I am trying to convert the following string '1.12.22 14:16UTC+01:00' in Pandas to December 1st 2022
my_date = '1.12.22 14:16UTC+01:00'
new_date = pd.to_datetime(my_date)
Timestamp('2022-01-12 14:16:00-0100', tz='pytz.FixedOffset(-60)')
It inverts month with day only in specific cases. I am trying to use format="%d.%m.%Y %H:%M%z" but it says that the string is not matching the format.
time data '1.12.22 14:16UTC+01:00' does not match format '%d.%m.%Y %H:%M%z' (match)
Thanks for your help.
>>> pd.to_datetime('01.12.22 14:16UTC', format='%d.%m.%y %H:%M%Z')
Timestamp('2022-12-01 14:16:00+0000', tz='UTC')
I am not sure if this is what you are looking for, but your placeholders are wrong, check this page to know what they stand for.
have you tried adding a zero?
my_date = '01.12.22 14:16UTC+01:00'
I am trying to convert a string to datetime object using the strptime function.
I am encountering a ValueError that says format doesn't match, so I did double checking and confirmed that the format in the string matches the format I am passing as the parameter for strptime.
I have also referenced this question: time data does not match format but there the month and year were swapped.
So does this only work with the '%y-%m-%d %H:%M:%S' format or is it dynamic as per the user input like in my case '%y-%m-%d-%H:%M:%S' ?
input:-
from datetime import datetime
stg = "2022-10-31-01:17:46"
do = datetime.strptime(stg, '%y-%m-%d-%H:%M:%S')
output
ValueError: time data '2022-09-31-01:17:46' does not match format '%y-%m-%d-%H:%M:%S'
Expected output:
#while printing 'do'
2020-09-31-01:17:46
You're almost there. You need %Y instead of %y since you're providing the year with the century (2022 instead of 22).
Your code would be
from datetime import datetime
stg = "2022-10-31-01:17:46"
do = datetime.strptime(stg, '%Y-%m-%d-%H:%M:%S')
In my csv file, the values for a column are 1/1/2022 12:34:16am. However, the value in the cell only shows 34:16.02 (minutes, seconds, and milliseconds).
I would like to convert this column to 1/1/2022 12:34:16am in datetime format so I can subtract another similar column to get the time difference.
I have tried to use strptime but it gives me an error that 'values must be in string format'. So I tried to convert the values to 'str' but that still does not work.
df['start'] = df['start'].astype("str")
df['start'] = datetime.strptime(df['start'], '%d/%m/%y %H:%M:%S')
Anyone able to help? Thanks alot!
Try the following:
date_string = str(df['start'])
format = '%d/%m/%Y %H:%M:%S%p'
df['start'] = datetime.strptime(date_string, format)
I am trying to get user input for 2 different dates which i will pass on to another function.
def twodifferentdates():
print("Data between 2 different dates")
start_date = datetime.strptime(input('Enter Start Date in m/d/y format'), '%m&d&Y')
end_date = datetime.strptime(input('Enter end date in m/d/y format'), '%m&d&Y')
print(start_date)
twodifferentdates()
I have tried a lot of different ways to enter the dates but i keep getting
ValueError: time data '01/11/1987' does not match format '%m&d&Y'
I have used the same code which was discussed in:
how do I take input in the date time format?
Any help here would be appreciated.
Replace %m&d&Y with %m/%d/%Y as described in the referenced post.
datetime.strptime() requires you to specify the format, on a character-by-character basis, of the date you want to input. For the string '01/11/1987' you'd do
datetime.strptime(..., '%m/%d/%Y')
where %m is "two-digit month", %d is "two-digit day" and %Y is "four-digit year" (as opposed to two-digit year %y. These values are separated by slashes.
See also the datetime documentation which describes how to use strptime and strftime.
I'm not very experienced with the datetime module, but the error seems to be the way you're taking input. You should be taking it like this:
start_date = datetime.strptime(input('Enter Start Date in m/d/y format'), '%m &d &Y')
or
start_date = datetime.strptime(input('Enter Start Date in m/d/y format'), '%m/&d/&Y')
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)