我已经尝试了不同的循环，试图遍历这个JSON，但我不知道该怎么做。我有一个数字列表，并希望将其与每个"数据"对象(例如，Aatrox，Ahri，Akali等)下的"键"值进行比较，如果数字匹配，则存储"名称" 另一个列表中的值。
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。
我明白这个 JSON 主要是通过键值访问而不是索引的，所以我不确定我将如何遍历 for 循环中的所有"数据"对象。
JSON im 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
}
}
}
}
有很多解决方案 你也可以试试这个 我认为它将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)
您只需遍历字典的值，检查"键"项的值是否在您的列表中，如果是这种情况，则将"name"项的值附加到输出列表中。
让jsonObj成为您的问题中呈现的 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，你最好熟悉一下 Python 的dict.
假设您的完整 json 在 all_data 中，您将获得与键"data"对应的值为 all_data["data"]。这又是一个字典，您可以使用返回键，值对的项目对其进行迭代。
names = [value['name'] for name, value in all_data['data'].items() if value['key'] in listOfNumbers]
首先转换为将键映射到名称的字典，然后在其中搜索。
代码注释中的进一步解释：
import json
keys = [266, 166, 123, 283]
# First, we need to parse the JSON string into a Python dictionary
# Skip this if you already have a dictionary.
data = json.loads(raw_json)
# Then map keys to names
key_to_id = {int(obj["key"]): obj["id"] for obj in data["data"].values()}
# Lastly, extract the names we want
ids = [key_to_id[key] for key in keys]
另一种方法是将dict转换为json，转换为pandas数据帧并对其进行筛选。这是一个将扁平化任何嵌套json的函数：
from flatten_json import flatten
import pandas as pd
def flatten_nested_json_df(df):
df = df.reset_index()
s = (df.applymap(type) == list).all()
list_columns = s[s].index.tolist()
s = (df.applymap(type) == dict).all()
dict_columns = s[s].index.tolist()
while len(list_columns) > 0 or len(dict_columns) > 0:
new_columns = []
for col in dict_columns:
horiz_exploded = pd.json_normalize(df[col]).add_prefix(f'{col}.')
horiz_exploded.index = df.index
df = pd.concat([df, horiz_exploded], axis=1).drop(columns=[col])
new_columns.extend(horiz_exploded.columns) # inplace
for col in list_columns:
#print(f"exploding: {col}")
df = df.drop(columns=[col]).join(df[col].explode().to_frame())
new_columns.append(col)
s = (df[new_columns].applymap(type) == list).all()
list_columns = s[s].index.tolist()
s = (df[new_columns].applymap(type) == dict).all()
dict_columns = s[s].index.tolist()
return df
有了这个，你可以执行以下操作：
json = json.dumps(data)
df = pd.json_normalize(data)
flatten_nested_json_df(df)
返回数据帧：
index type format version data.Aatrox.version
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
.. ... ... ... ... ...
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
0 0 champion standAloneComplex 12.2.1 12.2.1
data.Aatrox.id data.Aatrox.key data.Aatrox.name data.Aatrox.title
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
.. ... ... ... ...
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
0 Aatrox 266 Aatrox the Darkin Blade
data.Aatrox.blurb ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
.. ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
0 Once honored defenders of Shurima against the ... ...
data.Akshan.stats.crit data.Akshan.stats.critperlevel
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
.. ... ...
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
data.Akshan.stats.attackdamage data.Akshan.stats.attackdamageperlevel
0 52 3.5
0 52 3.5
0 52 3.5
0 52 3.5
0 52 3.5
.. ... ...
0 52 3.5
0 52 3.5
0 52 3.5
0 52 3.5
0 52 3.5
data.Akshan.stats.attackspeedperlevel data.Akshan.stats.attackspeed
0 4 0.638
0 4 0.638
0 4 0.638
0 4 0.638
0 4 0.638
.. ... ...
0 4 0.638
0 4 0.638
0 4 0.638
0 4 0.638
0 4 0.638
data.Aatrox.tags data.Ahri.tags data.Akali.tags data.Akshan.tags
0 Fighter Mage Assassin Marksman
0 Fighter Mage Assassin Assassin
0 Fighter Mage Assassin Marksman
0 Fighter Mage Assassin Assassin
0 Fighter Mage Assassin Marksman
.. ... ... ... ...
0 Tank Assassin Assassin Assassin
0 Tank Assassin Assassin Marksman
0 Tank Assassin Assassin Assassin
0 Tank Assassin Assassin Marksman
0 Tank Assassin Assassin Assassin
[8192 rows x 160 columns]
然后，您可以根据需要对其进行过滤：
listOfNumbers = ['266', '166', '123', '283']
df_sub = df[df['data.Aatrox.key'].isin(listOfNumbers)]
这是一个适用于您的 json 的 json，我更新了它以使其有效。
d = json.loads(j)
nums = [266, 166, 123, 283]
names = [v["name"] for v in d["data"].values() if int(v["key"]) in nums]
如果要匹配数字，还必须int值。
输出
['Aatrox', 'Akshan']