I have a datetime in format = '2020-05-01'. I want the output to be 2020-05-01 00:00:00.
I use this simple code to achieve this but I am missing the time part.
datetime.strptime(month, "%Y-%m-%d")
I am using python2. Does anyone have any idea on how to achieve this. This seems like a really simple problem but for some reason I am not able to achieve it.
Your one line of code already generates a datetime value which should be set to midnight (the default value for no explicit time component). If you want to view this data with its time component, then use strftime with an appropriate format mask:
month = '2020-05-01'
dt = datetime.strptime(month, "%Y-%m-%d")
dt_out = dt.strftime("%Y-%m-%d %H:%M:%S")
print(dt_out)
This prints:
2020-05-01 00:00:00
x = datetime.strptime(month, "%Y-%m-%d")
x.strftime(month, "%Y-%m-%d %H:%M:%S")
Source:
https://www.programiz.com/python-programming/datetime/strftime
Related
What is the best way to initialize a Date to midnight using AWS AppSync utility.
I need to know if we have something like this
var d = (new Date()).setUTCHours(0,0,0,0)
By using $util.time.nowEpochSeconds() , I am getting the epoch time but how do i identify the time difference that i need to add to set as midnight time
AppSync doesn't offer that capability through utils only yet and this is good feedback, I'll make sure the team sees this.
In the meantime, as a workaround, you could modify the date string to achieve what you need.
#set($time = $util.time.nowFormatted("yyyy-MM-dd/HH:mm:ss/Z"))
#set ($split = $time.split("/"))
#set ($midnight = $split[0] + " 00:00:00" + $split[2])
time: $time
midnight: $midnight
midnight epoch seconds: $util.time.parseFormattedToEpochMilliSeconds($midnight, "yyyy-MM-dd HH:mm:ssZ")
will print:
time: 2019-07-15/22:33:57/+0000
midnight: 2019-07-15 00:00:00+0000
midnight epoch seconds: 1563148800000
I'm trying to investigate the Python time striptime method to decompose a time represented as 11:49:57.74. The standard %H, %M, %S are able to decompose the hour , minute , second. However, since the data is a string ( which is taken in python pandas column as datatype object, the Milliseconds after the decimal second is left uninterpreted. Hence, I get an error. Could someone please advise how to parse the example so that the seconds and microseconds are correctly interpreted from the time string ?
I would then use them to find the time delta between two time stamps.
I don't know if I had correctly understood your question.
So, to convert that string time to datetime and calculate the timedelta between two times you need to do as follow:
timedelta = str() #declare an empty string where save the timedelta
my_string = '11:49:57.74' # first example time
another_example_time = '13:49:57.74' #second example time, invented by me for the example
first_time = datetime.strptime(my_string, "%H:%M:%S.%f") # extract the first time
second_time = datetime.strptime(another_example_time , "%H:%M:%S.%f") # extract the second time
#calculate the time delta
if(first_time > second_time):
timedelta = first_time - second_time
else:
timedelta = second_time - first_time
print "The timedelta between %s and %s is: %s" % (first_time, second_time, timedelta)
Here obviusly you don't have any date, so the datetime library as default use 1900-01-01 as you can see in the result of the print:
The timedelta between 1900-01-01 11:49:57.740000 and 1900-01-01 13:49:57.740000 is: 2:00:00
I hope this solution is what you need. Next time provide a little bit more information please, or share an example with the code that you have tried to write.
I'm trying to do some time math, but I'm not sure how I could do this. I'd like to subtract a specific date & time I have in a string (e.g.: 15:54:00 2017-5-20) from current GMT time (e.g: 20:06:27 2017-12-22).
Any thoughts on how I could do this?
