Question

我有一个列表：

[{"name":"john","age":40,"class":10},
 {"name":"john","age":45,"class":12},
 {"name":"sam","age":10,"class":5},
 {"name":"sam","age":11,"class":9},
 {"name":"mark","age":50,"class: 8"}]

我想获得唯一名字的最大年龄。我怎么能用python做到这一点？

我的预期答案：

[{"name":"john","age":45,"class":12},
{"name":"sam","age":11,"class":9},
{"name":"mark","age":50,"class: 8"}]

Answer 1

您可以使用itertools.groupby执行此操作：

from itertools import groupby

data = [{"name":"john","age":40,"class":10},
        {"name":"john","age":45,"class":12},
        {"name":"sam","age":10,"class":5},
        {"name":"sam","age":11,"class":9},
        {"name":"mark","age":50,"class": 8}]

# groups dictionary by "name"
# takes max dictionary of each group by "age"
result = [max(list(g), key=lambda x: x['age']) for _, g in groupby(sorted(data, key=lambda x: x['name']), key=lambda x: x['name'])]

print(result)

哪个输出：

[{'name': 'john', 'age': 45, 'class': 12}, 
 {'name': 'sam', 'age': 11, 'class': 9}, 
 {'name': 'mark', 'age': 50, 'class': 8}]

Answer 2

一个简单的for循环解决方案是

var _productServiceMock = new Mock< IProductService >();

结果：

A = [{"name":"john","age":40},
 {"name":"john","age":45},
 {"name":"sam","age":10},
 {"name":"sam","age":11},
 {"name":"mark","age":50}]


res = {}
for i in A:
    if i['name'] not in res:
        res[i['name']] = i['age']
    else:
        if res[i['name']] < i['age']:
            res[i['name']] = i['age']

print res

Answer 3

您能否进一步指明您的具体目标？你想保留这个结构吗？你想要字典中所有名字的列表吗？你可以写一个这样的函数：

def maxdictlist(name,dictlist):
    maxlst = []
    for x in dictlist:
        if x["name"] == name:
            maxlst.append(x["age"])
    return max(maxlst)

print(maxdictlist("john",A))

返回给定名称的最大值。如果您可以保存您感兴趣的名称列表并希望保留结构，那么您也可以使用最大功能键来执行此类操作：

def ke(v,name):
    if v["name"] == name: return v["age"]
    else: return 0 #0 as default value

lst = []
names = ["john","sam"]
res = [max(A,key= lambda v:ke(v,name))) for name in names]

print(lst)

我会建议不要使用第二种方法。

使用python从dict数组中查找最大键值对

3 个答案: