Getting start date of week from week number - pandas

I have a list like this lst = [25,26,27]
numbers: 25, 26, 27 are the week number.
For each number from list I would like to have a start date, e.x. for week 25 the start date is 2022-06-21 (week starts on Monday).
Please, can someone help?
I have tried sth like
import datetime
from dateutil.relativedelta import relativedelta
week = 25
year = 2021
date = datetime.date(year, 1, 1) + relativedelta(weeks=+week)
print(date)
but it doesn;t work.

IIUC, you can use to_datetime:
lst = [25,26,27]
year = 2021
out = pd.to_datetime(pd.Series(lst).astype(str)+str(year)+'Mon', format='%W%Y%a')
output:
0 2021-06-21
1 2021-06-28
2 2021-07-05
dtype: datetime64[ns]
intermediate:
pd.Series(lst).astype(str)+str(year)+'Mon'
0 252021Mon
1 262021Mon
2 272021Mon
dtype: object

Related

Pandas rolling window cumsum

I have a pandas df as follows:
YEAR MONTH USERID TRX_COUNT
2020 1 1 1
2020 2 1 2
2020 3 1 1
2020 12 1 1
2021 1 1 3
2021 2 1 3
2021 3 1 4
I want to sum the TRX_COUNT such that, each TRX_COUNT is the sum of TRX_COUNTS of the next 12 months.
So my end result would look like
YEAR MONTH USERID TRX_COUNT TRX_COUNT_SUM
2020 1 1 1 5
2020 2 1 2 7
2020 3 1 1 8
2020 12 1 1 11
2021 1 1 3 10
2021 2 1 3 7
2021 3 1 4 4
For example TRX_COUNT_SUM for 2020/1 is 1+2+1+1=5 the count of the first 12 months.
Two areas I am confused how to proceed:
I tried various variations of cumsum and grouping by USERID, YR, MONTH but am running into errors with handling the time window as there might be MONTHS where a user has no transactions and these have to be accounted for. For example in 2020/1 the user has no transactions for months 4-11, hence a full year of transaction count would be 5.
Towards the end there will be partial years, which can be summed up and left as is (like 2021/3 which is left as 4).
Any thoughts on how to handle this?
Thanks!
I was able to accomplish this using a combination of numpy arrays, pandas, and indexing
import pandas as pd
import numpy as np
#df = your dataframe
df_dates = pd.DataFrame(np.arange(np.datetime64('2020-01-01'), np.datetime64('2021-04-01'), np.timedelta64(1, 'M'), dtype='datetime64[M]').astype('datetime64[D]'), columns = ['DATE'])
df_dates['YEAR'] = df_dates['DATE'].apply(lambda x : str(x).split('-')[0]).apply(lambda x : int(x))
df_dates['MONTH'] = df_dates['DATE'].apply(lambda x : str(x).split('-')[1]).apply(lambda x : int(x))
df_merge = df_dates.merge(df, how = 'left')
df_merge.replace(np.nan, 0, inplace=True)
df_merge.reset_index(inplace = True)
for i in range(0, len(df_merge)):
max_index = df_merge['index'].max()
if(i + 11 < max_index):
df_merge.at[i, 'TRX_COUNT_SUM'] = df_merge.iloc[i:i + 12]['TRX_COUNT'].sum()
elif(i != max_index):
df_merge.at[i, 'TRX_COUNT_SUM'] = df_merge.iloc[i:max_index + 1]['TRX_COUNT'].sum()
else:
df_merge.at[i, 'TRX_COUNT_SUM'] = df_merge.iloc[i]['TRX_COUNT']
final_df = pd.merge(df_merge, df)
Try this:
# Set the Dataframe index to a time series constructed from YEAR and MONTH
ts = pd.to_datetime(df.assign(DAY=1)[["YEAR", "MONTH", "DAY"]])
df.set_index(ts, inplace=True)
df["TRX_COUNT_SUM"] = (
# Reindex the dataframe with every missing month in-between
# Also reverse the index so that rolling(12) means 12 months
# forward instead of backward
df.reindex(pd.date_range(ts.min(), ts.max(), freq="MS")[::-1])
# Roll and sum
.rolling(12, min_periods=1)
["TRX_COUNT"].sum()
)

