I want to add a blank column of date of format "%Y-%m-%d" to a dataframe. I tried datetime.datetime.strptime('0000-00-00',"%Y-%m-%d")
But I get an error ValueError: time data '0000-00-00' does not match format '%Y-%m-%d'
How can I create a column of blank date of format "%Y-%m-%d"?
In R following works.
df$date =""
class(df$date) = "Date"
How can I achieve this in Python?
Thank you.
I don't think that's possible with datetime module. The oldest you can go to is answered here:
What is the oldest time that can be represented in Python?
datetime.MINYEAR
The smallest year number allowed in a date or datetime object. MINYEAR is 1.
datetime.MAXYEAR
The largest year number allowed in a date or datetime object. MAXYEAR is 9999.
source: datetime documentation
initial_date = request.GET.get('data') or datetime.min # datetime.min is 1
end_date = request.GET.get('data_f') or datetime.max # datetime.max is 9999
Related
I have a df with dates in a column converted to a datetime. the current format is YYYYDDMM. I need this converted to YYYYMMDD. I tried the below code but it does not change the format and still gives me YYYYDDMM. the end goal is to subtract 1 business day from the effective date but the format needs to be in YYYYMMDD to do this otherwise it subtracts 1 day from the M and not D. can someone help?
filtered_df['Effective Date'] = pd.to_datetime(filtered_df['Effective Date'])
# Effective Date = 20220408 (4th Aug 2022 for clarity)
filtered_df['Effective Date new'] = filtered_df['Effective Date'].dt.strftime("%Y%m%d")
# Effective Date new = 20220408
desired output -- > Effective Date new = 20220804
By default, .to_datetime will interpret the input YYYYDDMM as YYYYMMDD, and therefore print the same thing with %Y%m%d as the format. You can fix this and make it properly parse days in the month greater than 12 by adding the dayfirst keyword argument.
filtered_df['Effective Date'] = pd.to_datetime(filtered_df['Effective Date'], dayfirst=True)
I like to use the datetime library for this purpose. You can use strptime to convert a string into the datetime object and strftime to convert your datetime object to the new string.
from datetime import datetime
def change_date(row):
row["Effective Date new"] = datetime.strptime(row["Effective Date"], "%Y%d%m").strftime("%Y%m%d")
return row
df2 = df.apply(change_date, axis=1)
The output df2 will have Effective Date new as your new column.
Using a Python script, I need to read a CVS file where dates are formated as DD/MM/YYYY, and convert them to YYYY-MM-DD before saving this into a SQLite database.
This almost works, but fails because I don't provide time:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%Y-%m-%d")
#ValueError: time data did not match format: data=21/12/2008 fmt=%Y-%m-%d
print lastconnection
I assume there's a method in the datetime object to perform this conversion very easily, but I can't find an example of how to do it. Thank you.
Your example code is wrong. This works:
import datetime
datetime.datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
The call to strptime() parses the first argument according to the format specified in the second, so those two need to match. Then you can call strftime() to format the result into the desired final format.
you first would need to convert string into datetime tuple, and then convert that datetime tuple to string, it would go like this:
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime('%Y-%m-%d')
I am new to programming. I wanted to convert from yyyy-mm-dd to dd/mm/yyyy to print out a date in the format that people in my part of the world use and recognise.
The accepted answer above got me on the right track.
The answer I ended up with to my problem is:
import datetime
today_date = datetime.date.today()
print(today_date)
new_today_date = today_date.strftime("%d/%m/%Y")
print (new_today_date)
The first two lines after the import statement gives today's date in the USA format (2017-01-26). The last two lines convert this to the format recognised in the UK and other countries (26/01/2017).
You can shorten this code, but I left it as is because it is helpful to me as a beginner. I hope this helps other beginner programmers starting out!
Does anyone else else think it's a waste to convert these strings to date/time objects for what is, in the end, a simple text transformation? If you're certain the incoming dates will be valid, you can just use:
>>> ddmmyyyy = "21/12/2008"
>>> yyyymmdd = ddmmyyyy[6:] + "-" + ddmmyyyy[3:5] + "-" + ddmmyyyy[:2]
>>> yyyymmdd
'2008-12-21'
This will almost certainly be faster than the conversion to and from a date.
