I have a dataframe with a date column. The duration is 365 days starting from 02/11/2017 and ending at 01/11/2018.
Date
02/11/2017
03/11/2017
05/11/2017
.
.
01/11/2018
I want to add an adjacent column called Day_Of_Year as follows:
Date Day_Of_Year
02/11/2017 1
03/11/2017 2
05/11/2017 4
.
.
01/11/2018 365
I apologize if it's a very basic question, but unfortunately I haven't been able to start with this.
I could use datetime(), but that would return values such as 1 for 1st january, 2 for 2nd january and so on.. irrespective of the year. So, that wouldn't work for me.
First convert column to_datetime and then subtract datetime, convert to days and add 1:
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')
df['Day_Of_Year'] = df['Date'].sub(pd.Timestamp('2017-11-02')).dt.days + 1
print (df)
Date Day_Of_Year
0 02/11/2017 1
1 03/11/2017 2
2 05/11/2017 4
3 01/11/2018 365
Or subtract by first value of column:
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')
df['Day_Of_Year'] = df['Date'].sub(df['Date'].iat[0]).dt.days + 1
print (df)
Date Day_Of_Year
0 2017-11-02 1
1 2017-11-03 2
2 2017-11-05 4
3 2018-11-01 365
Using strftime with '%j'
s=pd.to_datetime(df.Date,dayfirst=True).dt.strftime('%j').astype(int)
s-s.iloc[0]
Out[750]:
0 0
1 1
2 3
Name: Date, dtype: int32
#df['new']=s-s.iloc[0]
Python has dayofyear. So put your column in the right format with pd.to_datetime and then apply Series.dt.dayofyear. Lastly, use some modulo arithmetic to find everything in terms of your original date
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')
df['day of year'] = df['Date'].dt.dayofyear - df['Date'].dt.dayofyear[0] + 1
df['day of year'] = df['day of year'] + 365*((365 - df['day of year']) // 365)
Output
Date day of year
0 2017-11-02 1
1 2017-11-03 2
2 2017-11-05 4
3 2018-11-01 365
But I'm doing essentially the same as Jezrael in more lines of code, so my vote goes to her/him
Related
I have a data frame with a column month_year which is a string. Where the year is only a two digits. I would like to convert month_year column to a datetime column.
df:
month_year
Jan-98
Feb-98
Mar-99
Apr-99
May-99
Oct-00
Nov-00
Jun-01
Aug-03
Sep-08
Dec-21
Jul-22
Expected Output:
month_year
01-01-1998
01-02-1998
01-03-1999
01-04-1999
01-05-1999
01-10-2000
01-11-2000
01-06-2001
01-08-2003
01-09-2008
01-12-2021
01-07-2022
I have tried below code.
df['month_year'] = pd.to_datetime(df['month_year'])
and
df['month_year'] = pd.to_datetime(df['month_year']).dt.strftime('%m-%Y')
unfortunately both did not work.
You are close, need format with b for months first 3 letters and y for YY format of years:
df['month_year'] = pd.to_datetime(df['month_year'], format='%b-%y')
print (df)
month_year
0 1998-01-01
1 1998-02-01
2 1999-03-01
3 1999-04-01
4 1999-05-01
5 2000-10-01
6 2000-11-01
7 2001-06-01
8 2003-08-01
9 2008-09-01
10 2021-12-01
11 2022-07-01
I am using a csv with an accumulative number that changes daily.
Day Accumulative Number
0 9/1/2020 100
1 11/1/2020 102
2 18/1/2020 98
3 11/2/2020 105
4 24/2/2020 95
5 6/3/2020 120
6 13/3/2020 100
I am now trying to find the best way to aggregate it and compare the monthly results before a specific date. So, I want to check the balance on the 11th of each month but for some months, there is no activity for the specific day. As a result, I trying to get the latest day before the 12th of each Month. So, the above would be:
Day Accumulative Number
0 11/1/2020 102
1 11/2/2020 105
2 6/3/2020 120
What I managed to do so far is to just get the latest day of each month:
dateparse = lambda x: pd.datetime.strptime(x, "%d/%m/%Y")
df = pd.read_csv("Accumulative.csv",quotechar="'", usecols=["Day","Accumulative Number"], index_col=False, parse_dates=["Day"], date_parser=dateparse, na_values=['.', '??'] )
df.index = df['Day']
grouped = df.groupby(pd.Grouper(freq='M')).sum()
print (df.groupby(df.index.month).apply(lambda x: x.iloc[-1]))
which returns:
Day Accumulative Number
1 2020-01-18 98
2 2020-02-24 95
3 2020-03-13 100
Is there a way to achieve this in Pandas, Python or do I have to use SQL logic in my script? Is there an easier way I am missing out in order to get the "balance" as per the 11th day of each month?
