In List of Dicts, find min() value of a common Dict field

Question

I have a list of dictionaries like so:

[{\'price\': 99, \'barcode\': \'2342355\'}, {\'price\': 88, \'barcode\': \'2345566\'}]

I want to fin

Answer 1

One answer would be mapping your dicts to the value of interest inside a generator expression, and then applying the built-ins min and max.

myMax = max(d['price'] for d in myList)
myMin = min(d['price'] for d in myList)

Answer 2

There are several options. Here is a straight-forward one:

seq = [x['the_key'] for x in dict_list]
min(seq)
max(seq)

[Edit]

If you only wanted to iterate through the list once, you could try this (assuming the values could be represented as ints):

import sys

lo,hi = sys.maxint,-sys.maxint-1
for x in (item['the_key'] for item in dict_list):
    lo,hi = min(x,lo),max(x,hi)

Answer 3

lst = [{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]

maxPricedItem = max(lst, key=lambda x:x['price'])
minPricedItem = min(lst, key=lambda x:x['price'])

This tells you not just what the max price is but also which item is most expensive.

Answer 4

can also use this:

from operator import itemgetter

lst = [{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]  
max(map(itemgetter('price'), lst))

Answer 5

I think the most direct (and most Pythonic) expression would be something like:

min_price = min(item['price'] for item in items)

This avoids the overhead of sorting the list -- and, by using a generator expression, instead of a list comprehension -- actually avoids creating any lists, as well. Efficient, direct, readable... Pythonic!