python - 如果字符串包含重复的单词，则仅保留第一个单词

Question

customer_name                                               ANDY
number_of_product_variants                                      2
number_of_channels                                              1
number_of_discount_codes                                        1
order_count                                                     1
order_name                                            #1100,#1100
discount_code                        Christmas2020, Christmas2020
channel                                      Instagram, Instagram
product_variant                    Avengers Set A, Avengers Set B

仅当字符串包含重复项时，我才想删除重复的单词。
预期输出:

customer_name                                                ANDY
number_of_product_variants                                      2
number_of_channels                                              1
number_of_discount_codes                                        1
order_count                                                     1
order_name                                                  #1100
discount_code                                       Christmas2020
channel                                                 Instagram
product_variant                    Avengers Set A, Avengers Set B

我试过的代码:

def unique_string(l):
    ulist = []
    [ulist.append(x) for x in l if x not in ulist]
    return ulist

customer_df['channel_2']=customer_df['channel']
customer_df['channel_2'].apply(unique_string)

仅将以下代码用于 channel列返回:

0                                   [S, e, a, r, c, h, ,]
1                    [P, a, i, d,  , A, s, :, S, o, c, l]
2                 [P, a, i, d,  , A, s, :, S, o, c, l, ,]
3                                      [U, n, k, o, w, ,]
```

Answer 1

您可以使用 set用 , 分割值如果有多个值，则顺序不重要。
如果订单很重要，请使用 dict 和 .keys() :

customer_df = pd.DataFrame({"channel_2":['Instagram, Instagram',
                                         'Instagram, Instagram1, Instagram, Instagram2']})
    
f1 = lambda x: ', '.join(set(y for y in x.split(', ')))
f2 = lambda x: ', '.join(dict.fromkeys(y for y in x.split(', ')).keys())

customer_df['channel_2_1'] = customer_df['channel_2'].apply(f1)
customer_df['channel_2_2'] = customer_df['channel_2'].apply(f2)
print (customer_df)
                                      channel_2  \
0                          Instagram, Instagram   
1  Instagram, Instagram1, Instagram, Instagram2   

                         channel_2_1                        channel_2_2  
0                          Instagram                          Instagram  
1  Instagram2, Instagram1, Instagram  Instagram, Instagram1, Instagram2