#case_date= 03/31/2020
#Above is the value stored in case_date in format(mm/dd/yyyy )
demo=case_date.split("/")
new_case_date = demo[1]+"-"+demo[0]+"-"+demo[2]
#new format of date is (dd/mm/yyyy) test by printing it
print(new_case_date)
If you need to convert an entire column (from pandas DataFrame), first convert it (pandas Series) to the datetime format using to_datetime and then use .dt.strftime:
def conv_dates_series(df, col, old_date_format, new_date_format):
df[col] = pd.to_datetime(df[col], format=old_date_format).dt.strftime(new_date_format)
return df
Sample usage:
import pandas as pd
test_df = pd.DataFrame({"Dates": ["1900-01-01", "1999-12-31"]})
old_date_format='%Y-%m-%d'
new_date_format='%d/%m/%Y'
conv_dates_series(test_df, "Dates", old_date_format, new_date_format)
Dates
0 01/01/1900
1 31/12/1999
The most simplest way
While reading the csv file, put an argument parse_dates
df = pd.read_csv("sample.csv", parse_dates=['column_name'])
This will convert the dates of mentioned column to YYYY-MM-DD format
Convert date format DD/MM/YYYY to YYYY-MM-DD according to your question, you can use this:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
print(lastconnection)
df is your data frame
Dateclm is the column that you want to change
This column should be in DateTime datatype.
df['Dateclm'] = pd.to_datetime(df['Dateclm'])
df.dtypes
#Here is the solution to change the format of the column
df["Dateclm"] = pd.to_datetime(df["Dateclm"]).dt.strftime('%Y-%m-%d')
print(df)
I know I should import datetime to have actual date. But the rest is black magic for me right now.
ex.
dates = ['2019-010-11', '2013-05-16', '2011-06-16', '2000-04-22']
actual_date = datetime.datetime.now()
How can I subtract this and as a result have new list with days that passed by from dates to actual_date?
If I'm understanding correctly, you need to find the current date, and then find the number of days between the current date and the dates in your list?
If so, you could try this:
from datetime import datetime, date
dates = ['2019-10-11', '2013-05-16', '2011-06-16', '2000-04-22']
actual_date = date.today()
days = []
for date in dates:
date_object = datetime.strptime(date, '%Y-%m-%d').date()
days_difference = (actual_date - date_object).days
days.append(days_difference)
print(days)
What I am doing here is:
Converting the individual date strings to a "date" object
Subtracting the this date from the actual date. This gets you the time as well, so to strip that out we add .days.
Save the outcome to a list, although of course you could do whatever you wanted with the output.
I am extracting the date from a filename and storing it in a string variable. Suppose the filename is CRM_DATA_PUBLIC_20201120_052035.txt, I have extracted the date as 20201120. Now I want to get the previous month's date from this, like 20201020 or just 202010.
I tried using date functions but it is giving error for me.
Could you please help me out here ?
Thanks in anticipation.
Try this: (changes based on a comment)
import datetime
from dateutil.relativedelta import relativedelta
filename = 'CRM_DATA_PUBLIC_20201120_052035.txt'
date = filename.split('_')[3]
#If you want the output to include the day of month as well
date = datetime.datetime.strptime(date, '%Y%m%d')
#If you want only the month
date = datetime.datetime.strptime(date, '%Y%m')
date = date - relativedelta(months=1)
date = str(date.date()).replace('-','')
print(date)
Output:
20201020
You can find your answer here https://stackoverflow.com/a/9725093/10334833
What I get from your question is already answered here but if you are still confused let me know I can help you :)
I've been looking through every thread that I can find, and the only one that is relevant to this type of formatting issue is here, but it's for java...
How parse 2013-03-13T20:59:31+0000 date string to Date
I've got a column with values like 201604 and 201605 that I need to turn into date values like 2016-04-01 and 2016-05-01. To accomplish this, I've done what is below.
#Create Number to build full date
df['DAY_NBR'] = '01'
#Convert Max and Min date to string to do date transformation
df['MAXDT'] = df['MAXDT'].astype(str)
df['MINDT'] = df['MINDT'].astype(str)
#Add the day number to the max date month and year
df['MAXDT'] = df['MAXDT'] + df['DAY_NBR']
#Add the day number to the min date month and year
df['MINDT'] = df['MINDT'] + df['DAY_NBR']
#Convert Max and Min date to integer values
df['MAXDT'] = df['MAXDT'].astype(int)
df['MINDT'] = df['MINDT'].astype(int)
#Convert Max date to datetime
df['MAXDT'] = pd.to_datetime(df['MAXDT'], format='%Y%m%d')
#Convert Min date to datetime
df['MINDT'] = pd.to_datetime(df['MINDT'], format='%Y%m%d')
To be honest, I can work with this output, but it's a little messy because the unique values for the two columns are...
