[基础] 在Python中获得字典列表中最大值与最小值

假设有字典列表:

dict_list = [{'price': 99, 'barcode': '2342355'}, {'price': 88, 'barcode': '2345566'}, {'price': 77, 'barcode': '2342377'}]

要求price的最大值与最小值

最原始的方法:

min_price, max_price = 100, 0
for item in dict_list:
    if item['price'] < min_price:
        min_price = item['price']
    if item['price'] > max_price:
        max_price = item['price']
print(max_price, min_price)

刚学Python时会这么写...

先使用列表推导式(list comprehension), 再使用内置函数求最大值最小值:

seq = [item['price'] for item in dict_list]
print(max(seq), min(seq))

这种方式要遍历列表多次.

使用生成器表达式(generator expression)

min_price, max_price = 100, 0
for x in (item['the_key'] for item in dict_list):
    min_price,max_price = min(x,min_price),max(x,max_price)
print(max_price, min_price)

更简单的写法:

max_price = max(item['price'] for item in dict_list)
min_price = min(item['price'] for item in dict_list)
print(max_price, min_price)

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!

返回整个dict, 不仅仅是price:

max_priced_item = max(dict_list, key=lambda x:x['price'])
min_priced_item = min(dict_list, key=lambda x:x['price'])
print(max_priced_item, min_priced_item)
# {'price': 99, 'barcode': '2342355'} {'price': 77, 'barcode': '2342377'}

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

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