Finding day of the week for a datetime64

user_9112585 0
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Apologies if this is an easy one but I can't see anything in the numpy documentation. I have a datetime64 and I'd like to find out which day of the week it is.

Unlike python datetime, datetime64 doesn't seem to have a .weekday() function. However it does have a busday_offset() function, which implies it must know behind-the-scenes what day of the week a given date is. Any help greatly appreciated.

numpy datetime python
5 Answers

Please look at the following related SO question:

Get weekday/day-of-week for Datetime column of DataFrame

I understand that your data is a numpy array, it may be useful to convert your data into a dataframe format via pandas. This is shown below.

import pandas as pd
df = pd.DataFrame(numpyArray)

Provided your data is now in a dataframe, then the below code (taken from the above link), should provide you with the weekday you are after.

df['weekday'] = df['Timestamp'].dt.dayofweek

Sample Output

Additionally this reference may also assist you.

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.dayofweek.html

user_45446990

I think you could do in this way:

import numpy as np
import datetime
t = '2018-09-19'
t = np.datetime64(t)
day = t.astype(datetime.datetime).isoweekday()
print day

Output:

3
user_51789050

If you're looking for a genuinely vectorized numpy approach, you could do:

def day_of_week_num(dts):
    return (dts.astype('datetime64[D]').view('int64') - 4) % 7

It's a bit hacky, and just takes advantage of the fact that numpy.datetime64s are relative to the unix epoch, which was a Thursday.

(I don't know if this is an implementational detail that could change without notice, but you can always check with assert np.zeros(1).astype('datetime64[D]') == np.datetime64('1970-01-01', 'D').)

user_32917440

Something like this could help. It takes a pandas series or data frame column as an input and returns a dictionary having the desired attributes for all unique DATETIME values in the target series/column.

def compute_dateparts(fld):
  fld_dtype = type(fld[0])
  fld = fld.unique()
  if isinstance(fld_dtype, pd.core.dtypes.dtypes.DatetimeTZDtype):
    fld_dtype = np.datetime64
  if not np.issubdtype(fld_dtype, np.datetime64):
    fld = pd.to_datetime(fld, infer_datetime_format=True)

  out = dict()
  for attr in ['Dayofweek', 'Is_month_end', 'Is_month_start','Is_quarter_end', 'Is_quarter_start', 'Is_year_end', 'Is_year_start']:
    out[attr] = pd.Series(getattr(fld, attr.lower()),index=fld).to_dict()

  return out
user_123370850

Here is a simple and practial way to get the day of the week and the start and end date of any week in a year, directly using np.datetime64 and np.timedelta64

def dayofweek(d):
    """Return dayofweek (0..7)."""
    d = np.datetime64(d) # d can be string or np.datetime64
    return  int((d - np.datetime64('1969-12-29', 'D')) / np.timedelta64(1, 'D') % 7)

def week(yr, wk):
    """return start and end date of given week in given year."""
    if wk > 53:
        yr, wk = wk, yr # in case you forgot order of yr and wk
    d = np.datetime64(str(yr), 'D') + (wk - 1) * np.timedelta64(7, 'D')
    week_start = d - dayofweek(d)
    week_end = week_start + np.timedelta64(6, 'D')
    return week_start, week_end
user_44022670
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