You can do groupby with factorize
n = 12
df = df.sort_values('Day')
m = df.groupby(df.Day.dt.strftime('%Y-%m')).Day.transform(lambda x :x.factorize()[0])==n
df_sub = df[m].copy()
You can try filtering the dataframe where the days are less than 12 , then take last of each group(grouped by month) :
df['Day'] = pd.to_datetime(df['Day'],dayfirst=True)
(df[df['Day'].dt.day.lt(12)]
.groupby([df['Day'].dt.year,df['Day'].dt.month],sort=False).last()
.reset_index(drop=True))
Day Accumulative_Number
0 2020-01-11 102
1 2020-02-11 105
2 2020-03-06 120
I would try:
# convert to datetime type:
df['Day'] = pd.to_datetime(df['Day'], dayfirst=True)
# select day before the 12th
new_df = df[df['Day'].dt.day < 12]
# select the last day in each month
new_df.loc[~new_df['Day'].dt.to_period('M').duplicated(keep='last')]
Output:
Day Accumulative Number
1 2020-01-11 102
3 2020-02-11 105
5 2020-03-06 120
Here's another way using expanding the date range:
# set as datetime
df2['Day'] = pd.to_datetime(df2['Day'], dayfirst=True)
# set as index
df2 = df2.set_index('Day')
# make a list of all dates
dates = pd.date_range(start=df2.index.min(), end=df2.index.max(), freq='1D')
# add dates
df2 = df2.reindex(dates)
# replace NA with forward fill
df2['Number'] = df2['Number'].ffill()
# filter to get output
df2 = df2[df2.index.day == 11].reset_index().rename(columns={'index': 'Date'})
print(df2)
Date Number
0 2020-01-11 102.0
1 2020-02-11 105.0
2 2020-03-11 120.0
I am facing a situation where I need to group-by a dataframe by a column 'ID' and also calculate the total time frame depicted for that particular ID to complete. I only want to calculate the difference between the date_open and data_closed for the particular ID with the ID count.
We only need to focus on the date open and the date closed field. So it needs to do something taking the max closing date and the min open date and subtracting the two
The dataframe looks as follows:
ID Date_Open Date_Closed
1 01/01/2019 02/01/2019
1 07/01/2019 09/01/2019
2 10/01/2019 11/01/2019
2 13/01/2019 19/01/2019
3 10/01/2019 11/01/2019
The output should look like this :
ID Count_of_ID Total_Time_In_Days
1 2 8
2 2 9
3 1 1
How should I achieve this ?
Using GroupBy with named_aggregation and the min and max of the dates:
df[['Date_Open', 'Date_Closed']] = (
df[['Date_Open', 'Date_Closed']].apply(lambda x: pd.to_datetime(x, format='%d/%m/%Y'))
)
dfg = df.groupby('ID').agg(
Count_of_ID=('ID','size'),
Date_Open=('Date_Open','min'),
Date_Closed=('Date_Closed','max')
)
dfg['Total_Time_In_Days'] = dfg['Date_Closed'].sub(dfg['Date_Open']).dt.days
dfg = dfg.drop(columns=['Date_Closed', 'Date_Open']).reset_index()
ID Count_of_ID Total_Time_In_Days
0 1 2 8
1 2 2 9
2 3 1 1
Now we have Total_Time_In_Days as int:
print(dfg.dtypes)
ID int64
Count_of_ID int64
Total_Time_In_Days int64
dtype: object
This can also be used:
df['Date_Open'] = pd.to_datetime(df['Date_Open'], dayfirst=True)
df['Date_Closed'] = pd.to_datetime(df['Date_Closed'], dayfirst=True)
df_grouped = df.groupby(by='ID').count()
df_grouped['Total_Time_In_Days'] = df.groupby(by='ID')['Date_Closed'].max() - df.groupby(by='ID')['Date_Open'].min()
df_grouped = df_grouped.drop(columns=['Date_Open'])
df_grouped.columns=['Count', 'Total_Time_In_Days']
print(df_grouped)
Count Total_Time_In_Days
ID
1 2 8 days
2 2 9 days
3 1 1 days
I'll try first to create the a column depicting how much time passed from Date_open to Date_closed for each instance of the dataframe. Like this:
df['Total_Time_In_Days'] = df.Date_closed - df.Date_open
Then you can use groupby:
df.groupby('id').agg({'id':'count','Total_Time_In_Days':'sum'})
If you need any help with the .agg function you can refer to it's official documentation here.
