在python中使用pandas映射列上匹配的字数

我有一个df

Name    Step     Description
Ram        1     Ram is oNe of the good cricketer
Ram        2     gopal one
Sri        1     Sri is one of the member
Sri        2     ravi good 
Kumar      1     Kumar is a keeper
Madhu      1     good boy
Vignesh    1     oNe little
Pechi      1     one book
mario      1     good randokm
Roger      1     one milita good
bala       1     looks good
raj        1     more one
venk       1     likes good

和一份清单

my_list=["one","good"]

我正在尝试从我的_列表中获取至少有一个关键字的行

我试过了， mask=df[“Description”].str.contains（“|”） 我得到了输出_df

Name    Description
Ram     Ram is one of the good cricketer
Sri     Sri is one of the member        

我还想在单独的列中添加“描述”中的关键字及其计数

当df[“Name”]不是第一次出现时，即使“Description”包含一个关键字，也不应复制keys列中的关键字我想要的输出是

 Name   Step    Description                          keys        count
 Ram     1     Ram is one of the good cricketer      one,good    2
 Ram     2     gopal one
 Sri     1     Sri is one of the member              one         1
 Sri     2     ravi good
 Kumar   1     Kumar is a keeper
 Madhu   1     good boy                              good        1
 Vignesh 1     oNe little                            oNe         1
 Pechi   1     one book                              one         1 
 mario   1     good randokm good                     good        1
 Roger   1     one milita good                       one,good    2
 bala    1     looks good                            good        1
 raj     1     more one                              one         1
 venk    1     likes good                            good        1

我想要的输出是

 Name   Step    Description                          keys        count
 Ram     1     Ram is one of the good cricketer      one,good    2
 Ram     2     gopal one
 Sri     1     Sri is one of the member              one         1
 Sri     2     ravi good
 Kumar   1     Kumar is a keeper
 Madhu   1     good boy                              good        1
 Vignesh 1     oNe little                            oNe         1
 Pechi   1     one book                              one         1 
 mario   1     good randokm good                     good        1
 Roger   1     one milita good                       one,good    2
 bala    1     looks good                            good        1
 raj     1     more one                              one         1
 venk    1     likes good                            good        1

---

创建新的遮罩并应用它：

my_list=["one","good"]

mask=df["Description"].str.contains("|".join(my_list),na=False,flags=re.IGNORECASE ) & \
     (df.groupby('Name').cumcount() == 0)
print (mask)
0      True
1     False
2      True
3     False
4     False
5      True
6      True
7      True
8      True
9      True
10     True
11     True
12     True
dtype: bool

---

编辑：

---

---

---

我不想考虑步骤栏，我想在“名称”栏上应用逻辑。当名称值首次出现时。正如你在索引＝1中看到的那样，RAM发生了第二次，所以我们不应该考虑索引为1OK的行上的关键字，你认为 Cucunt< /Cord>为计数精度吗？并且只匹配第一个值？代码>打印（df.groupby（'Name'）.cumcount（））等于

0

？`0 0 1 1 2 0 3 1 4 0`好的，但我做了，

fillna（“”）

然后

s=df.groupby（'Name'）['Description']）。转换（'，'.join）

。是否需要将其更改为

s=df.groupby（'Name'）['Description'].transform（lambda x:'，'.join（x.astype（str）））

#transform all values if need same size of original
s = df.groupby('Name')['Description'].transform(','.join)
print (s)
0     Ram is oNe of the good cricketer,gopal one
1     Ram is oNe of the good cricketer,gopal one
2            Sri is one of the member,ravi good 
3            Sri is one of the member,ravi good 
4                              Kumar is a keeper
5                                       good boy
6                                     oNe little
7                                       one book
8                              good randokm good
9                                one milita good
10                                    looks good
11                                      more one
12                                    likes good
Name: Description, dtype: object

#for mask use new Series s
mask=s.str.contains("|".join(my_list),na=False,flags=re.IGNORECASE ) & \
     (df.groupby('Name').cumcount() == 0)
print (mask)
0      True
1     False
2      True
3     False
4     False
5      True
6      True
7      True
8      True
9      True
10     True
11     True
12     True
dtype: bool

#extract from new Series s
extracted = s.str.findall('(' + '|'.join(my_list) + ')', flags=re.IGNORECASE).apply(set)
df.loc[mask, 'keys'] = extracted.str.join(',')
df.loc[mask, 'count'] = extracted.str.len()
print (df)
       Name  Step                       Description          keys  count
0       Ram     1  Ram is oNe of the good cricketer  good,oNe,one    3.0
1       Ram     2                         gopal one           NaN    NaN
2       Sri     1          Sri is one of the member      good,one    2.0
3       Sri     2                        ravi good            NaN    NaN
4     Kumar     1                 Kumar is a keeper           NaN    NaN
5     Madhu     1                          good boy          good    1.0
6   Vignesh     1                        oNe little           oNe    1.0
7     Pechi     1                          one book           one    1.0
8     mario     1                 good randokm good          good    1.0
9     Roger     1                   one milita good      good,one    2.0
10     bala     1                        looks good          good    1.0
11      raj     1                          more one           one    1.0
12     venk     1                        likes good          good    1.0