英文:
how to get the smallest value of a key in multiple dictionaries nested in a list?
问题
我有类似这样的东西
data = [{'name': 'abc', 'id': 'abc123','price': '$10'},
{'name': 'def', 'id': 'efg123','price': '$15'},
{'name': 'ghi', 'id': 'ghi123','price': '$5'}, ]
我想要做的是首先从所有字典的'price'键的值中删除'$'符号，然后比较所有字典中的值并打印其中最小的值。
我用以下方式去除符号
for i in data:
data_price = i['price']
new_price = data_price.strip('$')
print(new_price)
但我不知道如何比较所有的值，然后获取最小的值，我还需要知道最小值属于哪个字典，因为我还需要它的'id'。
我该如何做呢？
英文:
I have something like this
data = [{'name': 'abc', 'id': 'abc123','price': '$10'},
{'name': 'def', 'id': 'efg123','price': '$15'},
{'name': 'ghi', 'id': 'ghi123','price': '$5'}, ]
what I want to do is first strip the '$' sign from the value of 'price' key from all the dictionaries and then compare the values in all the dictionaries and print the smallest one in them.
I did this for removing the symbol
for i in data:
data_price = i['price']
new_price = data_price.strip('$')
print(new_price)
But I don't know how I compare all the values and then get the smallest one, and I also then need to know in which dictionary that smallest value belongs to because I also need its id.
how do I do that?
答案1
得分: 2
你可以按照以下方式获取它
```python
min_data = min(data, key=lambda dic: float(dic['price'].lstrip('$')))
使用for循环采用相同的方法，以简单的方式实现
def get_min(arr):
store_min = {}
for record in arr:
val = float(record['price'].lstrip('$'))
if val not in store_min:
store_min[val] = []
store_min[val].append(record)
minimum_prices = store_min.keys()
min_price = min(minimum_prices)
return store_min[min_price][0]
print(get_min(data))
# 输出 -> {'name': 'ghi', 'id': 'ghi123', 'price': '$5'}
<details>
<summary>英文:</summary>
you can get it as this
min_data = min(data, key=lambda dic: float(dic['price'].lstrip('$')))
same approach with `for` loop in simple way
>>> def get_min(arr):
store_min = {}
for record in arr:
val = float(record['price'].lstrip('$'))
if val not in store_min:
store_min[val] = []
store_min[val].append(record)
minimum_prices = store_min.keys()
min_price = min(minimum_prices)
return store_minmin_price
print(get_min(data))
output -> {'name': 'ghi', 'id': 'ghi123', 'price': '$5'}
</details>
# 答案2
**得分**: 1
你可以使用[`min`][0]函数与一个关键函数：
```python
data = [{'name': 'abc', 'id': 'abc123', 'price': '$10'},
{'name': 'def', 'id': 'efg123', 'price': '$15'},
{'name': 'ghi', 'id': 'ghi123', 'price': '$5'}, ]
m = min(data, key=lambda d: int(d["price"][1:]))
# {'name': 'ghi', 'id': 'ghi123', 'price': '$5'}
请注意，你必须将price转换为数值类型，以便不会按字典顺序比较（例如，"10" < "5"）。
英文:
You can use min with a key function:
data = [{'name': 'abc', 'id': 'abc123','price': '$10'},
{'name': 'def', 'id': 'efg123','price': '$15'},
{'name': 'ghi', 'id': 'ghi123','price': '$5'}, ]
m = min(data, key=lambda d: int(d["price"][1:]))
# {'name': 'ghi', 'id': 'ghi123', 'price': '$5'}
Note that you have to cast price
to a numerical type, so that you won't compare lexicographically (where e.g. "10" < "5")
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