Python: 返回嵌套字典的值

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英文:

Python: return nested dict of values

问题

def task_2(data_in):
    '''
        返回每个期间的文档数量(收入、支出、总计)。
        例如:{
            '2020-04': {
                'incomes': 2480,
                'expenses': 2695,
                'total': 5175
            },
            '2020-05': {
                'incomes': 2673,
                'expenses': 2280,
                'total': 4953
            }
        }
    '''
    period = []
    for term in data_in:
        for dict_val in term['summary']:
            if dict_val['period'] in period:
                continue
            else:
                period.append(dict_val['period'])

    incomes = []
    for term in data_in:
        for dict_val in term['summary']:
            incomes.append(dict_val['documents']['incomes'])

    expenses = []
    for term in data_in:
        for dict_val in term['summary']:
            expenses.append(dict_val['documents']['expenses'])

    return {per: {"incomes": inc, "expenses": exp, "total": inc + exp} for per, inc, exp in
            zip(period, incomes, expenses)}
英文:

I'm typing 3 list of loops, and I need return dict like in docstring. And it's wrong. How can I solve problem?

For each period you have to count incomes and expenses and total occurrence.

  1. Changes out of functions body are not allowed.
  2. Additional imports are not allowed.
  3. Data could have wholes in periods range ex. 2022-01,2022-03 (missing February), then we assume that item has 0 documents in period

Sample of dict looks like below:

   [
    {
        "package": "FLEXIBLE",
        "created": "2020-03-10T00:00:00",
        "summary": [
            {
                "period": "2019-12",
                "documents": {
                    "incomes": 63,
                    "expenses": 13
                }
            },
            {
                "period": "2020-02",
                "documents": {
                    "incomes": 45,
                    "expenses": 81
                }
            }
        ]
    },
    {
        "package": "ENTERPRISE",
        "created": "2020-03-19T00:00:00",
        "summary": [
            {
                "period": "2020-01",
                "documents": {
                    "incomes": 15,
                    "expenses": 52
                }
            },
            {
                "period": "2020-02",
                "documents": {
                    "incomes": 76,
                    "expenses": 47
                }
            }
        ]
    }
   ]

Code for the task is here:

def task_2(data_in):
    '''
        Return number of documents per period (incomes, expenses, total).
        ex. {
            '2020-04': {
                'incomes': 2480,
                'expenses': 2695,
                'total': 5175
            },
            '2020-05': {
                'incomes': 2673,
                'expenses': 2280,
                'total': 4953
            }
        }
    '''
    period = []
    for term in data_in:
        for dict_val in term['summary']:
            if dict_val['period'] in period:
                continue
            else:
                period.append(dict_val['period'])

    incomes = []
    for term in data_in:
        for dict_val in term['summary']:
            incomes.append(dict_val['documents']['incomes'])

    expenses = []
    for term in data_in:
        for dict_val in term['summary']:
            expenses.append(dict_val['documents']['expenses'])

    # dict_keys = ['incomes', 'expenses', 'total']
    # list1 = [{k: v for k, v in zip(dict_keys, range(len(incomes)))}]

    # return {k: v for k, v in zip(period, list1)}
    return {per: {"incomes": inc, "expenses": exp, "total": inc + exp} for per, inc, exp in
            zip(period, range(len(incomes)), range(len(expenses)))}

I'm typing loops in dict but returns:

{'2019-12': {'incomes': 1, 'expenses': 1, 'total': 2},
 '2020-02': {'incomes': 2, 'expenses': 2, 'total': 4}}

答案1

得分: 3

你需要追踪期间并保存收入、支出和总计。

data = [
    {
        "package": "FLEXIBLE",
        "created": "2020-03-10T00:00:00",
        "summary": [
            {
                "period": "2019-12",
                "documents": {
                    "incomes": 63,
                    "expenses": 13
                }
            },
            {
                "period": "2020-02",
                "documents": {
                    "incomes": 45,
                    "expenses": 81
                }
            }
        ]
    },
    {
        "package": "ENTERPRISE",
        "created": "2020-03-19T00:00:00",
        "summary": [
            {
                "period": "2020-01",
                "documents": {
                    "incomes": 15,
                    "expenses": 52
                }
            },
            {
                "period": "2020-02",
                "documents": {
                    "incomes": 76,
                    "expenses": 47
                }
            }
        ]
    }
]

records = [record for records in data for record in records['summary']]

result =  {}
template = {
    "incomes": 0,
    "expenses": 0,
    "total": 0
}
for record in records:
    period = record['period']
    if period not in result:
        result[period] = {"incomes": 0, "expenses": 0, "total": 0}
    income = record['documents']['incomes']
    expense = record['documents']['expenses']
    total = income + expense

    result[period]['incomes'] += income
    result[period]['expenses'] += expense
    result[period]['total'] += total

print(result)

