Question

从此表中，我尝试通过数据框中可用的最小/最大每周日期来插入缺失的日期。然后，计算每个类别的0销售发生。

df=pd.DataFrame({'category_id': ['aaa','aaa','aaa','aaa','bbb','bbb','bbb','ccc','ccc'],
                 'week': ['2015-01-05', '2015-01-12', '2015-01-19', '2015-01-26','2015-01-12', '2015-01-19', '2015-01-26','2015-01-05', '2015-01-12'],
                 'sales': [0,20,30,10,45,0,47,0,10]})

第一步：将缺失的每周日期添加到所有类别中，并在缺失的日期中填充0（第一季度：我不确定如何获得此df_add_missing_dates结果）

# expected dates interpolation output
df_add_missing_dates=pd.DataFrame({'category_id': ['aaa','aaa','aaa','aaa','bbb','bbb','bbb','bbb','ccc','ccc','ccc','ccc'],
                                   'week': ['2015-01-05', '2015-01-12', '2015-01-19', '2015-01-26',
                                            '2015-01-05', '2015-01-12', '2015-01-19', '2015-01-26',
                                            '2015-01-05', '2015-01-12', '2015-01-19', '2015-01-26'],
                                   'sales': [0,20,30,10,
                                             0,45,0,47,
                                             0,10,0,0]})

第二步：计算每周销售量为0（第二季度：如何汇总每个类别的销售量= 0？）

# expected final output
category_id | sales_0_count
aaa         | 1
bbb         | 2
ccc         | 3

当前代码和逻辑：

# convert string to datetime and set as index
df['week'] = pd.to_datetime(df['week'], format='%Y-%m-%d')
# find min/max weekly dates in the dataframe --> I couldn't add missing dates with 0 sales though
idx = pd.period_range(start=df.week.min(),end=df.week.max(),freq='W')
df = df.reindex(idx, fill_value=0).reset_index(drop=True)
df_add_missing_dates = df
# group by category to count how many times weekly sales is 0

Answer 1

IIUC，您可以将pd.MultiIndex.from_products与reindex和fill_value = 0一起使用，然后将布尔矩阵和groupby与sum一起使用：

idx = pd.MultiIndex.from_product([df['category_id'].unique(), 
                                  df['week'].unique()], 
                                 names=['category_id', 'week'])
df_missing = (df.set_index(['category_id', 'week'])
                .reindex(idx, fill_value=0)
                .reset_index())
df_missing

输出：

   category_id        week  sales
0          aaa  2015-01-05      0
1          aaa  2015-01-12     20
2          aaa  2015-01-19     30
3          aaa  2015-01-26     10
4          bbb  2015-01-05      0
5          bbb  2015-01-12     45
6          bbb  2015-01-19      0
7          bbb  2015-01-26     47
8          ccc  2015-01-05      0
9          ccc  2015-01-12     10
10         ccc  2015-01-19      0
11         ccc  2015-01-26      0

现在，分组并求和：

(df_missing == 0).groupby(df_missing['category_id'])['sales'].sum()

输出：

category_id
aaa    1.0
bbb    2.0
ccc    3.0
Name: sales, dtype: float64

Answer 2

不确定重新索引部分是什么用途，但是在

之后

df['week'] = pd.to_datetime(df['week'], format='%Y-%m-%d')

您可以这样做：

groupedDf = df.groupby(['category_id', pd.Grouper(key='week', freq='W-MON')])['sales'].sum().reset_index().sort_values('week')

zeroSalesWeek = groupedDf[groupedDf.sales == 0]

输出：

zeroSalesWeek

    category_id   week        sales
0   aaa           2015-01-05    0
4   bbb           2015-01-05    0
8   ccc           2015-01-05    0
6   bbb           2015-01-19    0
10  ccc           2015-01-19    0
11  ccc           2015-01-26    0

要选择特定的category_id，您可以尝试：

df[(df.sales == 0) & (df.category_id=='bbb')]

这会给你

    category_id   week        sales
4   bbb           2015-01-05    0
6   bbb           2015-01-19    0

此外，如果您认为这可能会花费一些时间，则可以始终创建一个快速函数来选择特定的category_id，例如：

def zeroGroupedDf(df, category_id):
    category_id = str(category_id)
    tempDf = df[(df.sales == 0) & (df.category_id==category_id)]
    return tempDf

并调用要创建新df的任何category_id，例如：

test = zeroGroupedDf(df, 'bbb')
test

    category_id   week        sales
4   bbb           2015-01-05    0
6   bbb           2015-01-19    0

Answer 3

这将以粗略的方式为您提供预期的输出：

df_add_missing_dates[df_add_missing_dates.sales.eq(0)].groupby('category_id')['sales'].count()

如果您希望获得期望的实际数据帧（尽管可以做得更好）：

expected_output = df_add_missing_dates[df_add_missing_dates.sales.eq(0)].\
    groupby('category_id',as_index=False)['sales'].count().\
    rename({'sales':'sales_0_count'},axis=1)

Answer 4

我是这样做的：

fitted

熊猫groupby然后计数发生0

4 个答案: