获取 Python 中 JSON 个键的平均总和

Question

我对 Python 和 JSON 很陌生。我正在构建一个工具，可以让我快速分析玩家列表。

我正在尝试解析以下 JSON 并从所有条目中获取“百分位数”键的平均总和。

[   
    {
    "encounterID": 663,
    "encounterName": "Lucifron",
    "class": "Priest",
    "spec": "Healer",
    "rank": 1471,
    "outOf": 4463,
    "duration": 48171,
    "startTime": 1597001801681,
    "reportID": "74LJdPNCZahwmxyW",
    "fightID": 9,
    "difficulty": 3,
    "characterID": 42962953,
    "characterName": "Insanepriest",
    "server": "Lakeshire",
    "percentile": 67.03123111833183,
    "ilvlKeyOrPatch": 70,
    "azeritePowers": [],
    "total": 225.509,
    "estimated": true
  },
  {
    "encounterID": 664,
    "encounterName": "Magmadar",
    "class": "Priest",
    "spec": "Healer",
    "rank": 3205,
    "outOf": 5055,
    "duration": 67599,
    "startTime": 1597001988440,
    "reportID": "74LJdPNCZahwmxyW",
    "fightID": 11,
    "difficulty": 3,
    "characterID": 42962953,
    "characterName": "Insanepriest",
    "server": "Lakeshire",
    "percentile": 36.597235868212486,
    "ilvlKeyOrPatch": 70,
    "azeritePowers": [],
    "total": 130.623,
    "estimated": true
  },
  {
    "encounterID": 667,
    "encounterName": "Shazzrah",
    "class": "Priest",
    "spec": "Healer",
    "rank": 2655,
    "outOf": 12663,
    "duration": 27960,
    "startTime": 1597003660411,
    "reportID": "74LJdPNCZahwmxyW",
    "fightID": 32,
    "difficulty": 3,
    "characterID": 42962953,
    "characterName": "Insanepriest",
    "server": "Lakeshire",
    "percentile": 79.02778951070532,
    "ilvlKeyOrPatch": 70,
    "azeritePowers": [],
    "total": 441.345,
    "estimated": true
  },
]

这是我正在使用的 python 代码，我只设法对所有项目求和。

response = requests.get('https://classic.warcraftlogs.com/v1/parses/character/Insanepriest/Lakeshire/EU?metric=dps&zone=1000&partition=2&api_key=6436362d91e87781841b4b031f7c0a6c')
    response.raise_for_status()
    # access JSOn content
    jsonResponse = response.json()
    parse = round(jsonResponse[0]["percentile"], 1)
    sum = 0
    for parse in jsonResponse:
        if parse and "percentile" in parse.keys():
            sum += parse["percentile"]
        print(sum)

如有任何帮助，我们将不胜感激。很抱歉，如果我在这里问了一个愚蠢的问题，但我在网上找不到任何东西。

此致

Answer 1

percentiles = [round(i['percentile'], 1) for i in jsonResponse]
percentilesSum = sum(percentiles)
#I dont know what you mean by average sum, I count it as average
percentilesAvg = percentilesSum / len(percentiles)

获取 Python 中 JSON 个键的平均总和

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