how to create monthly and season 24 hours average table using pandas

I have a dataframe with 2 columns: Date and LMP and there are totals of 8760 rows. This is the dummy dataframe:
import pandas as pd
import numpy as np
df = pd.DataFrame({'Date': pd.date_range('2023-01-01 00:00', '2023-12-31 23:00', freq='1H'), 'LMP': np.random.randint(10, 20, 8760)})
I extract month from the date and then created the season column for the specific dates. Like this
df['month'] = pd.DatetimeIndex(df['Date']).month
season = []
for i in df['month']:
if i <= 2 or i == 12:
season.append('Winter')
elif 2 < i <= 5:
season.append('Spring')
elif 5 < i <= 8:
season.append('Summer')
else:
season.append('Autumn')
df['Season'] = season
df2 = df.groupby(['month']).mean()
df3 = df.groupby(['Season']).mean()
print(df2['LMP'])
print(df3['LMP'])
Output:
**month**
1 20.655113
2 20.885532
3 19.416946
4 22.025248
5 26.040606
6 19.323863
7 51.117965
8 51.434093
9 21.404680
10 14.701989
11 20.009590
12 38.706160
**Season**
Autumn 18.661426
Spring 22.499365
Summer 40.856845
Winter 26.944382
But I want the output to be in 24 hour average for both monthly and seasonal.
Desired Output:
for seasonal 24 hours average
For monthyl 24 hours average
Note: in the monthyl 24 hour average columns are months(1,2,3,4,5,6,7,8,9,10,11,12) and rows are hours(starting from 0).
Can anyone help?
try:
df['hour']=pd.DatetimeIndex(df['Date']).hour
dft = df[['Season', 'hour', 'LMP']]
dftg = dft.groupby(['hour', 'Season'])['LMP'].mean()
dftg.reset_index().pivot(index='hour', columns='Season')
result:

Pandas 1.0 create column of months from year and date

I have a dataframe df with values as:
df.iloc[1:4, 7:9]
Year Month
38 2020 4
65 2021 4
92 2022 4
I am trying to create a new MonthIdx column as:
df['MonthIdx'] = pd.to_timedelta(df['Year'], unit='Y') + pd.to_timedelta(df['Month'], unit='M') + pd.to_timedelta(1, unit='D')
But I get the error:
ValueError: Units 'M' and 'Y' are no longer supported, as they do not represent unambiguous timedelta values durations.
Following is the desired output:
df['MonthIdx']
MonthIdx
38 2020/04/01
65 2021/04/01
92 2022/04/01
So you can pad the month value in a series, and then reformat to get a datetime for all of the values:
month = df.Month.astype(str).str.pad(width=2, side='left', fillchar='0')
df['MonthIdx'] = pd.to_datetime(pd.Series([int('%d%s' % (x,y)) for x,y in zip(df['Year'],month)]),format='%Y%m')
This will give you:
Year Month MonthIdx
0 2020 4 2020-04-01
1 2021 4 2021-04-01
2 2022 4 2022-04-01
You can reformat the date to be a string to match exactly your format:
df['MonthIdx'] = df['MonthIdx'].apply(lambda x: x.strftime('%Y/%m/%d'))
Giving you:
Year Month MonthIdx
0 2020 4 2020/04/01
1 2021 4 2021/04/01
2 2022 4 2022/04/01

pandas to_datetime convert 6PM to 18

is there a nice way to convert Series data, represented like 1PM or 11AM to 13 and 11 accordingly with to_datetime or similar (other, than re)
data:
series
1PM
11AM
2PM
6PM
6AM
desired output:
series
13
11
14
18
6
pd.to_datetime(df['series']) gives the following error:
OutOfBoundsDatetime: Out of bounds nanosecond timestamp: 1-01-01 11:00:00
You can provide the format you want to use, with as format %I%p:
pd.to_datetime(df['series'], format='%I%p').dt.hour
The .dt.hour [pandas-doc] will thus obtain the hour for that timestamp. This gives us:
>>> df = pd.DataFrame({'series': ['1PM', '11AM', '2PM', '6PM', '6AM']})
>>> pd.to_datetime(df['series'], format='%I%p').dt.hour
0 13
1 11
2 14
3 18
4 6
Name: series, dtype: int64

week number from given date in pandas

I have a data frame with two columns Date and value.
I want to add new column named week_number that basically is how many weeks back from the given date
import pandas as pd
df = pd.DataFrame(columns=['Date','value'])
df['Date'] = [ '04-02-2019','03-02-2019','28-01-2019','20-01-2019']
df['value'] = [10,20,30,40]
df
Date value
0 04-02-2019 10
1 03-02-2019 20
2 28-01-2019 30
3 20-01-2019 40
suppose given date is 05-02-2019.
Then I need to add a column week_number in a way such that how many weeks back the Date column date is from given date.
The output should be
Date value week_number
0 04-02-2019 10 1
1 03-02-2019 20 1
2 28-01-2019 30 2
3 20-01-2019 40 3
how can I do this in pandas
First convert column to datetimes by to_datetime with dayfirst=True, then subtract from right side by rsub, convert timedeltas to days, get modulo by 7 and add 1:
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
df['week_number'] = df['Date'].rsub(pd.Timestamp('2019-02-05')).dt.days // 7 + 1
#alternative
#df['week_number'] = (pd.Timestamp('2019-02-05') - df['Date']).dt.days // 7 + 1
print (df)
Date value week_number
0 2019-02-04 10 1
1 2019-02-03 20 1
2 2019-01-28 30 2
3 2019-01-20 40 3