By : User1991
Date : January 12 2021, 08:33 AM
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seems to work fine I have a data frame where one of the columns is Date (i.e. 5/29/2007) and then there is col A and col B and a few other columns. The date and the value of col A and col B uniquely identify the row. I want to add columns to the data frame which are the historical values of other columns (not A or B). For example I have some row. Call it row i and it has some value for col D. And I want to add a column to the data frame which corresponds to the value of col D but for a previous day. That is the value of col D where the row, call it row j, has the same values for col A and col B as row i but the date is the date of row i minus one day. How can I do this efficiently? , I would create a shifted copy, then merge with the original.  code :
# shifted copy
shiftInterval = pd.Timedelta('-1 days') # your X days interval
dfShifted = df.copy()
dfShifted['Date'] = dfShifted['Date'] - shiftInterval
dfShifted.drop(['col-C'], axis=1, inplace=True)
# merge, keeping only observations where -1 lag is present
df2 = pd.merge(df,
dfShifted,
on=['Date','col-A','col-B'],
how = 'inner', # use 'left' to keep observations without lags
suffixes=['','-1'])
df2
# Date col-A col-B col-C col-D col-D-1
#0 2007-05-30 A B 10 1 1
#1 2007-05-30 AA Bf 3 1 124
#2 2007-05-30 AV Bf 1 8 4
from io import StringIO
import pandas as pd
from datetime import datetime
df = pd.read_csv(StringIO("""Date,col-A,col-B,col-C,col-D
5/29/2007,A,B,0,1
5/29/2007,AA,Bf,7,124
5/29/2007,AV,Bf,1,4
5/30/2007,A,B,10,1
5/30/2007,AA,Bf,3,1
5/30/2007,AV,Bf,1,8"""))
df['Date'] = df['Date'].apply(lambda x: datetime.strptime(x, '%m/%d/%Y'))
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 VB.Net 2005, how can I subtract a date from the (current date minus 7 days) and produce a number of days?
By : k r saini
Date : March 29 2020, 07:55 AM
With these it helps I have a date in the future e.g. 13/10/2008 I need to subtract the current date (today is the 28/09/2010) minus 7 days, so thats 21/09/2010 minus 13/10/2008, which would equal erm, 720 something ? code :
Sub Main()
Dim dt As DateTime = New DateTime(2008, 10, 13)
' be careful what you are subtracting from what
' the date you have is not in the future (year 2008)
' if the date is in the future: (dt.Subtract(DateTime.Now.AddDays(-7))).TotalDays
' or simply take the absolute value
Dim days As Double = (DateTime.Now.AddDays(-7).Subtract(dt)).TotalDays
Console.WriteLine(days)
End Sub
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 Android Date Picker - Actual Date minus X number of days
By : Sonia
Date : March 29 2020, 07:55 AM
To fix this issue I can get the current date with the following code to show in DatePicker dialog for example. , Try using: code :
c.add(Calendar.YEAR, -18);
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 Bash Shell Current Date Minus Number of Days
By : Andreas Kitzing
Date : March 29 2020, 07:55 AM
Any of those help I am new to bash and shell but I am running a debian install and I am trying to make a script which can find a date in the past without having to install any additional packages. From tutorials I have got to this stage: , Try doing this : code :
#!/bin/sh
#
# BACKUP DB TO S3
#
# VARIABLES
TYPE="DATABASE"
DAYS="30"
# GET CURRENT DATETIME
CURRENTDATE="$(date +%Y%m%d%H%M%S)"
# GENERATE PAST DATE FROM DAYS CONSTANT
OLDERDATE="$(date "+%Y%m%d%H%M%S" -d "$DAYS days ago")"
# CALL PYTHON SCRIPT WITH OLDERDATE ARGUMENT
python script.py "$OLDERDATE"
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 How to calculate if the difference from current date and another date is equal or minus of a number of days
By : user3147456
Date : March 29 2020, 07:55 AM
will help you I use this query to calculate if the current date minus another date is minus or equal an amount of days. Unfortunately this query only works if the calculation of days is inside the same month. Let's take for example these datas: , This is the right code for my goal: code :
SELECT * FROM condominio WHERE TIMESTAMPDIFF(DAY,NOW(), DATE_ADD(revisione, INTERVAL 1 YEAR)) <= 30;
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 Get a new Date based on today's date minus or plus a specific number of days in one line
By : user3480169
Date : March 29 2020, 07:55 AM
wish of those help Simple question: , Try this: code :
console.log(new Date(new Date().setDate(new Date().getDate() - 90)));
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