遍歷嵌套字典並在python中查找字典值中的關鍵字

Question

我有以下格式的數據。

data = {"policy": {"1": {"ID": "ML_0", "URL": "www.a.com", "Text": "my name is Martin and here is my code"} "2": {"ID": "ML_1", "URL": "www.b.com", "Plain_Text" my name is Mikal and here is my code"}}}


keywords = ['is', 'my']

以下是我想在 python 中處理我的數據的幾件事。

首先遍歷我的字典並在“1”和“2”中的“Text”值中查找並計算上述關鍵字，最后一件事是用關鍵字計數更新當前字典（在“1”和“2”如下所示。

{"policy": {"1": {"ID": "ML_0", "URL": "www.a.com", "Text": "my name is Martin and here is my code", "is": "2", "my": "2"} "2": {"ID": "ML_1", "URL": "www.b.com", "Plain_Text: "my name is Mikal and here is my code", "is": "2", "my": "2"}}}

如果有人可以幫助我，將不勝感激。

Answer 1

你可以使用collections. Counter collections. Counter ：

from collections import Counter
import json  # Only for pretty printing `data` dictionary.


def get_keyword_counts(text: str, keywords: list[str]) -> dict[str, int]:
    return {
        word: count for word, count in Counter(text.split()).items()
        if word in set(keywords)
    }


def main() -> None:
    data = {
        "policy": {
            "1": {
                "ID": "ML_0",
                "URL": "www.a.com",
                "Text": "my name is Martin and here is my code"
            },
            "2": {
                "ID": "ML_1",
                "URL": "www.b.com",
                "Text": "my name is Mikal and here is my code"
            }
        }
    }
    keywords = ['is', 'my']
    for policy in data['policy'].values():
        policy |= get_keyword_counts(policy['Text'], keywords)
    print(json.dumps(data, indent=4))


if __name__ == '__main__':
    main()

輸出：

{
    "policy": {
        "1": {
            "ID": "ML_0",
            "URL": "www.a.com",
            "Text": "my name is Martin and here is my code",
            "my": 2,
            "is": 2
        },
        "2": {
            "ID": "ML_1",
            "URL": "www.b.com",
            "Text": "my name is Mikal and here is my code",
            "my": 2,
            "is": 2
        }
    }
}

注意：使用|=來合並 dicts 是 Python 3.10 的一個特性。 如果您使用的是舊版本，應該不難谷歌如何做到這一點。

Answer 2

（我沒有足夠的聲譽發表評論，所以我將作為答案發布）首先，我認為您的 dict 結構一開始就不正確。 語法似乎不正確