英文:
How to replace digits in 1 column in dataframe inplace?
问题
df中有大约200,000行数据。其中一列是电话号码。我需要将数字8替换为+7。我不想创建另一个df。我想要原地进行替换。
df.head()client_id contact_number0 +777600135051 +777792614332 +770710610473 +777140324014 877877636215 87787763621
我想要将那些以8开头的数字替换为+7，原地操作。
starts_8_length_11 = df[(df['contact_number'].str.startswith('8')) & (df['contact_number'].str.len() == 11)]
但我不想从主df中创建starts_8_length_11，然后对其进行更改，然后再放回df中。如何原地操作呢？
英文:
The df has some 200000 rows. One column is telephone number. I need to replace 8 with +7. I dont want to create another df. I want to do it inplace.
df.head()client_id contact_number0 +777600135051 +777792614332 +770710610473 +777140324014 877877636215 87787763621
I want to replace those with 8 to +7 inplace.
starts_8_length_11 = df[(df['contact_number'].str.startswith('8')) & (df['contact_number'].str.len()==11)]
But I dont want to create starts8_length11 from the main df, change it then put it back into the df. How to do it inplace?
答案1
得分: 2
解决方案应该是使用正则表达式^来替换字符串开头的8，但仅针对长度为11的行：
m = (df['contact_number'].str.len()==11)df.loc[m, 'contact_number'] = df.loc[m, 'contact_number'].str.replace('^8', '+7')print (df)client_id contact_number0 0 +777600135051 1 +777792614332 2 +770710610473 3 +777140324014 4 +777877636215 5 +77787763621
英文:
Solution should be replace 8 with regex ^ for start of string, but only for rows with lenght 11:
m = (df['contact_number'].str.len()==11)df.loc[m, 'contact_number'] = df.loc[m, 'contact_number'].str.replace('^8', '+7')print (df)client_id contact_number0 0 +777600135051 1 +777792614332 2 +770710610473 3 +777140324014 4 +777877636215 5 +77787763621
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