CODEWITHSUNDEEP
pythonpandas
user10508414
user10508414

Reputation:

How to convert different date format to one

My Data frame is below

I am trying to find the age

customer_Id DOB
0   268408  1970-02-01
1   268408  1970-02-01
2   268408  1970-02-01
3   268408  1970-02-01
4   268408  1970-02-01

shape is of 207518

while converting the data i got ValueError: unconverted data remains: 5

code is below to convert in to age

def cal_age(dob=str):
    x = dt.datetime.strptime(dob, "%Y-%d-%m")
    y = dt.date.today()
    age = y.year - x.year - ((y.month, x.day) < (y.month, x.day))
    return age

df_n_4['DOB'] = pd.to_datetime(df_n_4['DOB'],errors='coerce')

df_n_4['DOB'] = df_n_4['DOB'].astype('str')

df_n_4['Age'] = df_n_4.DOB.apply(lambda z: cal_age(z))

Error is below

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-113-90ef9d4c7002> in <module>
----> 1 df_n_4['Age'] = df_n_4.DOB.apply(lambda z: cal_age(z))

~\Anaconda3\lib\site-packages\pandas\core\series.py in apply(self, func, convert_dtype, args, **kwds)
   4040             else:
   4041                 values = self.astype(object).values
-> 4042                 mapped = lib.map_infer(values, f, convert=convert_dtype)
   4043 
   4044         if len(mapped) and isinstance(mapped[0], Series):

pandas\_libs\lib.pyx in pandas._libs.lib.map_infer()

<ipython-input-113-90ef9d4c7002> in <lambda>(z)
----> 1 df_n_4['Age'] = df_n_4.DOB.apply(lambda z: cal_age(z))

<ipython-input-111-98546f386b50> in cal_age(dob)
      1 def cal_age(dob=str):
----> 2     x = dt.datetime.strptime(dob, "%Y-%d-%m")
      3     y = dt.date.today()
      4     age = y.year - x.year - ((y.month, x.day) < (y.month, x.day))
      5     return age

~\Anaconda3\lib\_strptime.py in _strptime_datetime(cls, data_string, format)
    575     """Return a class cls instance based on the input string and the
    576     format string."""
--> 577     tt, fraction, gmtoff_fraction = _strptime(data_string, format)
    578     tzname, gmtoff = tt[-2:]
    579     args = tt[:6] + (fraction,)

~\Anaconda3\lib\_strptime.py in _strptime(data_string, format)
    360     if len(data_string) != found.end():
    361         raise ValueError("unconverted data remains: %s" %
--> 362                           data_string[found.end():])
    363 
    364     iso_year = year = None

ValueError: unconverted data remains: 5

Upvotes: 0

Views: 53

Answers (2)

jezrael
jezrael

Reputation: 863761

You can change your function for working with datetimes, also convert to strings is not necessary:

def cal_age(x):
    y = dt.date.today()
    age = y.year - x.year - ((y.month, x.day) < (y.month, x.day))
    return age

df_n_4['DOB'] = pd.to_datetime(df_n_4['DOB'],errors='coerce')
df_n_4['Age'] = df_n_4.DOB.apply(cal_age)

Upvotes: 1

Kallol
Kallol

Reputation: 2189

try this code,

df['current_date']=pd.datetime.now()
df['age']=(df.current_date-pd.to_datetime(df.DOB)).dt.days/365

Upvotes: 0

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