I want to change the format of "Date" column from 10/15/2019 to m/d/y format.
tax['AsOfdate']= pd.to_datetime(tax['date'])
How do I do it?
like this, and here is the documentation.
tax['AsOfdate']= pd.to_datetime(tax['date'], format="%m/%d/%Y" )
Here is an example with today's date formatting:
from datetime import date
today = date.today()
new_format = today.strftime("%m/%d/%y")
print(today, new_format)
Pandas to_datetime function accepts a format command which accepts strftime notation.
Pandas docs: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.to_datetime.html
Strftime docs: https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior
m/d/y notation would be:
tax['AsOfdate']= pd.to_datetime(tax['date'], format='%m/%d/%y)
Assuming you want everything zero padded with two digits like 01/01/19 for January first 2019. If you need something else, the strftime formatting link shows all the codes that let you choose padding or not, four-digit year or two-digit, and so on.
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've been trying to convert a timestamp that is a string to a datetime object. The problem is the timestamps formatting. I haven't been able to properly parse the timestamp using datetime.datetime.strptime. I could write my own little parser as its a simple problem but I was hoping to use strptime function, I just need help on the formatting.
Example
import datetime
formater = "%y-%m-%dT%H:%M:%SZ"
str_timestamp = "2021-03-13T18:27:37.60918Z"
timestamp = datetime.datetime.strptime(str_timestamp, formater)
print (timestamp)
Output
builtins.ValueError: time data '2021-03-13T18:27:37.60918Z' does not match format '%y-%m-%dT%H:%M:%SZ'
I'm clearly not symbolizing the formatter properly, the T and Z parts are what I can't account for.
y should be Y. y is for 2 digits year.
You should also take care for the milliseconds with .%f:
%Y-%m-%dT%H:%M:%S.%fZ
This format works:
formater = "%Y-%m-%dT%H:%M:%S.%fZ"
output:
2021-03-13 18:27:37.609180
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')
I am trying to set a variable to equal today's date.
I looked this up and found a related article:
Set today date as default value in the model
However, this didn't particularly answer my question.
I used the suggested:
dt.date.today
But after
import datetime as dt
date = dt.date.today
print date
<built-in method today of type object at 0x000000001E2658B0>
Df['Date'] = date
I didn't get what I actually wanted which as a clean date format of today's date...in Month/Day/Year.
How can I create a variable of today's day in order for me to input that variable in a DataFrame?
You mention you are using Pandas (in your title). If so, there is no need to use an external library, you can just use to_datetime
>>> pandas.to_datetime('today').normalize()
Timestamp('2015-10-14 00:00:00')
This will always return today's date at midnight, irrespective of the actual time, and can be directly used in pandas to do comparisons etc. Pandas always includes 00:00:00 in its datetimes.
Replacing today with now would give you the date in UTC instead of local time; note that in neither case is the tzinfo (timezone) added.
In pandas versions prior to 0.23.x, normalize may not have been necessary to remove the non-midnight timestamp.
If you want a string mm/dd/yyyy instead of the datetime object, you can use strftime (string format time):
>>> dt.datetime.today().strftime("%m/%d/%Y")
# ^ note parentheses
'02/12/2014'
Using pandas: pd.Timestamp("today").strftime("%m/%d/%Y")
pd.datetime.now().strftime("%d/%m/%Y")
this will give output as '11/02/2019'
you can use add time if you want
pd.datetime.now().strftime("%d/%m/%Y %I:%M:%S")
this will give output as '11/02/2019 11:08:26'
strftime formats
You can also look into pandas.Timestamp, which includes methods like .now and .today.
Unlike pandas.to_datetime('now'), pandas.Timestamp.now() won't default to UTC:
import pandas as pd
pd.Timestamp.now() # will return California time
# Timestamp('2018-12-19 09:17:07.693648')
pd.to_datetime('now') # will return UTC time
# Timestamp('2018-12-19 17:17:08')
i got the same problem so tried so many things
but finally this is the solution.
import time
print (time.strftime("%d/%m/%Y"))
simply just use pd.Timestamp.now()
for example:
input: pd.Timestamp.now()
output: Timestamp('2022-01-12 14:43:05.521896')
I know all you want is Timestamp('2022-01-12') you don't anything after
thus we could use replace to remove hour, minutes , second and microsecond
here:
input: pd.Timestamp.now().replace(hour=0, minute=0, second=0, microsecond=0)
output: Timestamp('2022-01-12 00:00:00')
but looks too complicated right, here is a simple way use normalize
input: pd.Timestamp.now().normalize()
output: Timestamp('2022-01-12 00:00:00')
Easy solution in Python3+:
import time
todaysdate = time.strftime("%d/%m/%Y")
#with '.' isntead of '/'
todaysdate = time.strftime("%d.%m.%Y")
import datetime
def today_date():
'''
utils:
get the datetime of today
'''
date=datetime.datetime.now().date()
date=pd.to_datetime(date)
return date
Df['Date'] = today_date()
this could be safely used in pandas dataframes.
There are already quite a few good answers, but to answer the more general question about "any" period:
Use the function for time periods in pandas. For Day, use 'D', for month 'M' etc.:
>pd.Timestamp.now().to_period('D')
Period('2021-03-26', 'D')
>p = pd.Timestamp.now().to_period('D')
>p.to_timestamp().strftime("%Y-%m-%d")
'2021-03-26'
note: If you need to consider UTC, you can use: pd.Timestamp.utcnow().tz_localize(None).to_period('D')...
From your solution that you have you can use:
import pandas as pd
pd.to_datetime(date)
using the date variable that you use
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)