i have data frame which contains fields casenumber , count and credated date .here created date is months which are in numerical i want to make dataframe as arrenge the ranges to the count acoording to createddate column
Here i used below code but i didnot match my requirement.i have data frame which contains fields casenumber , count and credated date .here created date is months which are in numerical i want to make dataframe as arrenge the ranges to the count acoording to createddate column
i have data frame as below
casenumber count CREATEDDATE
3820516 1 jan
3820547 1 jan
3820554 2 feb
3820562 1 feb
3820584 1 march
4226616 1 april
4226618 2 may
4226621 2 may
4226655 1 june
4226663 1 june
Here i used below code but i didnot match my requirement.i have data frame which contains fields casenumber , count and credated date .here created date is months which are in numerical i want to make dataframe as arrenge the ranges to the count acoording to createddate column
import pandas as pd
import numpy as np
df = pd.read_excel(r"")
bins = [0, 1 ,4,8,15, np.inf]
names = ['0-1','1-4','4-8','8-15','15+']
df1 = df.groupby(pd.cut(df['CREATEDDATE'],bins,labels=names))['casenumber'].size().reset_index(name='No_of_times_statuschanged')
CREATEDDATE No_of_times_statuschanged
0 0-1 2092
1 1-4 9062
2 4-8 12578
3 8-15 3858
4 15+ 0
I got the above data as out put but my expected should be range for month on month based on the cases per month .
expected output should be like
CREATEDDATE jan feb march april may june
0-1 1 2 3 4 5 6
1-4 3 0 6 7 8 9
4-8 4 6 3 0 9 2
8-15 0 3 4 5 8 9
I got the above data as out put but my expected should be range for month on month based on the cases per month .
expected output should be like
Use crosstab with change CREATEDDATE to count for pd.cut and change order of column by subset by list of columns names:
#add another months if necessary
months = ["jan", "feb", "march", "april", "may", "june"]
bins = [0, 1 ,4,8,15, np.inf]
names = ['0-1','1-4','4-8','8-15','15+']
df1 = pd.crosstab(pd.cut(df['count'],bins,labels=names), df['CREATEDDATE'])[months]
print (df1)
CREATEDDATE jan feb march april may june
count
0-1 2 1 1 1 0 2
1-4 0 1 0 0 2 0
Another idea is use ordered categoricals:
df1 = pd.crosstab(pd.cut(df['count'],bins,labels=names),
pd.Categorical(df['CREATEDDATE'], ordered=True, categories=months))
print (df1)
col_0 jan feb march april may june
count
0-1 2 1 1 1 0 2
1-4 0 1 0 0 2 0
I have a pandas dataframe that looks like this:
ID Date Event_Type
1 01/01/2019 A
1 01/01/2019 B
2 02/01/2019 A
3 02/01/2019 A
I want to be left with:
ID Date
1 01/01/2019
2 02/01/2019
3 02/01/2019
Where my condition is:
If the ID is the same AND the dates are within 2 days of each other then drop one of the rows.
If however the dates are more than 2 days apart then keep both rows.
How do I do this?
I believe you need first convert values to datetimes by to_datetime, then get diff and get first values per groups by isnull() chained with comparing if next values are higher like timedelta treshold:
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')
s = df.groupby('ID')['Date'].diff()
df = df[(s.isnull() | (s > pd.Timedelta(2, 'd')))]
print (df)
ID Date Event_Type
0 1 2019-01-01 A
2 2 2019-02-01 A
3 3 2019-02-01 A
Check solution with another data:
print (df)
ID Date Event_Type
0 1 01/01/2019 A
1 1 04/01/2019 B <-difference 3 days
2 2 02/01/2019 A
3 3 02/01/2019 A
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')
s = df.groupby('ID')['Date'].diff()
df = df[(s.isnull() | (s > pd.Timedelta(2, 'd')))]
print (df)
ID Date Event_Type
0 1 2019-01-01 A
1 1 2019-01-04 B
2 2 2019-01-02 A
3 3 2019-01-02 A