Question

我有这样的input_data：

input_data = [
        {'vehicle': '001', 'store': 'foo1',  'qty': 100},
        {'vehicle': '001', 'store': 'foo1', 'qty': 200},
        {'vehicle': '001', 'store': 'baz1', 'qty': 300},
        {'vehicle': '001', 'store': 'baz1', 'qty': 400},
        {'vehicle': '002', 'store': 'foo2', 'qty': 500},
        {'vehicle': '002', 'store': 'baz2', 'qty': 600},
        {'vehicle': '002', 'store': 'baz2', 'qty': 700}]

在Python中获得out_put数据的最佳方法是如下所示：

output_data = [{'vehicle': '001',
            'store': [{'store': 'foo1', 'qty': [100, 200]},
                      {'store': 'baz1', 'qty': [300, 400]},
                      ]
            },
           {'vehicle': '002',
            'store': [{'store': 'foo2', 'qty': [500]},
                      {'store': 'baz2', 'qty': [600, 700]},
                      ]
            }]

Answer 1

您可以尝试一下，我敢肯定有一种优雅的方法，但目前还无法解决：

import pandas as pd

input_data = [
        {'vehicle': '001', 'store': 'foo1',  'qty': 100},
        {'vehicle': '001', 'store': 'foo1', 'qty': 200},
        {'vehicle': '001', 'store': 'baz1', 'qty': 300},
        {'vehicle': '001', 'store': 'baz1', 'qty': 400},
        {'vehicle': '002', 'store': 'foo2', 'qty': 500},
        {'vehicle': '002', 'store': 'baz2', 'qty': 600},
        {'vehicle': '002', 'store': 'baz2', 'qty': 700}]
df = pd.DataFrame(pd.DataFrame(input_data).groupby(['vehicle','store'])['qty'].apply(list))
dct = df.groupby(level=0).apply(lambda df: df.xs(df.name)['qty'].to_dict()).to_dict()
final_list = []
for k in dct:
    temp_dct = {}
    temp_dct['vehicle'] = k
    temp_l = [dict(zip(['store','qty'],[key,v])) for key,v in dct[k].items()]
    temp_dct['store'] = temp_l
    final_list.append(temp_dct)
print(final_list)

输出：

[{'vehicle': '001',
  'store': [{'qty': [300, 400], 'store': 'baz1'},
            {'qty': [100, 200], 'store': 'foo1'}]},
 {'vehicle': '002',
  'store': [{'qty': [600, 700], 'store': 'baz2'},
            {'qty': [500], 'store': 'foo2'}]}]

Answer 2

您可以使用itertools.groupby和operator.itemgetter

names = v, s, q = 'vehicle', 'store', 'qty'
v_key, s_key, q_key = map(itemgetter, names)
output_data = [{v: {s: [{s: sk, q: list(map(q_key, sv))}
                            for sk, sv in groupby(vv, s_key)]}}
                            for vk, vv in groupby(input_data, v_key)]

结果：

[{'vehicle': {'store': [{'qty': [100, 200], 'store': 'foo1'},
                        {'qty': [300, 400], 'store': 'baz1'}]}},
 {'vehicle': {'store': [{'qty': [500], 'store': 'foo2'},
                        {'qty': [600, 700], 'store': 'baz2'}]}}]

Answer 3

使用内置itertools - groupby method的一种方法：

# Input data
input_data = [
        {'vehicle': '001', 'store': 'foo1',  'qty': 100},
        {'vehicle': '001', 'store': 'foo1', 'qty': 200},
        {'vehicle': '001', 'store': 'baz1', 'qty': 300},
        {'vehicle': '001', 'store': 'baz1', 'qty': 400},
        {'vehicle': '002', 'store': 'foo2', 'qty': 500},
        {'vehicle': '002', 'store': 'baz2', 'qty': 600},
        {'vehicle': '002', 'store': 'baz2', 'qty': 700}]

# Main code
from itertools import groupby # For grouping a dataset

data = groupby(input_data, key = lambda x: x['vehicle']) # Grouping vehicles

final_data = []
for i in data:
    inner = {}
    outer = {}
    for j in list(i[1]):
        inner.setdefault(j['store'], []).append(j['qty'])
    outer["vehicle"] = i[0]
    outer["store"] = [{"store":i, "qty": j} for i, j in inner.items()]
    final_data.append(outer)

print (final_data)

# Output
# {'vehicle': '001', 'store': [{'store': 'foo1', 'qty': [100, 200]}, {'store': 'baz1', 'qty': [300, 400]}]}, {'vehicle': '002', 'store': [{'store': 'foo2', 'qty': [500]}, {'store': 'baz2', 'qty': [600, 700]}]}]

我希望这对您有所帮助！

Answer 4

下面的代码产生预期的输出：

result_list = [] def check_duplicate（车辆名称）：对于索引，d枚举（结果列表）：如果d.get（“ vehicle”）== vehicle_name：返回True，索引对于store_index，枚举（d.get（“ store”））中的store_value：如果store_value.get（“ store”）== vehicle_name：返回True，store_index

return False,None

对于input_data中的数据：

if isinstance(data,dict) and  data.has_key("vehicle"):
    #print data
    result= check_duplicate(data.get('vehicle'))
    if not result[0]:
        result_list.append({'vehicle':data.get('vehicle'),'store':[{'store':data.get('store') , 'qty': [data.get('qty')]}]})

    if result[0]:
        print "check if store is already present"
        store_check= check_duplicate(data.get('store'))

        if store_check[0]:
            result_list[result[1]].get("store")[store_check[1]].get("qty").append(data.get('qty'))
        if not store_check[0]:
            result_list[result[1]].get("store").append({'store':data.get('store') , 'qty': [data.get('qty')]})

打印result_list

字典列表到Python中的嵌套字典列表

4 个答案: