python - Pandas Dataframe 上的条件正则表达式函数

Question

我有以下df and function (see below) 。我可能把这件事复杂化了。如果有一双新的眼睛，我们将不胜感激。

df:

Site Name   Plan Unique ID  Atlas Placement ID
Affectv     we11080301      11087207850894
Mashable    we14880202      11087208009031
Alphr       uk10790301      11087208005229
Alphr       uk19350201      11087208005228

目标是:

1. 首先通过 df['Plan Unique ID'] ，如果有匹配，则搜索特定值( we_match 或 uk_match )

2. 检查字符串值是否大于该组中的某个值( we12720203 或 uk11350200 )

3. 如果该值大于则添加 we or uk value到新专栏df['Consolidated ID'] .

4. 如果值较小或没有匹配项，则搜索 df['Atlas Placement ID']与 new_id_search

5. 如果有匹配项，则将其添加到 df['Consolidated ID']

6. 如果不是，则返回 0 到 df['Consolidated ID]

当前的问题是它返回一个空列。

 def placement_extract(df="mediaplan_df", we_search="we\d{8}", uk_search="uk\d{8}", new_id_search= "(\d{14})"):

        if type(df['Plan Unique ID']) is str:
            we_match = re.search(we_search, df['Plan Unique ID'])
            if we_match:
                if we_match > "we12720203":
                    return we_match.group(0)
                else:
                    uk_match =  re.search(uk_search, df['Plan Unique ID'])
                    if uk_match:
                        if uk_match > "uk11350200":
                            return uk_match.group(0)
                        else:
                            match_new =  re.search(new_id_search, df['Atlas Placement ID'])
                            if match_new:
                                return match_new.group(0)

                            return 0


    mediaplan_df['Consolidated ID'] = mediaplan_df.apply(placement_extract, axis=1)

编辑:清理公式

我修改了following way (see below)中gzl的函数:首先看df1中是否有14个数字。如果有，请添加。

下一步，理想情况下是获取一列 MediaPlanUnique来自df2并把它变成一个系列filtered_placements :

we11080301  
we12880304  
we14880202  
uk19350201  
uk11560205  
uk11560305  

并查看 filtered_placements 中是否有任何值存在于 df['Plan Unique ID] 。如果匹配，则添加 df['Plan Unique ID]到我们的最后一栏= df[ConsolidatedID]

当前的问题是它的结果全为0。我认为这是因为比较是进行1对1( first result of new_match vs first result of filtered_placements )而不是1对多( first result of new_match vs all results of filtered_placements )

有什么想法吗？

def placement_extract(df="mediaplan_df", new_id_search="[a-zA-Z]{2}\d{8}", old_id_search= "(\d{14})"):

    if type(df['PlacementID']) is str:

        old_match =  re.search(old_id_search, df['PlacementID'])
        if old_match:
            return old_match.group(0)

        else:

            if type(df['Plan Unique ID']) is str:
                if type(filtered_placements) is str:


                    new_match = re.search(new_id_search, df['Plan Unique ID'])
                    if new_match:
                        if filtered_placements.str.contains(new_match.group(0)):
                            return new_match.group(0)          


        return 0

mediaplan_df['ConsolidatedID'] = mediaplan_df.apply(placement_extract, axis=1)

Answer 1

我建议不要使用如此复杂的嵌套 if 语句。正如菲尔指出的那样，每项检查都是相互排斥的。因此，您可以在同一缩进 if 语句中检查“we”和“uk”，然后退回到默认流程。

def placement_extract(df="mediaplan_df", we_search="we\d{8}", uk_search="uk\d{8}", new_id_search= "(\d{14})"):

    if type(df['Plan Unique ID']) is str:
        we_match = re.search(we_search, df['Plan Unique ID'])
        if we_match:
            if we_match.group(0) > "we12720203":
                return we_match.group(0)

        uk_match =  re.search(uk_search, df['Plan Unique ID'])
        if uk_match:
            if uk_match.group(0) > "uk11350200":
                return uk_match.group(0)


        match_new =  re.search(new_id_search, df['Atlas Placement ID'])

        if match_new:
            return match_new.group(0)

        return 0

测试:

In [37]: df.apply(placement_extract, axis=1)
Out[37]:
0    11087207850894
1        we14880202
2    11087208005229
3        uk19350201
dtype: object