如何使用条件将csv拆分为2个数据帧

Question

My idea is seperate both of the "String" then convert both dataframe into same datetime format. I try the code

data['date'] = pd.to_datetime(data['date'])
data['date'] = data['date'].dt.strftime('%Y-%m-%d')

but there are some error on the output. The 13/02/2020 will become 2020-02-13 that is what i want. But the 12/02/2020 will become 2020-12-02.

My dataframe have 2 type of date format. Which is YYYY-MM-DD and DD/MM/YYYY.

dataframe

I need to split it into 2 dataframe, all the row that have the date YYYY-MM-DD into the df1 .

The data type is object.

All all the row that have the date DD/MM/YYYY into the df2 .

Anyone know how to code it?

Answer 1

If dont need convert to datetimes use Series.str.contains with boolean indexing :

mask = df['date'].str.contains('-')
df1 = df[mask].copy()
df2 = df[~mask].copy()

If need datetimes you can use parameter errors='coerce' in to_datetime for missing values if not matching format, so last remove missing values:

df1 = (df.assign(date = pd.to_datetime(df['date'], format='%Y-%m-%d', errors='coerce')
        .dropna(subset=['date']))

df2 = (df.assign(date = pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')
         .dropna(subset=['date']))

EDIT: If need output column filled by correct datetimes you can replace missing values by another Series by Series.fillna :

date1 = pd.to_datetime(df['date'], format='%Y-%m-%d', errors='coerce')
date2 = pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')

df['date'] = date1.fillna(date2)

Answer 2

you can use the fact that the separation is different to find the dates.

If your dataframe is in this format:

 df = pd.DataFrame({'id' : [1,1,2,2,3,3], 
   "Date": ["30/8/2020","30/8/2021","30/8/2022","2019-10-24","2019-10-25","2020-10-24"] })

With either "-" or "/" to separate the data

you can use a function that finds this element and apply it to the date column:

   def find(string):
       if string.find('/')==2:
     return True 
       else:
     return False

   df[df['date'].apply(find)]