的Python：扁平化和解析JSON

的某些部分我有一個輸入JSON看起來像這樣：的Python：扁平化和解析JSON

> {"payment": {"payment_id": "AA340", "payment_amt": "20", "chk_nr": "321749", "clm_list": {"dtl": [{"clm_id": "1A2345", "name": "John", adj:{"adj_id":"W123","adj_cd":"45"}}, {"clm_id": "9999", "name": "Dilton", adj:{"adj_id":"X123","adj_cd":"5"}}]}}}

我需要的輸出看起來像這樣：

{"clm_id": "1A2345",adj:{"adj_id":"W123"},"payment_amt": "20", "chk_nr": "321749"} 
{"clm_id": "9999"adj:{"adj_id":"X123"},"payment_amt": "20", "chk_nr": "321749"}

因此，代碼發生在一個JSON文檔解析聲明數組部分，並通過向每個部分添加付款信息來對其進行標準化。甚至嵌套的JSON也被解析。

我能夠解析數據，但不確定如何正常化只有某些部分的數據。

下面的代碼將解析的數據，但不規範化

keep = ["payment","payment_id","payment_amt", "clm_list", "dtl", "clm_id","adj","adj_id"] 

old_dict={"payment": {"payment_id": "AA340", "payment_amt": "20", "chk_nr": "321749", "clm_list": {"dtl": [{"clm_id": "1A2345", "name": "John", "adj": {"adj_id": "W123", "adj_cd": "45"}}, {"clm_id": "9999", "name": "Dilton", "adj": {"adj_id": "X123", "adj_cd": "5"}}]}}} 

def recursively_prune_dict_keys(obj, keep): 
     if isinstance(obj, dict): 
       return dict([(k, recursively_prune_dict_keys(v, keep)) for k, v in obj.items() if k in keep]) 
     elif isinstance(obj, list): 
       return [recursively_prune_dict_keys(item, keep) for item in obj] 
     else: 
       return obj 

new_dict = recursively_prune_dict_keys(old_dict, keep) 
conv_json=new_dict["payment"] 
print json.dumps(conv_json)

來源

2016-10-11 user3646519

如果您需要幫助，您需要發佈[最小，完整和可驗證示例]（http://stackoverflow.com/help/mcve）你試過的代碼 – CAB

更新了代碼 – user3646519

它可能是整潔的方式是簡單地通過數據接，等;

new_dict = recursively_prune_dict_keys(old_dict, keep) 
payment = old_dict['payment'] 
claims = payment['clm_list']['dtl'] 
for claim in claims: 
    claim['payment_amt'] = payment['payment_amt'] 
    claim['chk_nr'] = payment['chk_nr'] 
print(json.dumps(claims))

這會產生;

[{"chk_nr": "321749", "clm_id": "1A2345", "payment_amt": "20", "adj": {"adj_id": "W123"}}, {"chk_nr": "321749", "clm_id": "9999", "payment_amt": "20", "adj": {"adj_id": "X123"}}]

這包含您要求的輸出，但不完全如您希望看到的那樣。

首先，您所需的輸出是不正確的JSON沒有方括號[]，這將使它成爲一個列表。但是，我們可以通過逐個傾銷每項索賠來擺脫這種情況;

new_dict = recursively_prune_dict_keys(old_dict, keep) 
payment = old_dict['payment'] 
claims = payment['clm_list']['dtl'] 
for claim in claims: 
    claim['payment_amt'] = payment['payment_amt'] 
    claim['chk_nr'] = payment['chk_nr'] 
    print(json.dumps(claim))

這給出了;

{"name": "John", "clm_id": "1A2345", "payment_amt": "20", "adj": {"adj_cd": "45", "adj_id": "W123"}, "chk_nr": "321749"} 
{"name": "Dilton", "clm_id": "9999", "payment_amt": "20", "adj": {"adj_cd": "5", "adj_id": "X123"}, "chk_nr": "321749"}

這是接近你想要的輸出，除了可能的排序。 Python字典不是固有的有序。不過，您可以對它們進行排序。因此，如果排序很重要，您需要閱讀How to Sort Python Dictionaries by Key or Value

來源

2016-10-12 02:37:53 CAB

的Python：扁平化和解析JSON

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