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Stack Overflow用户

提问于 2022-03-02 11:49:51

回答 1查看 42关注 0票数 0

我的dataset/datafame如下所示：

我想增加六列-一周，一周-2，周-3，销售-1，销售-2，销售-3在数据。第一周是本周减去2，第二周是前一周，第三周是下周。同样，销售-1是前一周的销售，2是前一周的销售，3是下一周的销售。就202201财政周而言，前一周和前一周缺勤，因此202201之后的3周变成1周、2周和3周。前一周的202202周缺席，因此202201,202203和202204成为第一周，第二周和第三周。在202205财政周，下周缺席，因此202202,202203,202204是三位候选人。同样，选择了sales-1、sales-2和sales-3的值。简单地说，选择4行，包括当前行，其中前2行是当前行的两行，最后一行是当前行之后的行。如果当前行的位置不存在任何位置行，则该窗口向下/向上移动，以得到一个包含当前窗口的4行窗口。

最后的dataframe应该是这样的：

目前，我正在使用2 for循环实现它：

df_final = pd.DataFrame()
for item in df['ItemNo'].unique() :
    df_temp = df[df.ItemNo==item].sort_values(by=['Fiscal_Week'])    
    sale_wk_minus2 = []
    sale_wk_minus1 = []
    sale_wk_plus1 = []
    wk_minus2 = []
    wk_minus1 = []
    wk_plus1 = []
    
    for i in range(len(df_temp)):
        if(i==0):
            wk_m2 = df_temp. Fiscal_Week.iloc[i+1]
            wk_m1 = df_temp. Fiscal_Week.iloc[i+2]
            wk_p1 = df_temp. Fiscal_Week.iloc[i+3]
            sale_m2 = df_temp.Sales.iloc[i+1]
            sale_m1 = df_temp.Sales.iloc[i+2]
            sale_plus2 = df_temp.Sales.iloc[i+3]

        if(i==1):
            wk_m2 = df_temp. Fiscal_Week.iloc[i-1]
            wk_m1 = df_temp. Fiscal_Week.iloc[i+1]
            wk_p1 = df_temp. Fiscal_Week.iloc[i+2]
            sale_m2 = df_temp.Sales.iloc[i-1]
            sale_m1 = df_temp.Sales.iloc[i+1]
            sale_plus2 = df_temp.Sales.iloc[i+2]
            
        if(i==( len(df_temp)-1)):
            wk_m2 = df_temp. Fiscal_Week.iloc[i-3]
            wk_m1 = df_temp. Fiscal_Week.iloc[i-2]
            wk_p1 = df_temp. Fiscal_Week.iloc[i-1]
            sale_m2 = df_temp.Sales.iloc[i-3]
            sale_m1 = df_temp.Sales.iloc[i-2]
            sale_plus2 = df_temp.Sales.iloc[i-1]
            
        if((i>1) and (i<len(df_temp)-1)):
            wk_m2 = df_temp. Fiscal_Week.iloc[i-2]
            wk_m1 = df_temp. Fiscal_Week.iloc[i-1]
            wk_p1 = df_temp. Fiscal_Week.iloc[i+1]
            sale_m2 = df_temp.Sales.iloc[i-2]
            sale_m1 = df_temp.Sales.iloc[i-1]
            sale_plus2 = df_temp.Sales.iloc[i+1]
            
    df_temp['week_minus_2'] = wk_minus2
    df_temp['week_minus_1'] = wk_minus1
    df_temp['week_plus_1'] = wk_plus1
    
    df_temp['sales_minus_2'] = sale_wk_minus2
    df_temp['sales_minus_1'] = sale_wk_minus1
    df_temp['sales_plus_1'] = sale_wk_plus1

    df_final = pd.concat( [df_final, df_temp])

我们能更有效地实施吗？

python

pandas

回答 1

Stack Overflow用户

发布于 2022-03-02 14:47:49

这大概是我的方法。您可以通过重新索引DataFrame来添加缺失的销售数据，这样就可以直接使用week和ItemNo的组合来查找销售情况。

另外，我还没有处理好最后/最高的一周。

import pandas as pd

df = pd.DataFrame({
    "ItemNo": [21, 21, 21, 21, 21, 29, 29, 29, 29, 29],
    "Fiscal_Week": ["202201", "202202", "202203", "202204", "202205"] * 2,
    "Sales": [45, 55, 65, 75, 85, 22, 32, 42, 52, 62],
})

df["Week"] = pd.to_datetime(df["Fiscal_Week"], format="%Y%m").dt.to_period('M')
df["Week-1"] = df["Week"] - 2
df["Week-2"] = df["Week"] - 1
df["Week-3"] = df["Week"] + 1

# Overwrite first week entries
week0idx = df[["Week-1"]].idxmin()
week0 = df.loc[week0idx,"Week"].iloc[0]
week0idxs = df["Week"][df["Week"] == week0].index
df.loc[week0idxs,["Week-1","Week-2","Week-3"]] = [week0 + 1, week0 + 2, week0 + 3]

# Overwrite second week entries
week1idx = df[["Week-1"]].idxmin()
week1 = df.loc[week1idx,"Week"].iloc[0]
week1idxs = df["Week"][df["Week"] == week1].index
df.loc[week1idx,["Week-1","Week-2","Week-3"]] = [week1 - 1, week1 + 1, week1 + 2]

这个任务有点奇怪，对我来说“周减3”、“负2”、“负1”和“加1”对我来说更有意义，让未知的那些空着.

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/71322208

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