Python在合并数据帧时使用条件逻辑/where

我有这些

df1

user_id     code     name     code_equivalence             name_equivalence
51          123    bi lovers            542                bi for marketing
51          123    bi lovers            545                i love bi
51          234    datascience          345                data and science
51          234    datascience          555                data lovers
51          255    antiquity history    429                roma
51          255    antiquity history    430                greece
52          123    bi lovers            542                bi for marketing
52          123    bi lovers            545                i love bi
52          256    modern history       500                france
52          256    modern history       501                germany
52          200    arts                 400                arts I
52          200    arts                 401                arts II

df2

user_id     code     name       status
51          123    bi lovers    ongoing
51          430    greece       ongoing
52          501    germany      ongoing
52          050    numbers      ongoing

我想通过检查df2代码是否与df1代码或df1代码_等价以及df2名称是否与df1名称或df1名称_等价来合并它们，以获得df2状态。 像这样：

合并df

user_id     code     name               code_equivalence    name_equivalence        status
51          123    bi lovers            542                 bi for marketing        ongoing
51          123    bi lovers            545                 i love bi               ongoing
51          234    datascience          345                 data and science        (null)
51          234    datascience          555                 data lovers             (null)
51          255    antiquity history    429                 roma                    (null)
51          255    antiquity history    430                 greece                  ongoing
52          123    bi lovers            542                 bi for marketing        (null)
52          123    bi lovers            545                 i love bi               (null)
52          256    modern history       500                 france                  (null)
52          256    modern history       501                 germany                 ongoing
52          200    arts                 400                 arts I                  (null)
52          200    arts                 401                 arts II                 (null)

user_id     code     name               code_equivalence    name_equivalence                    status
51          123    bi lovers            [542, 545]          [bi for marketing, i love bi]       ongoing
51          234    datascience          [345, 555]          [data and science, data lovers]     (null)
51          255    antiquity history    [429, 430]          [roma, greece]                      ongoing
52          123    bi lovers            [542, 545]          [bi for marketing, i love bi]       (null)
52          256    modern history       [500, 501]          [france, germany]                   ongoing
52          200    arts                 [400, 401]          [arts I, arts II]                   (null)

之后，我想转换数据以生成新的df，如下所示：

最终df

user_id     code     name               code_equivalence    name_equivalence        status
51          123    bi lovers            542                 bi for marketing        ongoing
51          123    bi lovers            545                 i love bi               ongoing
51          234    datascience          345                 data and science        (null)
51          234    datascience          555                 data lovers             (null)
51          255    antiquity history    429                 roma                    (null)
51          255    antiquity history    430                 greece                  ongoing
52          123    bi lovers            542                 bi for marketing        (null)
52          123    bi lovers            545                 i love bi               (null)
52          256    modern history       500                 france                  (null)
52          256    modern history       501                 germany                 ongoing
52          200    arts                 400                 arts I                  (null)
52          200    arts                 401                 arts II                 (null)

user_id     code     name               code_equivalence    name_equivalence                    status
51          123    bi lovers            [542, 545]          [bi for marketing, i love bi]       ongoing
51          234    datascience          [345, 555]          [data and science, data lovers]     (null)
51          255    antiquity history    [429, 430]          [roma, greece]                      ongoing
52          123    bi lovers            [542, 545]          [bi for marketing, i love bi]       (null)
52          256    modern history       [500, 501]          [france, germany]                   ongoing
52          200    arts                 [400, 401]          [arts I, arts II]                   (null)

---

有人能帮我吗？

不确定我的提问是否正确，但从我读到的内容来看，您进行了合并，现在您希望得到

最终结果

？如果是这样的话，考虑到

merged

是您的合并数据帧，这应该可以完成工作

 >>> merged.groupby(['user_id','code','name']).agg(list).reset_index()
   user_id  code               name code_equivalence                 name_equivalence              status
0       51   123          bi lovers       [542, 545]    [bi for marketing, i love bi]  [ongoing, ongoing]
1       51   234        datascience       [345, 555]  [data and science, data lovers]    [(null), (null)]
2       51   255  antiquity history       [429, 430]                   [roma, greece]   [(null), ongoing]
3       52   123          bi lovers       [542, 545]    [bi for marketing, i love bi]    [(null), (null)]
4       52   200               arts       [400, 401]                [arts I, arts II]       [(null), nan]
5       52   256     modern history       [500, 501]                [france, germany]   [(null), ongoing]

如果您只有

df1

和

df2

，那么这就是完整的解决方案：

 >>> (pd
     ...: .merge(df1,df2, left_on=['user_id','code','name'], right_on=['user_id','code','name'], how='left')
     ...: .groupby(['user_id','code','name'])
     ...: .agg(list)
     ...: .reset_index())

   user_id  code               name code_equivalence                 name_equivalence              status
0       51   123          bi lovers       [542, 545]    [bi for marketing, i love bi]  [ongoing, ongoing]
1       51   234        datascience       [345, 555]  [data and science, data lovers]          [nan, nan]
2       51   255  antiquity history       [429, 430]                   [roma, greece]          [nan, nan]
3       52   123          bi lovers       [542, 545]    [bi for marketing, i love bi]          [nan, nan]
4       52   200               arts       [400, 401]                [arts I, arts II]          [nan, nan]
5       52   256     modern history       [500, 501]                [france, germany]          [nan, nan]

---

这就是我如何通过三个步骤获得数据帧的方法：

在第一个条件下合并

在第二个条件下合并

用步骤2中的匹配项填充步骤1中缺失的匹配项

merge_df = pd.merge(df1, df2[["code","status"]], left_on=["code"], right_on=["code",], how="left")
merge_df2 = pd.merge(df1, df2[["code","status"]], left_on=["code_equivalence"], right_on=["code",], how="left")
merge_df["status"].fillna(merge_df2["status"], inplace=True)

---

然而，我想知道是否有一个简单的方法可以做到这一点（可能是的）。

可靠的要求，但到目前为止，您遇到了哪些问题

merge

和

groupby

是执行此任务所需的正确工具。即使使用left join，使用merge时，我也会丢失df1唯一的代码和名称。您可以检查merge\u df的第五行吗？基于df1和df2，我认为不应该有匹配项，所以状态列中不应该是（null）吗？我是指代码为255，名字为罗姆的那一行。@Michał89是的，你说得对。它是空的