being inexperienced with Python I am unable to properly deal with the datetime objects, when wanting to iterate over them. I imported timestamps from a csv file and parsed them into datetime objects. Now I am unable to perform functions on them, because I get one error after the other. Please see my code, what causes the "TypeError: 'datetime.datetime' object is not iterable".
If there is no simple solution to my problem, can someone tell me how to save the datetime objects into a list?
The function of my code is inspired by this post, which works on date time objects from a list: Getting the closest date to a given date
Thanks in advance.
from datetime import datetime, timedelta
import pandas as pd
from dateutil.parser import parse
csvFile = pd.read_csv('myFile.csv')
column = csvFile['timestamp']
column = column.str.slice(0, 19, 1)
dt1 = datetime.strptime(column[1], '%Y-%m-%d %H:%M:%S')
print("dt1", dt1) #output: dt1 2010-12-30 15:06:00
dt2 = datetime.strptime(column[2], '%Y-%m-%d %H:%M:%S')
print("dt2", dt2) #output: dt2 2010-12-30 16:34:00
dt3 = dt1 - dt2
print("dt3", dt3) #output: dt3 -1 day, 22:32:00
#parsing the timestamps as datetime objects works:
for row in range(len(column)):
timestamp = datetime.strptime(column[row], '%Y-%m-%d %H:%M:%S')
print("timestamp", timestamp) #output (excerpt): timestamp 2010-12-30 14:32:00 timestamp 2010-12-30 15:06:00
here error occurs:
base_date = dt1
def func(x):
d = x[0]
delta = d - base_date if d > base_date else timedelta.max
return delta
min(timestamp, key = func)
timestamp is an instance of datetime.datetime,you should put it into a list or a tuple.
after correct,it should be like this min([timestamp], key = func)
Related
How do I subtract/add days (integer) to a Pandas Timestamp object?
For example, my atomics and datatypes are (lifted from Pycharm):
startDate = {Timestamp} 2008-09-20 00:00:00
dayDistance = {int} 124
The code as pulled from the Internet returns None:
from datetime import timedelta
newDate = startDate - timedelta(days=dayDistance)
I am expecting an object of type Timestamp so it is compatible with the rest of the code downstream from here.
pandas has its own Timedelta data type:
start_date = pd.Timestamp("2008-09-20 00:00:00")
dayDistance = 124
new_date = start_date - pd.Timedelta(dayDistance, unit="d")
But Python's built-in timedelta works too:
from datetime import timedelta
new_date = start_date - timedelta(days=dayDistance)
Actually it works fine on my test. But you may use pd.Timedelta instead
import pandas as pd
from datetime import timedelta
ts = pd.Timestamp.now()
print(f"{type(ts)} {ts}")
dt_td_ts = ts - timedelta(days=10)
pd_td_ts = ts - pd.Timedelta(10, unit='d')
print(f"With datetime timedelta: ({type(dt_td_ts)}) {dt_td_ts}")
print(f"With pandas timedelta: ({type(pd_td_ts)}) {pd_td_ts}")
returns
<class 'pandas._libs.tslibs.timestamps.Timestamp'> 2022-05-15 22:20:04.596195
With datetime timedelta: (<class 'pandas._libs.tslibs.timestamps.Timestamp'>) 2022-05-05 22:20:04.596195
With pandas timedelta: (<class 'pandas._libs.tslibs.timestamps.Timestamp'>) 2022-05-05 22:20:04.596195
as you can see the type is Timestamp as expected even when using datetime.timedelta
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 the following string format (Python 3.6):
'2018-11-19T10:04:57.426872'
I get it as a parameter to my script.
I want to get the date as 'YYYY-MM-DD' and time as 'HH:MM'
I tried to convert it with:
from datetime import datetime
if __name__ == '__main__':
start_timestamp = sys.argv[1]
start_date = datetime.strptime(sys.argv[1], '%Y-%m-%d')
start_time = datetime.strptime(sys.argv[1], '%H:%M')
But this gives:
ValueError: unconverted data remains: T10:04:57.426872
In the above example I want to see:
start_date = '2018-11-19'
start_time = '10:04'
Since the date seems to be in ISO-Format, a simple
start = datetime.datetime.fromisoformat(text)
will parse it correctly. From there you can get your date and time with
start_date = start.strftime("%Y-%m-%d")
start_time = start.strftime("%H:%M")
Edit:
For Python < 3.7, you can use this format:
start = datetime.datetime.strptime(text, "%Y-%m-%dT%H:%M:%S.%f")
For the "duplicate" datetime confusion: I used import datetime. If you use from datetime import datetime, you can get rid of the additional datetime.
