如何从嵌套列表中获取一些值并将其添加到列表

Question

我有一个嵌套列表，可以在其中接收数据，但是该列表可以更改。 如果列表更改，如何获取一些值？

alltext=[{
    "name": "sample string 1",
    "relationships": [
        {
            "isBlocked": False,
            "orderIndex": 1,
            "relationship": {
                "debts": [
                    {
                        "currency": {
                            "id": "sample string2",
                            "isoCode": "string",
                            "isoNumber3": "string",
                            "name": "string"
                        },
                        #"sum"#: 0.0
                    }
                ],
                "deferment": 90,
                "id": "string",
                #"name"#: "string 1",
                "overdueDebts": [
                    {
                        "currency": {
                            "id": "string",
                            "isoCode": "string",
                            "isoNumber3": "string",
                            "name": "string"
                        },
                        #"sum"#: 1000.0
                    }
                ],
                "vendorId": "string"
            }
        },

        {
            "isBlocked": False,
            "orderIndex": 2,
            "relationship": {
                "debts": [
                    {
                        "currency": {
                            "id": "string",
                            "isoCode": "string",
                            "isoNumber3": "string",
                            "name": "string"
                        },
                        #"sum"#: 0.0
                    }
                ],
                "deferment": 90,
                "id": "string",
                "limit": 0.0,
                #"name"#: "string ",
                "overdueDebts": [
                    {
                        "currency": {
                            "id": "string",
                            "isoCode": "string",
                            "isoNumber3": "string",
                            "name": "string"
                        },
                        #"sum"#: 2000.0
                    }
                ],
                "vendorId": "string"
            }
        }
    ]
}
]

我正在尝试以它为例

for i in range(len(alltext)):
        ordersname.append(alltext[i]['relationships'][0]['relationship']['name'])

或喜欢它：

Debts.append([routeDict['sum'] for routeDict in alltext[i]['relationships'][0]['relationship']['debts']])

OverdueDebts.append([routeDict['sum'] for routeDict in alltext[i]['relationships'][0]['relationship']['overdueDebts']])

我了解到，我之所以这么认为，是因为[0]和我总是第一位的。 如果O delete [0]我收到错误TypeError: list indices must be integers or slices, not str

我想从所有关系中的债务和逾期债务和名称中获取总和（我标记了我想拥有的所有东西

Answer 1

我将此标记为地图减少问题。 Map-reduce是许多可用解决方案中的常见问题。 我鼓励您看一些例子。 将来，请发布您尝试过的内容。

这是映射减少字典的一种方法。 我实例化一个新的空数组，遍历每个关系，然后向该数组添加新的字典。

array_of_dicts = []

for relationship in main_dict['relationships']:
    r = relationship['relationship']
    debts = sum([debt['sum'] for debt in r['debts']])
    overdue = sum([debt['sum'] for debt in r['overdueDebts']])
    array_of_dicts.append({
        'name': r['name'],
        'debt': debts,
        'overdue': overdue
    })

Python为map-reduce提供了许多快捷方式，例如lambda函数和内置函数。 我正在使用一些列表理解，但也可以很容易地使用字典理解。

Answer 2

让我添加一些实用的方法（因为我们想学习多种策略来解决此问题）

import toolz.curried
from operator import itemgetter

get_name = toolz.curried.get_in(['relationship', 'name'])

get_debts = toolz.curried.get_in(['relationship', 'debts'])
get_overdue = toolz.curried.get_in(['relationship', 'overdueDebts'])

debt_sum = toolz.compose_left(get_debts, toolz.curried.map(itemgetter('sum')), sum)
overdue_sum = toolz.compose_left(get_overdue, toolz.curried.map(itemgetter('sum')), sum)

collect_info = toolz.juxt(get_name, debt_sum, overdue_sum)

list(map(collect_info, blubb['relationships']))

需要阅读代码的可怜的灵魂；-) ...因此，@ solbs的答案比这更具可读性。