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)
Related
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 have an excel file with a column date like these:
28.02.2022 00:00:00
What I want is to get it to datetime format without the hour.
I use this and this gives me an error "time data does not match data":
df['date'] = pd.to_datetime(df['date'], format='%d.%m.%Y %H:%M:%Y')
I can't find my error since the format seems to be right. I really appreciate every help.:)
I suspect your format= should be
'%d.%m.%Y %H:%M:%S'
and not
'%d.%m.%Y %H:%M:%Y'
(which has a duplicate %Y).
>>> pd.to_datetime('28.02.2022 00:00:00', format='%d.%m.%Y %H:%M:%S')
Timestamp('2022-02-28 00:00:00')
Whether or not you care about the time part in the data is another thing.
You can try:
pd.to_datetime(df['date']).date()
Something like:
str(pd.to_datetime('28.02.2022 00:00:00').date())
would return:
'2022-02-28'
I have a DataFrame with one columns that is a date and a time and is a string.
The format of the date and time is like this: 4/27/2021 12:39
This is what I have so far to try and convert the string into a datetime:
new_list = []
for i in range(len(open_times)):
date = df.iloc[i]['Open Datetime']
good_date = date.to_datetime()
# good_date = date.topydatetime()
new_list.append(good_date)
I have used to_pydatetime() in the past however the string was in a different format.
When I run the code from above I get this error: AttributeError: 'str' object has no attribute 'to_datetime' and I get the same error when I run the commented out line except with to_pydatetime.
Any thoughts on how to resolve this error? I think that this is happening because the format of the string is different than it typically is.
You need to use datetime.strptime(date_string, format) to convert a string to datetime type
from datetime import datetime
for i in range(len(open_times)):
date = df.iloc[i]['Open Datetime']
good_date = datetime.strptime(date, '%m/%d/%Y %H:%M')
But you could use pd.to_datetime directly
df['Open Datetime'] = pd.to_datetime(df['Open Datetime'])
# Convert a column to list
new_list = df['Open Datetime'].values.tolist()
I'm preprocessing data and one column represents dates such as '6/1/51'
I'm trying to convert the string to a date object and so far what I have is:
date = row[2].strip()
format = "%m/%d/%y"
datetime_object = datetime.strptime(date, format)
date_object = datetime_object.date()
print(date_object)
print(type(date_object))
The problem I'm facing is changing 2051 to 1951.
I tried writing
format = "%m/%d/19%y"
But it gives me a ValueError.
ValueError: time data '6/1/51' does not match format '%m/%d/19%y'
I couldn't easily find the answer online so I'm asking here. Can anyone please help me with this?
Thanks.
Parse the date without the century using '%m/%d/%y', then:
year_1900 = datetime_object.year - 100
datetime_object = datetime_object.replace(year=year_1900)
You should put conditionals around that so you only do it on dates that are actually in the 1900's, for example anything later than today.
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)