# import data into Python
with open(output_file2) as f:
reader = csv.reader(f, delimiter="\t")
d = list(reader)
# here, objects [0][4] and [0][5] would be, for instance: 15:54:00 and 2017-5-20
# , respectively
# UTC Time
os.system("date -u \"+%H:%M:%S %Y-%m-%d\" | gawk '{print \" UTC Date & Time: \", $1, \"\", $2}'")
# eg.: 20:06:27 2017-12-22
Any thoughts would be great! Thanks =)
Update: I've tried so far:
UTC_time = datetime.datetime.utcnow().strftime("%H:%M:%S %Y-%m-%d")
print ' UTC Date & Time: ', UTC_time
time1 = d[0][4]
date1 = d[0][5]
mytime = time1, date1
time_difference = datetime.datetime.utcnow() - mytime
print "HELLO", time_difference
but I keep getting an error:
TypeError: unsupported operand type(s) for -: 'datetime.datetime' and 'tuple'
Not sure what I am doing wrong...
Your mytime variable is not a datetime.datetime object and cannot be used to operate against one. Instead of mytime = time1, date1 you are looking for something more along the lines of
my_date_str = "{} {}".format(time1, date1)
mytime = datetime.datetime.strptime(my_date_str, "%H:%M:%S %Y-%m-%d")
Do realize that in spite of using utcnow() the resulting datetime.datetime object is NOT timezone aware, and neither is mytime. So any math between the two is pure clock math, no timezones taken into account.
If you require timezone support, look into pytz or Pendulum or perhaps find another if they don't suit you.
I have a script that processes an Excel file. The department that sends it has a system that generated it, and my script stopped working.
I suddenly got the error Can only use .str accessor with string values, which use np.object_ dtype in pandas for the following line of code:
df['DATE'] = df['Date'].str.replace(r'[^a-zA-Z0-9\._/-]', '')
I checked the type of the date columns in the file from the old system (dtype: object) vs the file from the new system (dtype: datetime64[ns]).
How do I change the date format to something my script will understand?
I saw this answer but my knowledge about date formats isn't this granular.
You can use apply function on the dataframe column to convert the necessary column to String. For example:
df['DATE'] = df['Date'].apply(lambda x: x.strftime('%Y-%m-%d'))
Make sure to import datetime module.
apply() will take each cell at a time for evaluation and apply the formatting as specified in the lambda function.
pd.to_datetime returns a Series of datetime64 dtype, as described here:
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.to_datetime.html
df['DATE'] = df['Date'].dt.date
or this:
df['Date'].map(datetime.datetime.date)
You can use pd.to_datetime
df['DATE'] = pd.to_datetime(df['DATE'])
I am importing data from a JSON file and it has the date in the following format 1/7/11 9:15
What would be the best variable type/format to define in order to accept this date as it is? If not what would be the most efficient way to accomplish this task?
Thanks.
"What would be the best variable type/format to define in order to accept this date as it is?"
The DateTimeField.
"If not what would be the most efficient way to accomplish this task?"
You should use the datetime.strptime method from Python's builtin datetime library:
>>> from datetime import datetime
>>> import json
>>> json_datetime = "1/7/11 9:15" # still encoded as JSON
>>> py_datetime = json.loads(json_datetime) # now decoded to a Python string
>>> datetime.strptime(py_datetime, "%m/%d/%y %I:%M") # coerced into a datetime object
datetime.datetime(2011, 1, 7, 9, 15)
# Now you can save this object to a DateTimeField in a Django model.
If you take a look at https://docs.djangoproject.com/en/dev/ref/models/fields/#datetimefield, it says that django uses the python datetime library which is docomented at http://docs.python.org/2/library/datetime.html.
Here is a working example (with many debug prints and step-by-step instructions:
from datetime import datetime
json_datetime = "1/7/11 9:15"
json_date, json_time = json_datetime.split(" ")
print json_date
print json_time
day, month, year = map(int, json_date.split("/")) #maps each string in stringlist resulting from split to an int
year = 2000 + year #be ceareful here! 2 digits for a year may cause trouble!!! (could be 1911 as well)
hours, minutes = map(int, json_time.split(":"))
print day
print month
print year
my_datetime = datetime(year, month, day, hours, minutes)
print my_datetime
#Generate a json date:
new_json_style = "{0}/{1}/{2} {3}:{4}".format(my_datetime.day, my_datetime.month, my_datetime.year, my_datetime.hour, my_datetime.minute)
print new_json_style