How to find the min/max value of a common key in a list of dicts?
Question:
I have a list of dictionaries like so:
[{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]
I want to find the min()
and max()
prices. Now, I can sort this easily enough using a key with a lambda expression (as found in another Stack Overflow post), so if there is no other way I’m not stuck. However, from what I’ve seen there is almost always a direct way in Python, so this is an opportunity for me to learn a bit more.
Answers:
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)
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!
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 int
s):
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)
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.
can also use this:
from operator import itemgetter
lst = [{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]
max(map(itemgetter('price'), lst))
I have a list of dictionaries like so:
[{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]
I want to find the min()
and max()
prices. Now, I can sort this easily enough using a key with a lambda expression (as found in another Stack Overflow post), so if there is no other way I’m not stuck. However, from what I’ve seen there is almost always a direct way in Python, so this is an opportunity for me to learn a bit more.
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)
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!
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 int
s):
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)
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.
can also use this:
from operator import itemgetter
lst = [{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}]
max(map(itemgetter('price'), lst))