Creating a column that takes a Week Number and Year and returns a Date

Question

I’m currently working with a dataframe where I created a Year and Week # column. I’m trying to create a new column Date that gives me the date for a from the Year and Week # columns.

This is what my dataframe looks like now

Year	Week #
2023	10
2023	11
2023	12

It should look like this

Year	Week #	Date
2023	10	3/6/23
2023	11	3/13/23
2023	12	3/20/23

I tried the following

from datetime import datetime

df['Date'] = datetime.strptime('{}-{}-1'.format(df['Year'], df['Week #']), '%Y-%W-%w').strftime('%Y-%m-%dT%H:%M:%S.%f')[:-3] + 'Z'

However, I got this error

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
Cell In[57], line 3
      1 from datetime import datetime
----> 3 df['Date'] = datetime.strptime('{}-{}-1'.format(df['Year'], df['Week #']), '%Y-%W-%w').strftime('%Y-%m-%dT%H:%M:%S.%f')[:-3] + 'Z'
      4 df

File /opt/anaconda3/lib/python3.9/_strptime.py:568, in _strptime_datetime(cls, data_string, format)
    565 def _strptime_datetime(cls, data_string, format="%a %b %d %H:%M:%S %Y"):
    566     """Return a class cls instance based on the input string and the
    567     format string."""
--> 568     tt, fraction, gmtoff_fraction = _strptime(data_string, format)
    569     tzname, gmtoff = tt[-2:]
    570     args = tt[:6] + (fraction,)

File /opt/anaconda3/lib/python3.9/_strptime.py:349, in _strptime(data_string, format)
    347 found = format_regex.match(data_string)
    348 if not found:
--> 349     raise ValueError("time data %r does not match format %r" %
    350                      (data_string, format))
    351 if len(data_string) != found.end():
    352     raise ValueError("unconverted data remains: %s" %
    353                       data_string[found.end():])

ValueError: time data '0        2020n1        2020n2        2020n3        2020n4        
2020n         ... n35913    2024n35914    2024n35915    2024n35916    
2024n35917    2024nName: Year, Length: 35918, dtype: int64-0        02n1        
03n2        04n3        05n4        06n         ..n35913    42n35914    
43n35915    44n35916    45n35917    46nName: Week #, Length: 35918, dtype: 
object-1' does not match format '%Y-%W-%w'

I also tried the following

from datetime import datetime
from isoweek import Week

df['Date'] = Week(df['Year'], df['Week #']).monday()

But I got the following error

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
Cell In[51], line 4
      1 from datetime import datetime
      2 from isoweek import Week
----> 4 df['Date'] = Week(df['Year'], df['Week #']).monday()

File /opt/anaconda3/lib/python3.9/site-packages/isoweek.py:34, in Week.__new__(cls, year, week)
     27 def __new__(cls, year, week):
     28     """Initialize a Week tuple with the given year and week number.
     29 
     30     The week number does not have to be within range.  The numbers
     31     will be normalized if not.  The year must be within the range
     32     1 to 9999.
     33     """
---> 34     if week < 1 or week > 52:
     35         return cls(year, 1) + (week - 1)
     36     if year < 1 or year > 9999:

File /opt/anaconda3/lib/python3.9/site-packages/pandas/core/generic.py:1527, in NDFrame.__nonzero__(self)
   1525 @final
   1526 def __nonzero__(self) -> NoReturn:
-> 1527     raise ValueError(
   1528         f"The truth value of a {type(self).__name__} is ambiguous. "
   1529         "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
   1530     )

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Asked By: Edd

||

Source

Answer 1

IIUC, you can use the format parameter to specify your date format. First build your date in string format, then turn it into a real date.

data = {"year": [2022, 2023], "week": [1,2]}
df = pd.DataFrame(data)
print(df)
#    year  week
# 0  2022     1
# 1  2023     2

# Create 'year-w-1' formatted dates
df['date'] = df['year'].astype(str).str.cat(others=[df['week'].astype(str), np.array(['1']*len(df))], sep='-')
print(df)
#    year  week      date
# 0  2022     1  2022-1-1
# 1  2023     2  2023-2-1

# format is year-week-weekday (1 = Monday)
df['date'] = pd.to_datetime(df['date'], format="%Y-%W-%w")
print(df)
#    year  week       date
# 0  2022     1 2022-01-03
# 1  2023     2 2023-01-09

Note that the -1 in the second dataframe is later interpreted as weekday!

Answered By: JarroVGIT

Answer 2

As i do not have your data frame to match i am using sample input

import pandas as pd
import datetime

# create a sample dataframe with week number and year columns
df = pd.DataFrame({"Year": [2016, 2016, 2016, 2017, 2017, 2017],"Week": [43, 44, 51, 2, 5, 12]})

# define a function that takes a week number and year and returns a date
def week_to_date(week, year):
  # get the first day of the week (Monday) for the given week and year
  date = datetime.date.fromisocalendar(year, week, 1)
  return date

# apply the function to the dataframe and create a new column with the date
df["Date"] = df.apply(lambda row: week_to_date(row["Week"], row["Year"]), axis=1)

# print the dataframe
print(df)

Output:

Year  Week        Date
0  2016    43  2016-10-24
1  2016    44  2016-10-31
2  2016    51  2016-12-19
3  2017     2  2017-01-09
4  2017     5  2017-01-30
5  2017    12  2017-03-20

Answered By: Avinash Chandravansi

Creating a column that takes a Week Number and Year and returns a Date

Question:

Answers: