如果你use nested list comprehension instead嵌套的for循环：

# import pandas as pd
# import ast ## I just needed this to parse the RESPONSE column from csv
# df = pd.read_csv('https://raw.githubusercontent.com/ajayvd/dataframe/main/data_sub.csv')
# df['RESPONSE'] = df['RESPONSE'].apply(ast.literal_eval) # maybe only after read_csv

k3List = ('provider', 'storeExternalId') 
get_e = lambda resp_v:resp_v['store-boundary-dsp']['estimates']
def get_separate_lists(data_frame,column='RESPONSE', k3List=k3List, get_l=get_e): 
    def get_k3(k3):
        return [e[k3] for resp in data_frame[column] for e in get_l(resp) if k3 in e]
      
    isList = isinstance(k3List, (list,tuple,set)) 
    lists = [get_k3(k) for k in (k3List if isList else [k3List])]
    return lists if isList else lists[0]

provider, store_id = get_separate_lists(df)
# provider = get_separate_lists(df, k3List='provider')
# store_id = get_separate_lists(df, 'RESPONSE', 'storeExternalId')

[k3List可以是单个键或键的列表（或元组或集合），get_l应该是一个函数。
如果你想要并行列表，你可以从一个元组列表开始，然后unpack和zip基本上将它们“解压缩”到单独的列表中：

# k3List, get_e = ... ## as before
def get_tuple_lists(data_frame,column='RESPONSE', k3List=k3List, get_l=get_e): 
    return [    tuple(e.get(k3) for k3 in k3List) 
                for resp in data_frame[column] for e in get_l(resp)    ]

provider_stores = get_tuple_lists(df)
provider, store_id = [list(t) for t in zip(*provider_stores)]
# provider, store_id = list(zip(*provider_stores)) ## 2 tuples instead of 2 lists

无论使用哪种函数，print(f'{store_id=}\n{provider=}')都应该打印

store_id=['1504', '1504', '9346', '9346', '1035', '4883', '3791', '5464', '5464', '3869', '3869', '7510', '6221', '5708', '5708', '3465']
provider=['Instacart', 'DoorDash', 'DoorDash', 'Uber', 'DoorDash', 'DoorDash', 'DoorDash', 'Postmates', 'DoorDash', 'Skipcart', 'DoorDash', 'DoorDash', 'DoorDash', 'Postmates', 'DoorDash', 'DoorDash']

但get_tuple_lists的直接输出如下所示：

provider_stores=[('Instacart', '1504'), ('DoorDash', '1504'), ('DoorDash', '9346'), ('Uber', '9346'), ('DoorDash', '1035'), ('DoorDash', '4883'), ('DoorDash', '3791'), ('Postmates', '5464'), ('DoorDash', '5464'), ('Skipcart', '3869'), ('DoorDash', '3869'), ('DoorDash', '7510'), ('DoorDash', '6221'), ('Postmates', '5708'), ('DoorDash', '5708'), ('DoorDash', '3465')]

如果你不确定所使用的外部键（如上面的 * store-boundary-dsp * 和 * estimates *）是否存在于每一行，你可以在get_l中使用try...except：

def get_b(resp_v):
    try: return resp_v['store-boundary-dsp']['boundaries']
    except: return [] 
boundary_names = set(get_separate_lists(df, k3List='name', get_l=get_b))
# --> # boundary_names={'9346 - Area - 1', '1504 - Primary', '1504-Primary'}

只是一些关于你问题中的片段的注解：

for index, row in data_frame.iterrows():
        for i in range(0,len(row[column])):
            for k,v in row[column]['store-boundary-dsp']['estimates'][i].items():
               # if k==....

• 这里实际上不需要使用.iterrows()、range或.items()-可以使用

for rc in data_frame[column]:
        for est in rc['store-boundary-dsp']['estimates']:
            if 'storeExternalId' in est: store_id.append(est['storeExternalId'])
            if 'provider' in est: provider.append(est['provider'])

• 即使在 * for k,v * 循环中，定义store_val或provider_val（作为v）也是多余的[除非您计划在其各自的if块之外使用它们，或者计划以某种方式修改v]，此时您只需**.append(v)**
• 您还可以添加/提取任意数量的列表[从store-boundary-dsp.estimates]，方法是将它们添加到下面的all_lists字典中[而不是在 * for est... * 中编写更多的if]

all_lists = {
        'storeExternalId': (store_id := []),
        'provider': (provider := []),
    }

    for rc in data_frame[column]:
          for est in rc['store-boundary-dsp']['estimates']:
              for k in all_lists:
                  if k in est: all_lists[k].append(est[k])

。

• 你也可以use nested list comprehension instead嵌套的for循环：

def get_k3(k3, k2='estimates', k1='store-boundary-dsp'):
        return [e[k3] for resp in data_frame[column] for e in resp[k1][k2] if k3 in e]

    k3List = ['provider', 'storeExternalId'] ## line them up EXACTLY
    provider, store_id = [get_k3(k) for k in k3List]

顺便说一句，你也可以使用.explode和json_normalize来完全扁平化DataFrame：

df1 = pd.concat([df[['CREATEDAT']], pd.json_normalize(df['RESPONSE'])], axis=1)
df1.columns = [c.split('.',1)[-1] for c in df1.columns]
lCols = ['value', 'errors', 'tags.1504']
dlCols = ['boundaries', 'distances', 'estimates', 'fulfillments']
for c in (lCols+dlCols): df1 = df1.explode(c)

df1 = pd.concat([df1.drop(dlCols, axis=1).reset_index(drop=True), *[
    pd.json_normalize(df1[c]).rename(columns=lambda cn: f'{c}.{cn}')
    for c in dlCols 
]],  axis=1)#.dropna(axis='columns', thresh=140) 
## dropna(axis='columns',thresh=N)--> only keep columns with < N empty cells

这将创建一个元组列表。每个元组将是store_id和provider对。apply函数用于迭代RESPONSE列，并将该单元格的值传递给extract_details函数，该函数处理主要的数据提取部分。

def extract_details(response):

    res = []

    # Check to ensure first two keys are in the response, otherwise quit early
    if not 'store-boundary-dsp' in response:
        return res

    if not 'estimates' in response['store-boundary-dsp']:
        return res

    # Iterate through each estimate
    for estimate in response['store-boundary-dsp']['estimates']:
        # Iterate through each key, value pair for the estimage
        store_id = ''
        providor = ''
        for k, v in estimate.items():
            
            if k == 'storeExternalId':
                store_id = v
            if k == 'provider':
                providor = v
        res.append((store_id, providor))
                
    return res

# For each cell in the 'RESPONSE' column, extract out the store_id, providor pair
response_values = data_frame['RESPONSE'].apply(extract_details).tolist()
response_values

pair_values = [val for sublist in response_values for val in sublist]
pair_values