CODEWITHSUNDEEP
pythonlistdictionarymin
helloworld
helloworld

Reputation: 181

getting minimum (key, value) in a list that holds a dictionary

I have a list that holds a dictionary row like this: 

queue = [{1: 0.39085439023582913, 2: 0.7138416909634645, 3: 0.9871959077954673}]

I'm tryin to get it to return the smallest value along with its key, so in this case it would return 

1,0.39085439023582913

I've tried using 

min(queue, key=lambda x:x[1])

but that just returns the whole row like this: any suggestions? thank you!

{1: 0.39085439023582913, 2: 0.7138416909634645, 3: 0.9871959077954673}

Upvotes: 3

Views: 4055

Answers (3)

J. Doe
J. Doe

Reputation: 3634

you can do it through 

mykey = min(queue[0], key=queue[0].get)

then just use this key and get your dictionary value

mykey, queue[0][mykey]

Upvotes: 0

pault
pault

Reputation: 43494

If you want the min for each dict in the list, you can use the following list comprehension:

[min(d.items(), key=lambda x: x[1]) for d in queue]

Which for your example returns: 

[(1, 0.39085439023582913)]

d.items() returns a list of tuples in the form (key, value) for the dictionary d. We then sort these tuples using the value (x[1] in this case).

If you always have your data in the form of a list with one dictionary, you could also call .items() on the first element of queue and find the min:

print(min(queue[0].items(), key=lambda x:x[1]))
#(1, 0.39085439023582913)

Upvotes: 3

Sagar B Hathwar
Sagar B Hathwar

Reputation: 536

Yes, min(queue) will find minimum in the list and not in the enclosed dictionary Do this:

key_of_min_val = min(queue[0], key=queue[0].get)
print((key_of_min_val , queue[0][key_of_min_val]))

Upvotes: 0

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