I want to extract the date (eg.2018-07-16) from strings (eg. 2018-07-16 10:17:53.460035).
The strings have two formats: "2018-07-16 10:17:53.460035" and "2018-05-20 14:37:21".
When I use strptime(d, "%Y-%m-%d %H:%M:%S.%f") to convert the strings before extracting the date, it pops this error:
ValueError: time data '2018-05-20 14:37:21' does not match format
%Y-%m-%d %H:%M:%S.%f'
How can I convert both time formats to DateTime type and extract date from it?
Use to_datetime from pandas.
import pandas as pd
a = "2018-07-16 10:17:53.460035"
b = "2018-05-20 14:37:21"
print(pd.to_datetime(a).date())
print(pd.to_datetime(b).date())
You don't need the .%f at the end for the first format, that is what is causing the format error.
t = "2018-05-20 14:37:21"
strptime(t, "%Y-%m-%d %H:%M:%S")
You need to create a second format for the other time string:
t = "2018-07-16 10:17:53.460035"
strptime(t, "%Y-%m-%d %H:%M:%S.%f")
Edit: Here is another example which excepts both
time_stamps = ["2018-05-20 14:37:21", "2018-07-16 10:17:53.460035"]
for stamp in time_stamps:
fmt = "%Y-%m-%d %H:%M:%S"
try:
time = datetime.datetime.strptime(stamp, fmt+".%f")
except ValueError:
time = datetime.datetime.strptime(stamp, fmt)
print(time)
Related
I have a date in format 2022-12-16T16-48-47" and I would like to change it to datetime using function pd.to_datetime.
My first idea was to create split the string to have it in more readable way:
string = "2022-12-16T16-48-47"
date, hour = string.split("T")
string = date + " " + hour
string
And now to use:
import pandas as pd
pd.to_datetime(string, format = "%Y-%M-%D %h-%m-%S")
But I have error:
ValueError: 'D' is a bad directive in format '%Y-%M-%D %h-%m-%S'
Do you know how it should be done properly?
Use Y, m, d and H, M, S instead:
>>> pd.to_datetime(string, format = "%Y-%m-%d %H-%M-%S")
Timestamp('2022-12-16 16:48:47')
>>>
You should check out the strftime format codes documentation for better understanding.
How can I convert string 2021-09-30_1 to datetime 2021/09/30 00:00, which means that from the last string we have to substract one to get the hour.
I tried datetime.strptime(date, '%Y %d %Y %I')
datetime.strptime if to define the timestamp from a string, the format should match the provided one. datetime.strftime (note the f) is to generate a string from a datetime object.
You can use:
datetime.strptime(date, '%Y-%m-%d_%H').strftime('%Y/%m/%d %H:%M')
output: '2021/09/30 01:00'
in case the _x defines a delta:
from datetime import datetime, timedelta
d, h = date.split('_')
d = datetime.strptime(d, '%Y-%m-%d')
h = timedelta(hours=int(h))
(d-h).strftime('%Y/%m/%d %H:%M')
output: '2021/09/29 23:00'
Considering the _1 is hour and appears in al of your data (The hour part takes value between [1, 24]), your format was wrong.
For reading the date from string you'll need format it correctly:
from datetime import datetime, timedelta
date = "2021-09-30_1"
date_part, hour_part = date.split("_")
date_object = datetime.strptime(date_part, '%Y-%m-%d') + timedelta(hours=int(hour_part) - 1)
Now you have the date object. And you can display it as:
print(date_object.strftime('%Y/%m/%d %H:%M'))
from datetime import datetime
raw_date = "2021-09-30_1"
date = raw_date.split("_")[0]
parsed_date = datetime.strptime(date, '%Y-%m-%d')
formated_date = parsed_date.strftime('%Y/%m/%d %H:%M')
strptime is used for parsing string and strftime for formating.
Also for date representation you should provide format codes for hours and minutes as in:
formated_date = parsed_date.strftime('%Y/%m/%d %H:%M')
I have two columns Date_x and Date_y. I would like to compare them (i.e Date_x + 1 hour < Date_y)
Format of the strings looks as follows "2020-01-29 11:31:32.754292 UTC"
I have tried converting it using datetime:
from datetime import datetime as dt
df["Date_x"] = [dt.strptime(x, '%Y-%m-%d %H:%M:%S.%f') for x in df['Date_x']]
However, it throws an error regarding the UTC part. I tried removing it with no avail.
Last traceback:
time data '2020-01-29 18:30:28' does not match format '%Y-%m-%d %H:%M:%S.%f'
How would you go about converting the string to hh:mm:ss only?
You could use an if statement:
df["Date_x"] = [dt.strptime(x, '%Y-%m-%d %H:%M:%S.%f') if '.' in x else dt.strptime(x, '%Y-%m-%d %H:%M:%S') for x in df['Date_x']]
But why not just pd.to_datetime:
df["Date_x"] = pd.to_datetime(df["Date_x"], infer_datetime_format=True)
To convert a string date to date format dropping the '00:00:00' I use :
import datetime
strDate = '2017-04-17 00:00:00'
datetime.datetime.strptime(strDate, '%Y/%m/%d %H:%M:%S').strftime('%Y-%m-%d')
Returns :
ValueError: time data '2017-04-17 00:00:00' does not match format '%Y/%m/%d %H:%M:%S'
Is %H:%M:%S not correct format ?
This is the correct way:
datetime.datetime.strptime(strDate, '%Y-%m-%d %H:%M:%S').strftime('%Y-%m-%d')
Notice the - instead of / in strptime. The date is converted to: 2017-04-17.
If you would like to have it displayed a different way, have a look here.
I have a datetime type mydate in %Y-%m-%dT%H:%M:%S format.
I want to replace the hours
I did this using mydate.replace() method
Now I want to comapre it with another specific date -> myNEWdate whose format is %Y-%m-%d %H:%M:%S :
newdate = mydate.replace(hour = islot)
print newdate
appointmentDict[mydate]['time_start'] = datetime.strptime(str(newdate),"%Y/%m/%d %H:%M:%S")
The date is printed as 2015-06-26 08:00:00
and I get the error
ValueError: time data '2015-06-26 08:00:00' does not match format '%Y/%m/%d %H:%M:%S'
What should I do to resolve this
You need to set the correct format
datetime.strptime(str(newdate),"%Y-%m-%d %H:%M:%S")
To solve the exception. Although converting from datetime 2 string and backwards doesn't make much sense, as mentioned in the comments.