Convert Year Month Day to Day of Year in Python

Question

I m using the datetime module  i e    gt  gt  gt  import datetime  gt  gt  gt  today   datetime datetime now    gt  gt  gt  print today  2009-03-06 13 24 58 857946  and I would like to compute the day of year that takes leap years into account  e g  today  March 6  2009  is the 65th day of 2009  I see a two options   Create a number of days in month    31  28       array  decide if it s a leap year and manually sum up the days   Use datetime timedelta to make a guess  amp  then binary search for the correct day of the year   gt  gt  gt  import datetime  gt  gt  gt  YEAR   2009  gt  gt  gt  DAY OF YEAR   62  gt  gt  gt  d   datetime date YEAR  1  1    datetime timedelta DAY OF YEAR - 1     These both feel pretty clunky  amp  I have a gut feeling that there s a more  quot Pythonic quot  way of calculating the day of the year  Any ideas suggestions

User · Answer

You could use strftime with a  j format string   gt  gt  gt  import datetime  gt  gt  gt  today   datetime datetime now    gt  gt  gt  today strftime   j    065   but if you wish to do comparisons or calculations with this number  you would have to convert it to int   because strftime   returns a string  If that is the case  you are better off using DzinX s answer

User · Answer

Use datetime timetuple   to convert your datetime object to a time struct time object then get its tm yday property  from datetime import datetime day of year   datetime now   timetuple   tm yday    returns 1 for January 1st

User · Answer

I want to present performance of different approaches  on Python 3 4  Linux x64  Excerpt from line profiler         Line        Hits         Time  Per Hit     Time  Line Contents                                                                                              823      1508        11334      7 5     41 6          yday   int period end strftime   j             824      1508         2492      1 7      9 1          yday   period end toordinal   - date period end year  1  1  toordinal     1          825      1508         1852      1 2      6 8          yday    period end - date period end year  1  1   days   1          826      1508         5078      3 4     18 6          yday   period end timetuple   tm yday                  So most efficient is  yday    period end - date period end year  1  1   days

User · Answer

DZinX s answer is a great answer for the question   I found this question and used DZinX s answer while looking for the inverse function  convert dates with the julian day-of-year into the datetimes  I found this to work  import datetime datetime datetime strptime  1936-077T13 14 15    Y- jT H  M  S     gt  gt  gt  gt  datetime datetime 1936  3  17  13  14  15   datetime datetime strptime  1936-077T13 14 15    Y- jT H  M  S   timetuple   tm yday   gt  gt  gt  gt  77  Or numerically  import datetime year julian    1936 77  datetime datetime year  1  1  datetime timedelta days julian -1    gt  gt  gt  gt  datetime datetime 1936  3  17  0  0   Or with fractional 1-based jdates popular in some domains  jdate frac    datetime datetime 1936  3  17  13  14  15 -datetime datetime 1936  1  1   total seconds   86400 1 display jdate frac    gt  gt  gt  gt  77 5515625  year julian    1936 jdate frac  display datetime datetime year  1  1  datetime timedelta days julian -1     gt  gt  gt  gt  datetime datetime 1936  3  17  13  14  15   I m not sure of etiquette around here  but I thought a pointer to the inverse functionality might be useful for others like me

User · Answer

If you have reason to avoid the use of the datetime module  then these functions will work   def is leap year year           if year is a leap year return True         else return False         if year   100    0          return year   400    0     return year   4    0  def doy Y M D           given year  month  day return day of year         Astronomical Algorithms  Jean Meeus  2d ed  1998  chap 7         if is leap year Y           K   1     else          K   2     N   int  275   M    9 0  - K   int  M   9    12 0    D - 30     return N  def ymd Y N           given year   Y and day of year   N  return year  month  day         Astronomical Algorithms  Jean Meeus  2d ed  1998  chap 7             if is leap year Y           K   1     else          K   2     M   int  9    K   N     275 0   0 98      if N  lt  32          M   1     D   N - int  275   M    9 0    K   int  M   9    12 0    30     return Y  M  D

User · Answer

Just subtract january 1 from the date   import datetime today   datetime datetime now   day of year    today - datetime datetime today year  1  1   days   1

[python] Convert Year/Month/Day to Day of Year in Python

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