Python数据框到按列分组的字典列表答案

【问题标题】：Python dataframe to list of dictionaries grouped by columnPython数据框到按列分组的字典列表
【发布时间】：2021-06-26 00:05:17
【问题描述】：

我有下表：

Pattern	tag	Responses
hi	greeting	Hey :-)
hey	greeting	Hello, thanks for visiting
how are you	greeting	Hi there, what can I do for you?
Is anyone there?	greeting	Hi there, how can I help?
Bye	goodbye	See you later, thanks for visiting
See you later	goodbye	Have a nice day
Goodbye	goodbye	Bye! Come back again soon

我想通过“标签”对其进行分组并获得以下json格式：

{
  "intents": [
    {
      "tag": "greeting",
      "patterns": [
        "Hi",
        "Hey",
        "How are you",
        "Is anyone there?"
      ],
      "responses": [
        "Hey :-)",
        "Hello, thanks for visiting",
        "Hi there, what can I do for you?",
        "Hi there, how can I help?"
      ]
    },
    {
      "tag": "goodbye",
      "patterns": ["Bye", "See you later", "Goodbye"],
      "responses": [
        "See you later, thanks for visiting",
        "Have a nice day",
        "Bye! Come back again soon"
      ]
    }
  ]
}

我尝试了很多东西，例如以下：

df.groupby('tag').apply(lambda x: list(x['Pattern']))

但这并不是我想要的……有人有什么建议吗？

【问题讨论】：

标签： python json dataframe dictionary

【解决方案1】：

您可以使用此示例如何将数据框格式化为您的 Json 格式：

import json

tmp = (
    df.groupby("tag")
    .apply(
        lambda x: dict(
            patterns=x["Pattern"].to_list(), responses=x["Responses"].to_list()
        )
    )
    .to_dict()
)

out = {"intents": []}
for k, v in tmp.items():
    out["intents"].append({})
    out["intents"][-1]["tag"] = k
    out["intents"][-1]["patterns"] = v["patterns"]
    out["intents"][-1]["responses"] = v["responses"]

print(json.dumps(out, indent=4))

打印：

{
    "intents": [
        {
            "tag": "goodbye",
            "patterns": [
                "Bye",
                "See you later",
                "Goodbye"
            ],
            "responses": [
                "See you later, thanks for visiting",
                "Have a nice day",
                "Bye! Come back again soon"
            ]
        },
        {
            "tag": "greeting",
            "patterns": [
                "hi",
                "hey",
                "how are you",
                "Is anyone there?"
            ],
            "responses": [
                "Hey :-)",
                "Hello, thanks for visiting",
                "Hi there, what can I do for you?",
                "Hi there, how can I help?"
            ]
        }
    ]
}

【讨论】：

【解决方案2】：

这里有一个解决方案：

def create_dic(x):
    
    d = {'tag': list(set(x['tag']))[0], 'patterns' : list(set(x['Pattern'])) , 'responses': list(set(x['Responses']))} 
    return d

result = {"intents"  : list( dict(df.groupby('tag').apply(lambda x: create_dic(x))).values()) }

输出：

print(result)
{'intents': [{'tag': 'goodbye',
   'patterns': ['Goodbye', 'Bye', 'See you later'],
   'responses': ['Bye! Come back again soon',
    'Have a nice day',
    'See you later, thanks for visiting']},
  {'tag': 'greeting',
   'patterns': ['Is anyone there?', 'how are you', 'hi', 'hey'],
   'responses': ['Hi there, how can I help?',
    'Hey :-)',
    'Hi there, what can I do for you?',
    'Hello, thanks for visiting']}]}

【讨论】：