将Python字典中的JSON值累积为数组

obadul024 • 4 年前 • 772 次点击

{ 
"links": [
{"source":"0","target":"1","weight":1,"color":"white"},
{"source":"0","target":"2","weight":1,"color":"yellow"},
{"source":"0","target":"3","weight":1,"color":"white"},
]
}

我想收集所有 target 为一个人 source 这样地:

{"source": 0, "neighbors": ["1","2","3"]} 哪里 neighbors 都收齐了吗

这是我的密码

import json

with open("linksGr.json") as file:
    data = json.load(file)

collectDict = {}
for obj in data["links"]:
    if (collectDict["source"] == obj["source"]):
        collectDict["neighbour"] = obj["target"]

我只需要一种方法来累积每个源的所有目标,而不是像我在这里所做的那样有多个源

collectDict["source"] = obj["source"]
collectDict["neighbour"] = obj["target"]

任何帮助都将不胜感激。我确信这里缺少一些基本的概念和简单的方法。谢谢你的帮助。

Python社区是高质量的Python/Django开发社区
本文地址：http://www.python88.com/topic/55681

772 次点击

文章 [ 3 ] | 最新文章 4 年前

• 1 楼

Riccardo Bucco 4 年前

import pandas as pd

result = [{"source": source, "neighbors": df["target"].tolist()}
          for source, df in pd.DataFrame(data["links"]).groupby("source")]

• 2 楼

Michael H. 4 年前

data = { 
    "links": [
        {"source":"0","target":"1","weight":1,"color":"white"},
        {"source":"0","target":"2","weight":1,"color":"yellow"},
        {"source":"0","target":"3","weight":1,"color":"white"},
        {"source":"5","target":"7","weight":1,"color":"white"},
        {"source":"5","target":"8","weight":1,"color":"yellow"},
        {"source":"6","target":"9","weight":1,"color":"white"},
    ]
}

collected = []
for obj in data["links"]:
    source_matches = [item for item in collected if item["source"] == obj["source"]]
    if len(source_matches) == 0:
        source_match = {"source": obj["source"], "neighbour": [obj["target"]]}
        collected.append(source_match)
    elif len(source_matches) == 1:
        source_matches[0]["neighbour"].append(obj["target"])
    else:
        raise BaseException()

print(collected)  # [{'source': '0', 'neighbour': ['1', '2', '3']}, {'source': '5', 'neighbour': ['7', '8']}, {'source': '6', 'neighbour': ['9']}]

不是很优雅,但能胜任工作。

如果你真的不需要格式 {"source": 0, "neighbors": ["1","2","3"]} 我建议使用上面的解决方案 defaultdict . 如果需要这种格式,还可以从 默认听写 解决方案。

• 3 楼

Adam.Er8 4 年前

如果我理解正确,你可以使用 collections.defaultdict ,从源映射到像这样的目标:

(我添加了一些数据以拥有多个源)

from collections import defaultdict

data = { 
"links": [
{"source":"0","target":"1","weight":1,"color":"white"},
{"source":"0","target":"2","weight":1,"color":"yellow"},
{"source":"0","target":"3","weight":1,"color":"white"},
{"source":"5","target":"7","weight":1,"color":"white"},
{"source":"5","target":"8","weight":1,"color":"yellow"},
{"source":"6","target":"9","weight":1,"color":"white"},
]
}

collectDict = defaultdict(list)
for obj in data["links"]:
    collectDict[obj["source"]].append(obj["target"])

print(dict(collectDict))

输出:

{'0': ['1', '2', '3'], '5': ['7', '8'], '6': ['9']}

这里有另一种方法 itertools.groupby ,

from itertools import groupby

collectDict = {k: [t["target"] for t in g] for k,g in groupby(data["links"], lambda obj: obj["source"])}

print(collectDict)

登录后回复