如何更改字典列表中的键名

Question

我有一个如下所示的df：

route_id client_lat client_lng
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0068   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0073   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066
 BR0074   -23.7424   -46.7066

如你所见，每个route_id都有许多纬度和经度。我想做的是生成一个字典列表。每个字典都会根据每个client_lat和client_lng存储route_id。因此，在我的最终列表中，我应该有n个字典，其中n是唯一route_id的数量。从上面的df中，我应该在我的列表中有3本字典，其中的条目应该如下所示：

{'start1': 'geo!-23.7424,-46.7066',
'start2': 'geo!-23.7424,-46.7066',
'start3': 'geo!-23.7424,-46.7066',
'start4': 'geo!-23.7424,-46.7066',
...}

每个字典的条目数量将根据每个client_lat和client_lng的数量而有所不同。

要生成具有上述结构的条目的字典列表，我将使用以下代码：

lista = []

for i in data.route_id.unique():
    start_points = {f"start{k + 1}": "geo!" + data.loc[k, 'client_lat'] +  "," 
                    + data.loc[k, 'client_lng'] for k in range(0, len(data)) if data.loc[k, "route_id"] == i} 
    
    lista.append(start_points)

我的清单是这样的：

lista = [{'start1': 'geo!-23.7424,-46.7066',
  'start2': 'geo!-23.7424,-46.7066',
  'start3': 'geo!-23.7424,-46.7066',
  'start4': 'geo!-23.7424,-46.7066',
  'start5': 'geo!-23.7424,-46.7066',
  'start6': 'geo!-23.7424,-46.7066',
  'start7': 'geo!-23.7424,-46.7066',
  'start8': 'geo!-23.7424,-46.7066',
  'start9': 'geo!-23.7424,-46.7066',
  'start10': 'geo!-23.7424,-46.7066'},
 {'start11': 'geo!-23.7424,-46.7066',
  'start12': 'geo!-23.7424,-46.7066',
  'start13': 'geo!-23.7424,-46.7066',
  'start14': 'geo!-23.7424,-46.7066',
  'start15': 'geo!-23.7424,-46.7066',
  'start16': 'geo!-23.7424,-46.7066',
  'start17': 'geo!-23.7424,-46.7066',
  'start18': 'geo!-23.7424,-46.7066',
  'start19': 'geo!-23.7424,-46.7066',
  'start20': 'geo!-23.7424,-46.7066'},
 {'start21': 'geo!-23.7424,-46.7066',
  'start22': 'geo!-23.7424,-46.7066',
  'start23': 'geo!-23.7424,-46.7066',
  'start24': 'geo!-23.7424,-46.7066',
  'start25': 'geo!-23.7424,-46.7066',
  'start26': 'geo!-23.7424,-46.7066',
  'start27': 'geo!-23.7424,-46.7066',
  'start28': 'geo!-23.7424,-46.7066',
  'start29': 'geo!-23.7424,-46.7066'}]

它将每个字典存储在我的列表中，但是每个字典中的键应该从1到m，其中m是每个route_id的lat和lng的数量。因此，我想要的输出应该如下所示：

[{'start1': 'geo!-23.7424,-46.7066',
  'start2': 'geo!-23.7424,-46.7066',
  'start3': 'geo!-23.7424,-46.7066',
  'start4': 'geo!-23.7424,-46.7066',
  'start5': 'geo!-23.7424,-46.7066',
  'start6': 'geo!-23.7424,-46.7066',
  'start7': 'geo!-23.7424,-46.7066',
  'start8': 'geo!-23.7424,-46.7066',
  'start9': 'geo!-23.7424,-46.7066',
  'start10': 'geo!-23.7424,-46.7066'},
 {'start1': 'geo!-23.7424,-46.7066',
  'start2': 'geo!-23.7424,-46.7066',
  'start3': 'geo!-23.7424,-46.7066',
  'start4': 'geo!-23.7424,-46.7066',
  'start5': 'geo!-23.7424,-46.7066',
  'start6': 'geo!-23.7424,-46.7066',
  'start7': 'geo!-23.7424,-46.7066',
  'start8': 'geo!-23.7424,-46.7066',
  'start9': 'geo!-23.7424,-46.7066',
  'start10': 'geo!-23.7424,-46.7066'},
 {'start1': 'geo!-23.7424,-46.7066',
  'start2': 'geo!-23.7424,-46.7066',
  'start3': 'geo!-23.7424,-46.7066',
  'start4': 'geo!-23.7424,-46.7066',
  'start5': 'geo!-23.7424,-46.7066',
  'start6': 'geo!-23.7424,-46.7066',
  'start7': 'geo!-23.7424,-46.7066',
  'start8': 'geo!-23.7424,-46.7066',
  'start9': 'geo!-23.7424,-46.7066'}]

如果有人知道怎么做的话，我非常感谢你的帮助。

Answer 1

使用enumerate总是很好地创建索引。您可以更改i开始使用参数start=1的位置。

这样做的工作是：

data = []
for route in df['route_id'].unique():
    data.append({f'start_{i}': f'geo!{x["client_lat"]},{x["client_lng"]}' for i, x in enumerate(df[df['route_id'] == route].to_dict('records'), start=1)})

打印data

[{'start_1': 'geo!-23.7424,-46.7066',
  'start_2': 'geo!-23.7424,-46.7066',
  'start_3': 'geo!-23.7424,-46.7066',
  'start_4': 'geo!-23.7424,-46.7066',
  'start_5': 'geo!-23.7424,-46.7066',
  'start_6': 'geo!-23.7424,-46.7066',
  'start_7': 'geo!-23.7424,-46.7066',
  'start_8': 'geo!-23.7424,-46.7066',
  'start_9': 'geo!-23.7424,-46.7066',
  'start_10': 'geo!-23.7424,-46.7066'},
 {'start_1': 'geo!-23.7424,-46.7066',
  'start_2': 'geo!-23.7424,-46.7066',
  'start_3': 'geo!-23.7424,-46.7066',
  'start_4': 'geo!-23.7424,-46.7066',
  'start_5': 'geo!-23.7424,-46.7066',
  'start_6': 'geo!-23.7424,-46.7066',
  'start_7': 'geo!-23.7424,-46.7066',
  'start_8': 'geo!-23.7424,-46.7066',
  'start_9': 'geo!-23.7424,-46.7066',
  'start_10': 'geo!-23.7424,-46.7066'},
 {'start_1': 'geo!-23.7424,-46.7066',
  'start_2': 'geo!-23.7424,-46.7066',
  'start_3': 'geo!-23.7424,-46.7066',
  'start_4': 'geo!-23.7424,-46.7066',
  'start_5': 'geo!-23.7424,-46.7066',
  'start_6': 'geo!-23.7424,-46.7066',
  'start_7': 'geo!-23.7424,-46.7066',
  'start_8': 'geo!-23.7424,-46.7066',
  'start_9': 'geo!-23.7424,-46.7066'}]