已解决。

Question

我有一个包含人们信息的DataFrame，但是地址重复的行有些不同。

如何基于模糊匹配或其他检测相似性的方式删除重复项，但确保只有在姓氏和名字也匹配的情况下，具有相似地址的行才会被删除？

示例数据：

function fn<T extends (...a: any[]) => (c: number) => number>(o: T) {
  return o;
}
const createSum5 = fn(() => c => c + 5)
const createMultiplyN = fn((n: number) => c => n * c);
const createWordsSum = fn((word: string) => c => word.length + c);

（故意在街上打字）

示例数据代码：

    First name | Last name | Address
0     John         Doe        ABC 9
1     John         Doe        KFT 2
2     Michael      John       ABC 9
3     Mary         Jane       PEP 9/2
4     Mary         Jane       PEP, 9-2
5     Gary         Young      verylongstreetname 1       
6     Gary         Young      1 verylongstretname

预期输出：

df = pd.DataFrame([
    ['John', 'Doe', 'ABC 9'],
    ['John', 'Doe', 'KFT 2'],
    ['Michael', 'John', 'ABC 9'],
    ['Mary', 'Jane', 'PEP 9/2'],
    ['Mary', 'Jane', 'PEP, 9-2'],
    ['Gary', 'Young', 'verylongstreetname 1'],
    ['Gary', 'Young', '1 verylongstretname']
], columns=['First name', 'Last name', 'Address'])

Answer 1

使用str.replace删除所有非单词字符，然后drop_duplicates

df['Address'] = df['Address'].str.replace(r'\W','')
temp_address = df['Address']
df.drop_duplicates(inplace=True)

输出

  First name Last name Address
0       John       Doe    ABC9
1       John       Doe    KFT2
2    Michael      John    ABC9
3       Mary      Jane   PEP92

替换原始地址

b['Address'] = b['Address'].apply(lambda x: [w for w in temp_address if w.split(' ')[0] in x][0])

输出

 First name Last name  Address
0       John       Doe    ABC 9
1       John       Doe    KFT 2
2    Michael      John    ABC 9
3       Mary      Jane  PEP 9/2

好的，这是一种方法

df['Address'] = df['Address'].str.replace(r'\W',' ') # giving a space

def check_simi(d):
    temp = []
    flag = 0
    for w in d:
        temp.extend(w.split(' '))
    temp = [t for t in temp if t]    
    flag = len(temp) / 2

    if len(set(temp)) == flag:
        return int(d.index[0])
    else:


indexes = df.groupby(['First name','Last name'])['Address'].apply(check_simi)
indexes = [int(i) for i in indexes if i >= 0]

df.drop(indexes)

  First name Last name               Address
0       John       Doe                 ABC 9
1       John       Doe                 KFT 2
2    Michael      John                 ABC 9
4       Mary      Jane              PEP  9 2
6       Gary     Young  1 verylongstreetname

PS-请研究https://github.com/seatgeek/fuzzywuzzy，以获得更简洁的方法，但由于网络不允许，我没有这样做

Answer 2

已解决。

基于@iamklaus anwser，我编写了这段代码：

def remove_duplicates_inplace(df, groupby=[], similarity_field='', similar_level=85):
    def check_simi(d):
        dupl_indexes = []
        for i in range(len(d.values) - 1):
            for j in range(i + 1, len(d.values)):
                if fuzz.token_sort_ratio(d.values[i], d.values[j]) >= similar_level:
                    dupl_indexes.append(d.index[j])

        return dupl_indexes

    indexes = df.groupby(groupby)[similarity_field].apply(check_simi)

    for index_list in indexes:
        df.drop(index_list, inplace=True)


remove_duplicates_inplace(df, groupby=['firstname', 'lastname'], similarity_field='address')

输出：

  firstname lastname               address
0      John      Doe                 ABC 9
1      John      Doe                 KFT 2
2   Michael     John                 ABC 9
3      Mary     Jane               PEP 9/2
5      Gary    Young  verylongstreetname 1

基于模糊匹配的大熊猫删除重复项

2 个答案:

已解决。