, 2021 at 17:21
What do you mean by first value? dict and set are not ordered, even though the dict in the latest versions of python3 is ordered by default. — Abdul Aziz Barkat
– Abdul Aziz Barkat, Commented Jan 14, 2021 at 17:22
The expected output would be "some value", because of the "foo" on the set and on the dict — Rodrigo
– Rodrigo, Commented Jan 14, 2021 at 17:22
Not just dict, set are also unordered. So you can never be sure what the first value will be. — Moinuddin Quadri
– Moinuddin Quadri, Commented Jan 14, 2021 at 17:30

Olvin Roght · Accepted Answer · 2021-01-14 17:37:53Z

7

You can apply next() with generator expression:

result = next(d[k] for k in labels if k in d)

Upd. Code above works but it doesn't fit 100%, because it iterates over labels and retrieve value of first key, which not always first occurrence.

To get value of first occurrence of any key from labels use next code:

result = next(v for k, v in d.items() if k in labels)

edited Jan 14, 2021 at 17:37

answered Jan 14, 2021 at 17:27

Olvin Roght

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3 Comments

flakes Over a year ago

maybe add None as the second argument to next to avoid a StopIteration error.

Olvin Roght Over a year ago

@flakes, I've thought about that, but decided to leave it like it is, because I don't know exact values and it could cause logical mistakes if actual value will be None.

flakes Over a year ago

Just for the sake of evil: next(filter(None, map(d.get, labels))) :^)

LightVillet · Accepted Answer · 2021-01-14 17:23:11Z

1

for key in d: if key in labels: print(d[key]) break

answered Jan 14, 2021 at 17:23

LightVillet

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Comments

Timur Shtatland · Accepted Answer · 2021-01-14 17:25:17Z

Use list comprehension and select the element with 0th index. You may want to wrap this in try...except to catch the case where no element is found in the dictionary.

labels = set(["foo", "bar"]) d = { "foo": "some value", "asdf": "another value", } print([d[k] for k in labels if k in d][0]) # some value

Would highly suggest using next here with a generator expression rather than creating a new list. That would avoid the possible index-out-of-bounds for no matches.

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