# Get max value index for a list of dicts

Tell max() how to calculate the maximum for a sequence of indices:

max(range(len(ld)), key=lambda index: ld[index]['size'])


This'll return the index for which the size key is the highest:

>>> ld = [{'prop': 'foo', 'size': 100}, {'prop': 'boo', 'size': 200}]
>>> max(range(len(ld)), key=lambda index: ld[index]['size'])
1
>>> ld[1]
{'size': 200, 'prop': 'boo'}


If you wanted that dictionary all along, then you could just use:

max(ld, key=lambda d: d['size'])


and to get both the index and the dictionary, you could use enumerate() here:

max(enumerate(ld), key=lambda item: item[1]['size'])


Some more demoing:

>>> max(ld, key=lambda d: d['size'])
{'size': 200, 'prop': 'boo'}
>>> max(enumerate(ld), key=lambda item: item[1]['size'])
(1, {'size': 200, 'prop': 'boo'})


The key function is passed each element in the input sequence in turn, and max() will pick the element where the return value of that key function is highest.

Using a separate list to extract all the size values then mapping that back to your original list is not very efficient (you now need to iterate over the list twice). list.index() cannot work as it has to match the whole dictionary, not just one value in it.

You can pass the enumerate(ld) to max function with a proper key :

>>> max(enumerate(ld),key=lambda arg:arg[1]['size'])[0]
1


If you just want the dictionary with max size value, as a Pythonic approach you could use operator.itemgetter function as the key:

In [10]: from operator import itemgetter

In [11]: ld = [{'prop': 'foo', 'size': 100}, {'prop': 'boo', 'size': 200}]

In [12]: fn = itemgetter('size')

In [13]: max(ld, key=fn)
Out[13]: {'prop': 'boo', 'size': 200}