Python函数返回嵌套字典错误：如何按周期统计收支及总文档数？

需要实现task_2函数，返回按周期统计的收支及总文档数的嵌套字典，要求：

• 按周期汇总incomes、expenses及总文档数（total = incomes + expenses）
• 禁止修改函数体外代码、禁止额外导入模块
• 若周期范围存在缺失月份，该周期的文档数视为0

示例输入数据：

[
    {
        "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
                }
            }
        ]
    }
]

当前实现代码：

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'])

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

错误输出：

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

1. 未按周期累加数据：当前代码仅收集了零散的周期和收支值，未对同一周期的收支进行汇总，返回时错误使用range(len(incomes))取索引值，导致输出数字而非实际收支金额。
2. 未处理缺失周期：没有生成完整的连续周期序列，缺失的月份未被纳入统计范围。

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
            }
        }
    '''
    # 1. 按周期累加收支数据
    period_data = {}
    for term in data_in:
        for summary_item in term['summary']:
            period = summary_item['period']
            docs = summary_item['documents']
            if period not in period_data:
                period_data[period] = {'incomes': 0, 'expenses': 0}
            period_data[period]['incomes'] += docs['incomes']
            period_data[period]['expenses'] += docs['expenses']
    
    if not period_data:
        return {}
    
    # 2. 生成完整的连续周期序列
    # 将周期转换为(year, month)元组用于排序和生成连续月份
    sorted_periods = sorted([(int(p.split('-')[0]), int(p.split('-')[1])) for p in period_data.keys()])
    start_year, start_month = sorted_periods[0]
    end_year, end_month = sorted_periods[-1]
    
    full_periods = []
    current_year, current_month = start_year, start_month
    while (current_year, current_month) <= (end_year, end_month):
        full_periods.append(f"{current_year}-{current_month:02d}")
        # 月份进位处理
        if current_month == 12:
            current_year += 1
            current_month = 1
        else:
            current_month += 1
    
    # 3. 构建最终结果，缺失周期填0
    result = {}
    for period in full_periods:
        incomes = period_data.get(period, {}).get('incomes', 0)
        expenses = period_data.get(period, {}).get('expenses', 0)
        result[period] = {
            'incomes': incomes,
            'expenses': expenses,
            'total': incomes + expenses
        }
    
    return result

1. 数据累加：用字典period_data按周期存储收支总和，遍历输入数据时自动累加同一周期的收支数值。
2. 生成完整周期：将所有周期转换为可排序的(year, month)元组，找到首尾周期后，生成中间所有连续月份的字符串格式。
3. 结果构建：遍历完整周期列表，从period_data中提取对应数据，缺失周期用0填充，同时计算total值。

内容的提问来源于stack exchange，提问作者Marek Grzesiak