python-从json对象中选择唯一键值 [英] python-select unique key values from json object

问题描述

我有一个 json 回复:

{
  "data": [
  {
     "id": "1",
     "name": "Tom",
     "age": "24",
  },
  {
     "id": "2",
     "name": "Nick",
     "age": "45",
 },
 {
     "id": "3",
     "name": "Harry",
     "age": "18",
 },
 {
     "id": "1",
     "name": "Tom",
     "age": "29",
 }
],
 "count": 4
}

我希望输出像这样:

output[
{
   "id": "1",
   "name": "Tom",
   "age": "24",
},
{
   "id": "2",
   "name": "Nick",
   "age": "45",
},
{
   "id": "3",
   "name": "Harry",
   "age": "18",
}
]

我想要的是获取所有具有唯一名称的字典对象. 我知道如何获取唯一名称，但是我也想获取id和age. 有两个对应于name Tom的字典对象.我想在输出中保留一个.

What I want is to fetch all dictionary objects having unique names. I know how to fetch unique names, but I want to fetch id and age too. There are two dictionary objects corresponding to name Tom. I want to keep one in my output.

这是用于获取唯一名称的工作代码:

size=len(data["data"])
uniqueNames = [];
for i in range(0,size,1):  
    if(data["data"][i]["name"] not in uniqueNames):
         uniqueNames.append(data["data"][i]["name"]); 
print uniqueNames

推荐答案

在这里，您应该修改代码:只需保留名称注册表，然后添加一个算法即可保留其余信息.对我来说，我创建了另一个数组来存储具有唯一名称的整个数据对象，这些名称称为returnValue.只要有一个唯一的名称，它就会将整个数据对象推送到returnValue上.然后，将其打印出来(如果将其转换为函数，则将其返回).

Here's how you should fix your code: Simply keep your registry of names and then add a algorithm for keeping the rest of the information. For me, I created another array for storing the whole data object that have unique names, called returnValue. And whenever there is a unique name, it pushes the entire data object onto returnValue. Then, it prints it out (or returns it, if you turn it into a function).

returnValue = []
size=len(data["data"])
uniqueNames = []
for i in range(0,size,1):  
    if(data["data"][i]["name"] not in uniqueNames):
         uniqueNames.append(data["data"][i]["name"]) 
         returnValue.append(data["data"][i])
print returnValue

由于作者的问题:

returnValue = []
badValues = []
size=len(data["data"])
uniqueNames = []
for i in range(0,size,1):  
    if(data["data"][i]["name"] not in uniqueNames):
         uniqueNames.append(data["data"][i]["name"]) 
         returnValue.append(data["data"][i])
    else:
         badValues.append(data["data"][i])
print "Good ones: " 
print returnValue
print "Bad ones: "
print badValues

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