迭代具有不等第二索引级别的多索引数据帧组

Question

我有一个 MultiIndex (Name, Date) DataFrame df，我需要通过 Date 对其进行迭代处理，以便分配一个基于当前和前一个日期组的值。

当我在不同日期有不同的Name 集时，我不知道该怎么做。我在下面的循环中用注释HELP!!! 标记了有问题的行：

    df = pd.DataFrame(data=[('A', '20210101', 5.0),  # 1 Jan
                            ('B', '20210101', 3.0),
                            ('C', '20210101', 2.0),
                            ('A', '20210102', 0.0),  # 2 Jan
                            ('C', '20210102', 0.0),
                            ('A', '20210103', 0.0),  # 3 Jan
                            ('C', '20210103', 0.0),
                            ('D', '20210103', 0.0)],
                      columns=('Name', 'Date', 'Dollars')).set_index(['Name', 'Date'])

    # Logic: Each day total wealth sums to $10.
    # Each day: Each person starts with what he had the previous day; 
    #           excess wealth gets allocated evenly to everyone
    
    dft = df.groupby(df.index.get_level_values('Date'))
    dates = list(dft.groups.keys())
    # Initialize first group:
    previous = dft.get_group(dates[0])

    # Loop over groups in order:
    for date in dates[1:]:
        current = dft.get_group(date)
        current.Dollars = previous.Dollars  # << HELP!!!
        excess = 10.0 - current.Dollars.sum()
        current.Dollars = current.Dollars + excess / current.Dollars.count()
        # Assign the calculated values back to the DataFrame:
        df.loc[current.index] = current
        # Prepare for next iteration:
        previous = current

在HELP!!! 点，我需要以某种方式完成以下操作：

1. 将current.Dollars 分配给previous.Dollars，忽略current 和previous 上的Date 索引。即使我执行.reset_index(level=1)，鉴于Name 索引每天都可能发生变化，我看不到如何完成此任务– 请注意，在示例df 中，第二天输了 em> 行名B，第三天有一个新行名D。

2. 在执行循环中的逻辑时忽略Date 级别后，我需要以某种方式恢复它以将结果分配回相应Date 上的主df。

---

在这个例子中，循环结束后 `df` 的值应该是：

               Dollars
Name Date             
A    20210101      5.0
B    20210101      3.0
C    20210101      2.0
A    20210102      6.5
C    20210102      3.5
A    20210103      6.5
C    20210103      3.5
D    20210103      0.0

Answer 1

如简化版here所示， 您必须操作索引才能进行分配，然后将值返回到原始 DataFrame：

# Initialize first group, but take Date out of its index:
previous = dft.get_group(dates[0]).reset_index(level=1)

for date in dates[1:]:
    # Take the Date out of the index:
    current = dft.get_group(date).reset_index(level=1)

    # Get the common keys
    commonKeys = current.index.intersection(previous.index)

    # Do the "left assignment"
    current.loc[commonKeys, 'Dollars'] = previous.loc[commonKeys].Dollars

    # Do your calculations:
    excess = 10.0 - current.Dollars.sum()
    current.Dollars = current.Dollars + excess / current.Dollars.count()

    # Restore the Date column to the index:
    current.set_index('Date', append=True, inplace=True)

    # Assign the calculated values back to the DataFrame:
    df.loc[current.index] = current

    # Prepare for next iteration by removing the Date index again:
    previous = current.reset_index(level=1)