Ale*_*Sam 5 python json nested
我尝试了不同的 for 循环尝试迭代这个 JSON，但我不知道如何做到这一点。我有一个数字列表，想要将其与每个“数据”对象（例如，Aatrox、Ahri、Akali 等）下的“键”值进行比较，如果数字匹配，则将“名称”值存储在另一个清单。
示例：数字列表 = [266, 166, 123, 283]
266 和 166 将分别匹配 Aatrox 和 Akshan 对象中的“密钥”，因此我想提取该名称并将其存储在列表中。
我知道这个 JSON 主要是通过键值访问而不是索引，所以我不确定如何迭代 for 循环中的所有“数据”对象。
JSON 我引用：
{
"type": "champion",
"format": "standAloneComplex",
"version": "12.2.1",
"data": {
"Aatrox": {
"version": "12.2.1",
"id": "Aatrox",
"key": "266",
"name": "Aatrox",
"title": "the Darkin Blade",
"blurb": "Once honored defenders of Shurima against the Void, Aatrox and his brethren would eventually become an even greater threat to Runeterra, and were defeated only by cunning mortal sorcery. But after centuries of imprisonment, Aatrox was the first to find...",
"info": {
"attack": 8,
"defense": 4,
"magic": 3,
"difficulty": 4
},
"image": {
"full": "Aatrox.png",
"sprite": "champion0.png",
"group": "champion",
"x": 0,
"y": 0,
"w": 48,
"h": 48
},
"tags": [
"Fighter",
"Tank"
],
"partype": "Blood Well",
"stats": {
"hp": 580,
"hpperlevel": 90,
"mp": 0,
"mpperlevel": 0,
"movespeed": 345,
"armor": 38,
"armorperlevel": 3.25,
"spellblock": 32,
"spellblockperlevel": 1.25,
"attackrange": 175,
"hpregen": 3,
"hpregenperlevel": 1,
"mpregen": 0,
"mpregenperlevel": 0,
"crit": 0,
"critperlevel": 0,
"attackdamage": 60,
"attackdamageperlevel": 5,
"attackspeedperlevel": 2.5,
"attackspeed": 0.651
}
},
"Ahri": {
"version": "12.2.1",
"id": "Ahri",
"key": "103",
"name": "Ahri",
"title": "the Nine-Tailed Fox",
"blurb": "Innately connected to the latent power of Runeterra, Ahri is a vastaya who can reshape magic into orbs of raw energy. She revels in toying with her prey by manipulating their emotions before devouring their life essence. Despite her predatory nature...",
"info": {
"attack": 3,
"defense": 4,
"magic": 8,
"difficulty": 5
},
"image": {
"full": "Ahri.png",
"sprite": "champion0.png",
"group": "champion",
"x": 48,
"y": 0,
"w": 48,
"h": 48
},
"tags": [
"Mage",
"Assassin"
],
"partype": "Mana",
"stats": {
"hp": 526,
"hpperlevel": 92,
"mp": 418,
"mpperlevel": 25,
"movespeed": 330,
"armor": 21,
"armorperlevel": 3.5,
"spellblock": 30,
"spellblockperlevel": 0.5,
"attackrange": 550,
"hpregen": 5.5,
"hpregenperlevel": 0.6,
"mpregen": 8,
"mpregenperlevel": 0.8,
"crit": 0,
"critperlevel": 0,
"attackdamage": 53,
"attackdamageperlevel": 3,
"attackspeedperlevel": 2,
"attackspeed": 0.668
}
},
"Akali": {
"version": "12.2.1",
"id": "Akali",
"key": "84",
"name": "Akali",
"title": "the Rogue Assassin",
"blurb": "Abandoning the Kinkou Order and her title of the Fist of Shadow, Akali now strikes alone, ready to be the deadly weapon her people need. Though she holds onto all she learned from her master Shen, she has pledged to defend Ionia from its enemies, one...",
"info": {
"attack": 5,
"defense": 3,
"magic": 8,
"difficulty": 7
},
"image": {
"full": "Akali.png",
"sprite": "champion0.png",
"group": "champion",
"x": 96,
"y": 0,
"w": 48,
"h": 48
},
"tags": [
"Assassin"
],
"partype": "Energy",
"stats": {
"hp": 500,
"hpperlevel": 105,
"mp": 200,
"mpperlevel": 0,
"movespeed": 345,
"armor": 23,
"armorperlevel": 3.5,
"spellblock": 37,
"spellblockperlevel": 1.25,
"attackrange": 125,
"hpregen": 9,
"hpregenperlevel": 0.9,
"mpregen": 50,
"mpregenperlevel": 0,
"crit": 0,
"critperlevel": 0,
"attackdamage": 62,
"attackdamageperlevel": 3.3,
"attackspeedperlevel": 3.2,
"attackspeed": 0.625
}
},
"Akshan": {
"version": "12.2.1",
"id": "Akshan",
"key": "166",
"name": "Akshan",
"title": "the Rogue Sentinel",
"blurb": "Raising an eyebrow in the face of danger, Akshan fights evil with dashing charisma, righteous vengeance, and a conspicuous lack of shirts. He is highly skilled in the art of stealth combat, able to evade the eyes of his enemies and reappear when they...",
"info": {
"attack": 0,
"defense": 0,
"magic": 0,
"difficulty": 0
},
"image": {
"full": "Akshan.png",
"sprite": "champion0.png",
"group": "champion",
"x": 144,
"y": 0,
"w": 48,
"h": 48
},
"tags": [
"Marksman",
"Assassin"
],
"partype": "Mana",
"stats": {
"hp": 560,
"hpperlevel": 90,
"mp": 350,
"mpperlevel": 40,
"movespeed": 330,
"armor": 26,
"armorperlevel": 3,
"spellblock": 30,
"spellblockperlevel": 0.5,
"attackrange": 500,
"hpregen": 3.75,
"hpregenperlevel": 0.65,
"mpregen": 8.175,
"mpregenperlevel": 0.7,
"crit": 0,
"critperlevel": 0,
"attackdamage": 52,
"attackdamageperlevel": 3.5,
"attackspeedperlevel": 4,
"attackspeed": 0.638
}
}
}
}
小智 6
有很多解决方案，您也可以尝试这个，我认为它非常适合您的解决方案
import json
#Load your json or manually declare your json here
with open('yourfile.json') as f:
Data= json.load(f)
#This the main code for accesing the value and get the result
Temp_Data=Data['data']
result_list=[]
listOfNumbers =[266,166,123,283]
for data,data_info in Temp_Data.items():
key_value=int(Temp_Data[data]['key'])
if key_value in listOfNumbers:
result_list.append(data)
print(result_list)
您只需迭代字典的值，检查“key”项的值是否在您的列表中，如果是这样，则将“name”项的值附加到输出列表中。
让jsonObj是您的问题中呈现的 JSON 对象。那么这段代码应该可以工作：
listOfNumbers = [266, 166, 123, 283]
names = []
for value in jsonObj['data'].values():
if value['key'] in listOfNumbers:
names.append(value['name'])
Python 中的 JSON 对象只是字典。所以，你最好熟悉Python的dict.