MAXDT Values
['2016-07-01T00:00:00.000000000' '2017-09-01T00:00:00.000000000'
'2018-06-01T00:00:00.000000000' '2017-07-01T00:00:00.000000000'
'2017-03-01T00:00:00.000000000' '2018-12-01T00:00:00.000000000'
'2017-12-01T00:00:00.000000000' '2019-01-01T00:00:00.000000000'
'2018-09-01T00:00:00.000000000' '2018-10-01T00:00:00.000000000'
'2016-04-01T00:00:00.000000000' '2018-03-01T00:00:00.000000000'
'2017-05-01T00:00:00.000000000' '2018-08-01T00:00:00.000000000'
'2017-02-01T00:00:00.000000000' '2016-12-01T00:00:00.000000000'
'2018-01-01T00:00:00.000000000' '2018-02-01T00:00:00.000000000'
'2017-06-01T00:00:00.000000000' '2018-11-01T00:00:00.000000000'
'2018-05-01T00:00:00.000000000' '2019-11-01T00:00:00.000000000'
'2016-06-01T00:00:00.000000000' '2017-10-01T00:00:00.000000000'
'2016-08-01T00:00:00.000000000' '2018-04-01T00:00:00.000000000'
'2016-03-01T00:00:00.000000000' '2016-10-01T00:00:00.000000000'
'2016-11-01T00:00:00.000000000' '2019-12-01T00:00:00.000000000'
'2016-09-01T00:00:00.000000000' '2017-08-01T00:00:00.000000000'
'2016-05-01T00:00:00.000000000' '2017-01-01T00:00:00.000000000'
'2017-11-01T00:00:00.000000000' '2018-07-01T00:00:00.000000000'
'2017-04-01T00:00:00.000000000' '2016-01-01T00:00:00.000000000'
'2016-02-01T00:00:00.000000000' '2019-02-01T00:00:00.000000000'
'2019-07-01T00:00:00.000000000' '2019-10-01T00:00:00.000000000'
'2019-09-01T00:00:00.000000000' '2019-03-01T00:00:00.000000000'
'2019-05-01T00:00:00.000000000' '2019-04-01T00:00:00.000000000'
'2019-08-01T00:00:00.000000000' '2019-06-01T00:00:00.000000000'
'2020-02-01T00:00:00.000000000' '2020-01-01T00:00:00.000000000']
MINDT Values
['2016-04-01T00:00:00.000000000' '2017-07-01T00:00:00.000000000'
'2016-02-01T00:00:00.000000000' '2017-01-01T00:00:00.000000000'
'2017-02-01T00:00:00.000000000' '2018-12-01T00:00:00.000000000'
'2017-08-01T00:00:00.000000000' '2018-04-01T00:00:00.000000000'
'2017-10-01T00:00:00.000000000' '2019-01-01T00:00:00.000000000'
'2018-05-01T00:00:00.000000000' '2018-09-01T00:00:00.000000000'
'2018-10-01T00:00:00.000000000' '2016-01-01T00:00:00.000000000'
'2016-03-01T00:00:00.000000000' '2017-11-01T00:00:00.000000000'
'2017-05-01T00:00:00.000000000' '2018-07-01T00:00:00.000000000'
'2018-06-01T00:00:00.000000000' '2017-12-01T00:00:00.000000000'
'2016-10-01T00:00:00.000000000' '2018-02-01T00:00:00.000000000'
'2017-06-01T00:00:00.000000000' '2018-08-01T00:00:00.000000000'
'2018-03-01T00:00:00.000000000' '2018-11-01T00:00:00.000000000'
'2016-08-01T00:00:00.000000000' '2016-06-01T00:00:00.000000000'
'2018-01-01T00:00:00.000000000' '2016-07-01T00:00:00.000000000'
'2016-11-01T00:00:00.000000000' '2016-09-01T00:00:00.000000000'
'2017-04-01T00:00:00.000000000' '2016-05-01T00:00:00.000000000'
'2017-09-01T00:00:00.000000000' '2016-12-01T00:00:00.000000000'
'2017-03-01T00:00:00.000000000']
I'm trying to build a loop that runs through these dates, and it works, but I don't want to have an index with all of these irrelevant zeros and a T in it. How can I convert these empty timestamp values to just the date that is in yyyy-mm-dd format?
Thank you!
Unfortunately, I believe Pandas always stores datetime objects as datetime64[ns], meaning the precision has to be like that. Even if you attempt to save as datetime64[D], it will be cast to datetime64[ns].
It's possible to just store these datetime objects as strings instead, but the simplest solution is likely to just strip the extra zeroes when you're looping through them (i.e, using df['MAXDT'].to_numpy().astype('datetime64[D]') and looping through the formatted numpy array), or just reformatting using datetime.