根据重复项更改数据框中的值 - python

Question

我有一个相当大的数据集，包含超过 100k 行，其中包含许多重复项和一些缺失或错误的值。尝试在下面的 sn-p 中简化问题。

sampleData = {
    'BI Business Name' : ['AAA', 'BBB', 'CCC', 'DDD','DDD'],
    'BId Postcode' : ['NW1 8NZ', 'NW1 8NZ', 'WC2N 4AA', 'CV7 9JY', 'CV7 9JY'],
    'BI Website' : ['www@1', 'www@1', 'www@2', 'www@3', np.nan],
    'BI Telephone' : ['999', '999', '666', np.nan, '12345']    
}
df = pd.DataFrame(sampleData)

我正在尝试根据重复行更改值，因此如果任何三个字段匹配，那么第四个字段也应该匹配。我应该得到这样的结果：

result = {
    'BI Business Name' : ['AAA', 'AAA', 'CCC', 'DDD','DDD'],
    'BId Postcode' : ['NW1 8NZ', 'NW1 8NZ', 'WC2N 4AA', 'CV7 9JY', 'CV7 9JY'],
    'BI Website' : ['www@1', 'www@1', 'www@2', 'www@3', 'www@3'],
    'BI Telephone' : ['999', '999', '666', '12345', '12345']    
}
df = pd.DataFrame(result)

我找到了非常冗长的方法 - 这里只显示更改名称的部分。

df['Phone_code_web'] = df['BId Postcode'] + df['BI Website'] + df['BI Telephone'] 
reference_name = df[['BI Business Name', 'BI Telephone', 'BId Postcode','BI Website']]
reference_name = reference_name.dropna()
reference_name['Phone_code_web'] = reference_name['BId Postcode'] + reference_name['BI Website'] + 
reference_name['BI Telephone'] 
duplicate_ref = reference_name[reference_name['Phone_code_web'].duplicated()]
reference_name = pd.concat([reference_name,duplicate_ref]).drop_duplicates(keep=False)
reference_name

def replace_name(row):
    try:
        old_name = row['BI Business Name']
        reference = row['Phone_code_web']     
        new_name = reference_name[reference_name['Phone_code_web']==reference].iloc[0,0]  
        print(new_name)

        return new_name
    except Exception as e:
        return old_name
df['BI Business Name']=df.apply(replace_name, axis=1)
df

有没有更简单的方法？

Answer 1

你可以试试这个：

import pandas as pd

sampleData = {
    'BI Business Name': ['AAA', 'BBB', 'CCC', 'DDD','DDD'],
    'BId Postcode': ['NW1 8NZ', 'NW1 8NZ', 'WC2N 4AA', 'CV7 9JY', 'CV7 9JY'],
    'BI Website': ['www@1', 'www@1', 'www@2', 'www@3', np.nan],
    'BI Telephone': ['999', '999', '666', np.nan, '12345']
}
df = pd.DataFrame(sampleData)
print(df)

def fill_gaps(_df, _x):  # _df and _x are local variables that represent the dataframe and one of its rows, respectively
    # pd.isnull(_x) = list of Booleans indicating which columns have NaNs
    # df.columns[pd.isnull(_x)] = list of columns whose value is a NaN
    for col in df.columns[pd.isnull(_x)]:
        # len(set(y) & set(_x)) = length of the intersection of the row being considered (_x) and each of the other rows in turn (y)
        # the mask is a list of Booleans which are True if:
        # 1) y[col] is not Null (e.g. for row 3 we need to replace (BI Telephone = NaN) with a non-NaN 'BI Telephone' value)
        # 2) and the length of the intersection above is at least 3 (as required)
        mask = df.apply(lambda y: pd.notnull(y[col]) and len(set(y) & set(_x)) == 3, axis=1)
        # if the mask has at least one "True" value, select the value in the corresponding column (if there are several possible values, select the first one)
        _x[col] = df[mask][col].iloc[0] if any(mask) else _x[col]
    return _x

# Apply the logic described above to each row in turn (x = each row)
df = df.apply(lambda x: fill_gaps(df, x), axis=1)

print(df)

输出：

  BI Business Name BId Postcode BI Website BI Telephone
0              AAA      NW1 8NZ      www@1          999
1              BBB      NW1 8NZ      www@1          999
2              CCC     WC2N 4AA      www@2          666
3              DDD      CV7 9JY      www@3          NaN
4              DDD      CV7 9JY        NaN        12345
  BI Business Name BId Postcode BI Website BI Telephone
0              AAA      NW1 8NZ      www@1          999
1              BBB      NW1 8NZ      www@1          999
2              CCC     WC2N 4AA      www@2          666
3              DDD      CV7 9JY      www@3        12345
4              DDD      CV7 9JY      www@3        12345