Week of a month pandas

Question

I\'m trying to get week on a month, some months might have four weeks some might have five. For each date i would like to know to which week does it belongs to. I\'m mostly

Answer 1

I've used the code below when dealing with dataframes that have a datetime index.

import pandas as pd
import math

def add_week_of_month(df):
    df['week_in_month'] = pd.to_numeric(df.index.day/7)
    df['week_in_month'] = df['week_in_month'].apply(lambda x: math.ceil(x))
    return df

If you run this example:

df = test = pd.DataFrame({'count':['a','b','c','d','e']},
                     index = ['2018-01-01', '2018-01-08','2018-01-31','2018-02-01','2018-02-28'])
df.index = pd.to_datetime(df.index)

you should get the following dataframe

               count  week_in_month

2018-01-01     a              1
2018-01-08     b              2
2018-01-31     c              5
2018-02-01     d              1
2018-02-28     e              4

Answer 2

See this answer and decide which week of month you want.

There's nothing built-in, so you'll need to calculate it with apply. For example, for an easy 'how many 7 day periods have passed' measure.

data['wom'] = data[0].apply(lambda d: (d.day-1) // 7 + 1)

For a more complicated (based on the calender), using the function from that answer.

import datetime
import calendar

def week_of_month(tgtdate):
    tgtdate = tgtdate.to_datetime()

    days_this_month = calendar.mdays[tgtdate.month]
    for i in range(1, days_this_month):
        d = datetime.datetime(tgtdate.year, tgtdate.month, i)
        if d.day - d.weekday() > 0:
            startdate = d
            break
    # now we canuse the modulo 7 appraoch
    return (tgtdate - startdate).days //7 + 1

data['calendar_wom'] = data[0].apply(week_of_month)

Answer 3

You can get it subtracting the current week and the week of the first day of the month, but extra logic is needed to handle first and last week of the year:

def get_week(s):
    prev_week = (s - pd.to_timedelta(7, unit='d')).dt.week
    return (
        s.dt.week
        .where((s.dt.month != 1) | (s.dt.week < 50), 0)
        .where((s.dt.month != 12) | (s.dt.week > 1), prev_week + 1)
    )

def get_week_of_month(s):
    first_day_of_month = s - pd.to_timedelta(s.dt.day - 1, unit='d')
    first_week_of_month = get_week(first_day_of_month)
    current_week = get_week(s)
    return  current_week - first_week_of_month

Answer 4

TL;DR

import pandas as pd

def weekinmonth(dates):
    """Get week number in a month.
    
    Parameters: 
        dates (pd.Series): Series of dates.
    Returns: 
        pd.Series: Week number in a month.
    """
    firstday_in_month = dates - pd.to_timedelta(dates.dt.day - 1, unit='d')
    return (dates.dt.day-1 + firstday_in_month.dt.weekday) // 7 + 1
    
    
df = pd.DataFrame(pd.date_range(' 1/ 1/ 2000', periods = 100, freq ='D'), columns=['Date'])
weekinmonth(df['Date'])

0     1
1     1
2     2
3     2
4     2
     ..
95    2
96    2
97    2
98    2
99    2
Name: Date, Length: 100, dtype: int64

Explanation

At first, calculate first day in month (from this answer: How floor a date to the first date of that month?):

df = pd.DataFrame(pd.date_range(' 1/ 1/ 2000', periods = 100, freq ='D'), columns=['Date'])
df['MonthFirstDay'] = df['Date'] - pd.to_timedelta(df['Date'].dt.day - 1, unit='d')
df

         Date MonthFirstDay
0  2000-01-01    2000-01-01
1  2000-01-02    2000-01-01
2  2000-01-03    2000-01-01
3  2000-01-04    2000-01-01
4  2000-01-05    2000-01-01
..        ...           ...
95 2000-04-05    2000-04-01
96 2000-04-06    2000-04-01
97 2000-04-07    2000-04-01
98 2000-04-08    2000-04-01
99 2000-04-09    2000-04-01

[100 rows x 2 columns]

Obtain weekday from first day:

df['FirstWeekday'] = df['MonthFirstDay'].dt.weekday
df

         Date MonthFirstDay  FirstWeekday
0  2000-01-01    2000-01-01             5
1  2000-01-02    2000-01-01             5
2  2000-01-03    2000-01-01             5
3  2000-01-04    2000-01-01             5
4  2000-01-05    2000-01-01             5
..        ...           ...           ...
95 2000-04-05    2000-04-01             5
96 2000-04-06    2000-04-01             5
97 2000-04-07    2000-04-01             5
98 2000-04-08    2000-04-01             5
99 2000-04-09    2000-04-01             5

[100 rows x 3 columns]

Now I can calculate with modulo of weekdays to obtain the week number in a month:

1. Get day of the month by df['Date'].dt.day and make sure that begins with 0 due to modulo calculation df['Date'].dt.day-1.
2. Add weekday number to make sure which day of month starts + df['FirstWeekday']
3. Be safe to use the integer division of 7 days in a week and add 1 to start week number in month from 1 // 7 + 1.

Whole modulo calculation:

df['WeekInMonth'] = (df['Date'].dt.day-1 + df['FirstWeekday']) // 7 + 1
df

         Date MonthFirstDay  FirstWeekday  WeekInMonth
0  2000-01-01    2000-01-01             5            1
1  2000-01-02    2000-01-01             5            1
2  2000-01-03    2000-01-01             5            2
3  2000-01-04    2000-01-01             5            2
4  2000-01-05    2000-01-01             5            2
..        ...           ...           ...          ...
95 2000-04-05    2000-04-01             5            2
96 2000-04-06    2000-04-01             5            2
97 2000-04-07    2000-04-01             5            2
98 2000-04-08    2000-04-01             5            2
99 2000-04-09    2000-04-01             5            2

[100 rows x 4 columns]

Answer 5

This seems to do the trick for me

df_dates = pd.DataFrame({'date':pd.bdate_range(df['date'].min(),df['date'].max())})
df_dates_tues = df_dates[df_dates['date'].dt.weekday==2].copy()
df_dates_tues['week']=np.mod(df_dates_tues['date'].dt.strftime('%W').astype(int),4)