Replacing a column value by another column value based on regex - Python

Question

This is an extract of my DataFrame

data = [
    ['Citroën Amillis', '20 Za Des Baliveaux - 77120 Amillis', '77120', 'ok'],
    ['Relat Paris 9e', 'Métro Opéra - 75009 Paris 9e', 'Paris', 'error'],
    ['Macif Avon', '49 Av Franklin Roosevelt - 77210 Avon', '77210', 'ok'],
    ['Atac La Chapelle-la-Reine', 'Za Rue De L\'avenir - 77760 La Chapelle-la-Reine', 'La', 'error'],
    ['Société Générale La Ferté-Gaucher', '42 Rue De Paris - 77320 La Ferté-Gaucher', 'La', 'error']
]

df = pd.DataFrame(data, columns=['nom_magasin', 'adresse', 'code_postal', 'is_code_postal'])

df

As you can see, there are mistakes in my dataframe. For some addresses, especially when the city name is composed (ex:"La Chapelle-la-Reine"), the column "code_postal" is wrong.

What I'm looking to do is the following: if the column "is_code_postal" is an "error", replace "code_postal" by the regex of the postal code appearing in the column "adresse".

I can't find the solution. To do I've try this df['is_code_postal'] = np.where(df.code_postal.str.match('^[a-zA-z]'), 'error', 'ok') . At first I was thinking about doing all changes within the same function. But I'm missing something.

And the important thing is that my dataframe is a little bit heavy (more than 250K rows) so I'd like to go for an effective solution.

Do you guys have any idea?

Answer 1

您可以忽略 code_postal 并使用 Quang 的代码直接从“地址”中提取它：

df['code_postal']=df['adresse'].str.extract('(\d{5})')