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yixiugegegege
V2EX  ›  Python

迫于逻辑实在理不清了, Python 求助

  •  
  •   yixiugegegege · 2021-01-16 23:11:19 +08:00 · 2658 次点击
    这是一个创建于 1167 天前的主题,其中的信息可能已经有所发展或是发生改变。
    {
    "child": [{
    "f_brandname": "奔驰",
    "f_pyfirstletter": "B",
    "seriesname": "载货车"
    }, {
    "f_brandname": "北奔重卡",
    "f_pyfirstletter": "B",
    "seriesname": "载货车"
    }, {
    "f_brandname": "福田欧曼",
    "f_pyfirstletter": "F",
    "seriesname": "载货车"
    }, {
    "f_brandname": "江铃汽车",
    "f_pyfirstletter": "J",
    "seriesname": "载货车"
    }]
    }

    变成

    {
    "child": {
    "B": [{
    "f_brandname": "奔驰",
    "f_pyfirstletter": "B",
    "seriesname": "载货车"
    }, {
    "f_brandname": "北奔重卡",
    "f_pyfirstletter": "B",
    "seriesname": "载货车"
    }],
    "F": [{
    "f_brandname": "福田欧曼",
    "f_pyfirstletter": "F",
    "seriesname": "载货车"
    }],
    "J": [{
    "f_brandname": "江铃汽车",
    "f_pyfirstletter": "J",
    "seriesname": "载货车"
    }]
    }
    }


    就是把每个小字典给归类了,还有好多个品牌,这个是少数
    语言是 python,求 demo,
    14 条回复    2021-01-17 00:05:15 +08:00
    taogen
        1
    taogen  
       2021-01-16 23:21:31 +08:00 via Android
    没写过 Python 。不过把对象数组变成哈希数组,不会很难吧
    yixiugegegege
        2
    yixiugegegege  
    OP
       2021-01-16 23:27:52 +08:00
    @taogen 我百度一手,谢谢
    nvkou
        3
    nvkou  
       2021-01-16 23:28:36 +08:00 via Android
    Java 的话第一反应是 stream. map. groupingby
    Python 的话我估计也会这么做
    大不了先拿 firstletter 的集合,再用这个集合过滤到不同组。也就是先建立键,再通过键过滤值
    eggshell
        4
    eggshell  
       2021-01-16 23:34:55 +08:00
    用 itertools.groupby
    xiaoming1992
        5
    xiaoming1992  
       2021-01-16 23:40:46 +08:00 via Android
    就用 for in 也行吧
    wzwwzw
        6
    wzwwzw  
       2021-01-16 23:43:55 +08:00
    from itertools import groupby

    before_data = {
    "child": [
    {"f_brandname": "奔驰", "f_pyfirstletter": "B", "seriesname": "载货车"},
    {"f_brandname": "北奔重卡", "f_pyfirstletter": "B", "seriesname": "载货车"},
    {"f_brandname": "福田欧曼", "f_pyfirstletter": "F", "seriesname": "载货车"},
    {"f_brandname": "江铃汽车", "f_pyfirstletter": "J", "seriesname": "载货车"},
    ]
    }

    after_data = {"child": {}}
    for j, i in groupby(before_data["child"], key=lambda x: x["f_pyfirstletter"]):
    if j not in after_data["child"]:
    after_data_example["child"][j] = []
    for k in i:
    after_data_example["child"][j].append(k)

    print(after_data_example)
    zyx199199
        7
    zyx199199  
       2021-01-16 23:46:16 +08:00   ❤️ 2
    ···
    from itertools import groupby

    {k: list(v) for k,v in groupby(data['child'], key=lambda x: x["f_pyfirstletter"])}
    ```
    yixiugegegege
        8
    yixiugegegege  
    OP
       2021-01-16 23:46:28 +08:00
    @eggshell 卧槽,解决了,老哥,谢谢!!!
    my8100
        9
    my8100  
       2021-01-16 23:46:43 +08:00 via iPhone
    from collections import defaultdict

    child_dict = defaultdict(list)
    for d in data["child"]:
    child_dict[d["f_pyfirstletter"]].append(d)

    assert {"child": child_dict} == target_data
    yixiugegegege
        10
    yixiugegegege  
    OP
       2021-01-16 23:46:55 +08:00
    @wzwwzw 谢谢老哥的代码,终于解决了哭😭
    yixiugegegege
        11
    yixiugegegege  
    OP
       2021-01-16 23:48:18 +08:00
    @zyx199199 谢谢老哥解决了,喜极而泣,祝福本月无 bug,无迫于
    yixiugegegege
        12
    yixiugegegege  
    OP
       2021-01-16 23:49:05 +08:00
    @my8100 谢谢老哥,❤️你,爱你😭!
    imn1
        13
    imn1  
       2021-01-16 23:58:51 +08:00
    new = {'child': {x["f_pyfirstletter"]: [] for x in d["child"]}}
    [new['child'][x["f_pyfirstletter"]].append(x) for x in d["child"]]

    下划线为空格
    d 为源字典,new 为结果
    imn1
        14
    imn1  
       2021-01-17 00:05:15 +08:00
    # 13 “下划线为空格”不需要
    原来是用 for 的,有缩进,后来改成列表表达式,不需要缩进了
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