# 输出 -> {'2019-12': {'incomes': 63, 'expenses': 13, 'total': 76}, '2020-02': {'incomes': 121, 'expenses': 128, 'total': 249}, '2020-01': {'incomes': 15, 'expenses': 52, 'total': 67}}
英文:

you need to track the period and save income, expenses, total

data = [
{
"package": "FLEXIBLE",
"created": "2020-03-10T00:00:00",
"summary": [
{
"period": "2019-12",
"documents": {
"incomes": 63,
"expenses": 13
}
},
{
"period": "2020-02",
"documents": {
"incomes": 45,
"expenses": 81
}
}
]
},
{
"package": "ENTERPRISE",
"created": "2020-03-19T00:00:00",
"summary": [
{
"period": "2020-01",
"documents": {
"incomes": 15,
"expenses": 52
}
},
{
"period": "2020-02",
"documents": {
"incomes": 76,
"expenses": 47
}
}
]}]
records = [record for records in data for record in records['summary']  ]
result =  {}
template = {
"incomes": 0,
"expenses": 0,
"total": 0
}
for record in records:
period = record['period']
if period not in result:
result[period] = {"incomes": 0,"expenses": 0,"total": 0}
income = record['documents']['incomes']
expense = record['documents']['expenses']
total = income + expense
result[period]['incomes'] += income
result[period]['expenses'] += expense
result[period]['total'] += total
print(result)
# output -> {'2019-12': {'incomes': 63, 'expenses': 13, 'total': 76}, '2020-02': {'incomes': 121, 'expenses': 128, 'total': 249}, '2020-01': {'incomes': 15, 'expenses': 52, 'total': 67}}

答案2

得分: 3

你可以通过创建一个字典,以期间作为键,并以'incomes'、'expenses'和'total'的和作为值来解决这个问题。结构将如下所示:

result = dict(
    '2019-12' = dict(
      'incomes' = 0,
      'expenses' = 0,
      'total' = 0,
    ),
    '2020-01' = dict(
      'incomes' = 0,
      'expenses' = 0,
      'total' = 0,
    )
)

但首先,你会将其实例化为空的dict,然后在迭代数据时填充它。

result = dict()
# 或者简单地:result = {}

你将遍历data_in中的每个term['summary']中的每个term,就像你已经在下面的代码中所做的那样:

for term in data_in:
    for dict_val in term['summary']:

一旦你有要处理的数据,你需要检查当前dict_val的期间是否已经存在于result字典中。如果不存在,则使用初始值创建它:

if dict_val['period'] not in result:
    result[dict_val['period']] = {'incomes': 0, 'expenses': 0, 'total': 0} 

然后,你可以继续对当前dict_val执行计算:

result[dict_val['period']]['incomes'] += dict_val['incomes']
result[dict_val['period']]['expenses'] += dict_val['expenses']
result[dict_val['period']]['total'] += dict_val['incomes'] + dict_val['expenses']

一旦迭代结束,你将得到你需要的数据。

英文:

You can solve that by creating a dict that has the period as key and a dict with the sums of incomes, expenses and total as value for every key. The structure will be this:

result = dict(
    '2019-12' = dict(
      'incomes' = 0,
      'expenses' = 0,
      'total' = 0,
    ),
    '2020-01' = dict(
      'incomes' = 0,
      'expenses' = 0,
      'total' = 0,
    )
)

But you will instantiate as an empty dict at first, and will populate it as you iterate over the data.

result = dict()
# or simply: result = {}

You will iterate over every term['summary'] in every term, in data_in, as you are already doing with

for term in data_in:
    for dict_val in term['summary']:

Once you have the data to be processed, you need to check if the period of the current dict_val already exists in the result dict. If not, create it with initial values:

if dict_val['period'] not in result:
    result[dict_val['period']] = {'incomes': 0, 'expenses': 0, 'total': 0} 

then, you proceed with doing the calculations for the current dict_val:

result[dict_val['period']]['incomes'] += dict_val['incomes']
result[dict_val['period']]['expenses'] += dict_val['expenses']
result[dict_val['period']]['total'] += dict_val['incomes'] + dict_val['expenses']

Once the iteration is over, you will have your result variable with the data you need

huangapple
  • 本文由 发表于 2023年3月8日 18:12:21
  • 转载请务必保留本文链接:https://go.coder-hub.com/75671716.html
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