以有效的方式将数据列表替换为 DataFrame

Question

I'm doing something like this to replace some number-strings to normal number-strings (I need a string because of front 0 in my dataframe )

char_to_repalce = [('１', "1"), ('２', "2"), ('３', "3"), ('４', "4"), ('５', "5"),
                                ('６', "6"), ('７', "7"), ('８', "8"), ('９', "9"), ('０', "0"), ('O', "0")]
            for col in ['phone_number', 'phone_number2']:
                for char in char_to_repalce:
                    try:
                        df_data[col]=df_data[col].replace(char[0], char[1])
                    except:
                        pass

Which is not very effective way I think... There are also a normal numbers as well, but everything is a string as I need strings

Answer 1

We have replace function which can accept dict

df[['phone_number', 'phone_number2']].replace(dict(char_to_repalce), inplace = True , regex=True)

My thought

df[['phone_number', 'phone_number2']].replace({' ':''}, inplace = True , regex=True)

Answer 2

Given your char_to_replace variable, I would do the following using map :

replacers = {x[0]:x[1] for x in char_to_replace}
df = pd.DataFrame({'N1':['１','２','２','３','８'],
                   'N2':['９','５','７','０','O']}) 
df['N1'] = df['N1'].map(replacers)
df['N2'] = df['N2'].map(replacers)
print(df)          

Output:

  N1 N2
0  1  9
1  2  5
2  2  7
3  3  0
4  8  0