CODEWITHSUNDEEP
pythondictionary
donbe
donbe

Reputation: 33

How to get the maximum value of a key in dict?

I would like to get the URL of a video with maximum resolution.

If I had the following dictionary, would it be easiest to get the URL of the video with the maximum size?

Is it best to split the size by partition, format the number into an int type, and then get the value?

videos = {
        'size_476_306': 'https://www.....',
        'size_560_360': 'https://www.....',
        'size_644_414': 'https://www.....',
        'size_720_480': 'https://www.....',
}

Solved

I couldn't figure out lambda, so I implemented it in a different way.

for size, v_url in videos.items():

    max_size_info = size.split('_')
    split_size = int(max_size_info[1])
    size_array.append(split_size)
    video_array.append(v_url)

max_size = size_array.index(max(size_array))
if str(size_array[max_size]) in v_url:
    print(size,v_url)

Upvotes: 1

Views: 73

Answers (1)

Samwise
Samwise

Reputation: 71562

Is it best to split the size by partition, format the number into an int type, and then get the value?

Yes, that's the way I'd do it:

>>> max(videos.items(), key=lambda i: int.__mul__(*map(int, i[0].split("_")[1:])))
('size_720_480', 'https://www.....')

Here's a slightly more verbose version with a named function:

>>> def get_resolution(key: str) -> int:
...     """
...     Gets the resolution (as pixel area) from 'size_width_height'.
...     e.g. get_resolution("size_720_480") -> 345600
...     """
...     _, width, height = key.split("_")
...     return int(width) * int(height)
...
>>> max(videos, key=get_resolution)
'size_720_480'

Given that expression that gives us the largest key, we can easily get the corresponding value:

>>> videos[max(videos, key=get_resolution)]
'https://www.....'

or we could get the key, value pair by taking the max of items(), here using a much simpler lambda that just translates key, value into get_resolution(key):

>>> max(videos.items(), key=lambda i: get_resolution(i[0]))
('size_720_480', 'https://www.....')

Upvotes: 1

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