Try this:We have one of the best package for parsing dates called dateutil.
from dateutil import parser
date1='2018-11-19T10:04:57.426872'
print 'Start_date:',parser.parse(date1).strftime("%Y-%m-%d")
print 'Start_time:',parser.parse(date1).strftime("%H:%M")
Result:Start_date:2018-11-19
Start_time:10:04
You need to parse the entire string into one datetime object and then extract your required values from that.
dt = datetime.datetime.strptime('2018-11-19T10:04:57.426872', '%Y-%m-%dT%H:%M:%S.%f')
d = dt.date()
t = dt.time()
print(d.strftime('%Y-%m-%d'))
print(t.strftime('%H:%M'))
Which outputs:
2018-11-19
10:04
Seems like this should be so simple but for the life of me, I can't find the answer. I pull two datetimes/timestamps from the database:
2015-08-10 19:33:27.653
2015-08-10 19:31:28.209
How do I subtract the first from the second, preferably the result being in milliseconds? And yes, I have the date in there, too, because I need it to work at around midnight, as well.
Parse your strings as datetime.datetime objects and subtract them:
from datetime import datetime
d1 = datetime.strptime("2015-08-10 19:33:27.653", "%Y-%m-%d %H:%M:%S.%f")
d2 = datetime.strptime("2015-08-10 19:31:28.209", "%Y-%m-%d %H:%M:%S.%f")
print(d1 - d2)
Gives me:
0:01:59.444000
Also check out timedelta documentation for all possible operations.
you can do subtraction on 2 datetime objects to get the difference
>>> import time
>>> import datetime
>>>
>>> earlier = datetime.datetime.now()
>>> time.sleep(10)
>>> now = datetime.datetime.now()
>>>
>>> diff = now - earlier
>>> diff.seconds
10
convert your strings to datetime objects with time.strptime
datetime.strptime("2015-08-10 19:33:27.653", "%Y-%m-%d %H:%M:%S.%f")
timedelta.seconds does not represent the total number of seconds in the timedelta, but the total number of seconds modulus 60.
Call the function timedelta.total_seconds() instead of accessing the timedelta.seconds property.
For python 3.4, first you'd need to convert the strings representing times into datetime objects, then the datetime module has helpful tools work with dates and times.
from datetime import datetime
def to_datetime_object(date_string, date_format):
s = datetime.strptime(date_string, date_format)
return s
time_1 = '2015-08-10 19:33:27'
time_2 = '2015-08-10 19:31:28'
date_format = "%Y-%m-%d %H:%M:%S"
time_1_datetime_object = to_datetime_object(time_1, date_format)
time_2_datetime_object = to_datetime_object(time_2, date_format)
diff_time = time_1_datetime_object - time_2_datetime_object
As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. I have a working version which is:
yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
There must be a more elegant way to do this, interested for educational purposes as much as anything else!
You Just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.
datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime in order to convert the result --which is a date object-- to string format based on your format of choice:
>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)
>>> type(yesterday)
>>> datetime.datetime
>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:
>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
As a function:
from datetime import datetime, timedelta
def yesterday(frmt='%Y-%m-%d', string=True):
yesterday = datetime.now() - timedelta(1)
if string:
return yesterday.strftime(frmt)
return yesterday
example:
In [10]: yesterday()
Out[10]: '2022-05-13'
In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)
An alternative answer that uses today() method to calculate current date and then subtracts one using timedelta(). Rest of the steps remain the same.
https://docs.python.org/3.7/library/datetime.html#timedelta-objects
from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)
Output:
2019-06-14
2019-06-13
>>> import datetime
>>> datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'
Calling .isoformat() on a date object will give you YYYY-MM-DD
from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()
I'm trying to use only import datetime based on this answer.
import datetime
oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday