This question already has answers here:
pandas datetime to unix timestamp seconds
(5 answers)
Closed 1 year ago.
I have a df with time column in ISO Format
df = pd.DataFrame({'time':['2021-05-01T16:08:59.094953+00:00','2021-05-01T16:08:56.675183+00:00','2021-05-01T16:08:56.675183+00:00']})
How can i convert this time column to epoch timestamp in Python?
Desired result:
TRY this approach
pd.to_datetime(df.time).astype(int) / 10**9
Related
This question already has answers here:
Python: give start and end of week data from a given date
(5 answers)
how to get all dates of week based on week number in python
(7 answers)
Adding days to a date in Python
(16 answers)
Closed 9 months ago.
good morning, I'm in the middle of a project and I'm stuck on finding a way to return every day of the current week in a list. It would be something like this:
week=[ '2022-06-07' , '2022-06-08' , '2022-06-09' , '' , '...']
I would be very grateful if you could explain to me a way to obtain this result,,, thanks
You can shorten this but I wrote it out like this for clarity.
from datetime import datetime, timedelta
date_obj = datetime(2022, 6, 7) # Today, Tuesday
monday = date_obj - timedelta(days=date_obj.weekday()) # assuming start of week is Monday
week = []
for i in range(7):
week.append((monday + timedelta(days=i)).strftime("%Y-%m-%d"))
print(week)
Output:
# Monday 6/6 to Sunday 6/12
['2022-06-06', '2022-06-07', '2022-06-08', '2022-06-09', '2022-06-10', '2022-06-11', '2022-06-12']
This question already has an answer here:
How can I get pandas Timestamp offset by certain amount of months?
(1 answer)
Closed 2 years ago.
I'd like to offset a date in a DatetimeIndex by a certain number of months using a string but can only find reference to the MonthEnd version.
Here is my (incorrect) attempt
rng = pd.date_range('2020-01-01','2020-07-03')
rng[-1] - pd.tseries.frequencies.to_offset('2m')
This will return Timestamp('2020-05-31 00:00:00') when instead I'm looking for Timestamp('2020-05-03 00:00:00')
Is there an alternative to '2m' here which will give me that result?
The DateOffset method in pandas works well:
rng[-1] - pd.DateOffset(months=2)
Output:
Timestamp('2020-05-03 00:00:00')
This question already has answers here:
removing time from date&time variable in pandas?
(3 answers)
Closed 2 years ago.
I have the time column in a df
accrual_time_x
1/4/19 20:32
trying to convert into
%m/%d/%y
df["accrual_time_x"] = pd.to_datetime(df["accrual_time_x"],format="%m/%d/%y")
yet still return
0 2019-01-04 20:32:00
How could I convert into 01-04-2019? thanks!
Is this what you want?
df['accrual_time_x'].apply(pd.to_datetime).dt.strftime('%m-%d-%Y')
This question already has answers here:
Convert string "Jun 1 2005 1:33PM" into datetime
(26 answers)
Closed 3 years ago.
i m currently working in Generating dynamic invoice on monthly basis but i m stuck in getting month name in python from user input date
Example: Given Date is :15/03/2020 is date and month is March
Any help would be appreciated
Assuming that you are receiving the date in the form of a string like "15/03/2020":
int(date.split('/')[1]) returns the value between the first and second / of date as an integer. This can then be used as the index for a list of month names to return the name of the month:
months = ["January","February","March",...]
print(months[int(date.split('/')[1])-1])
This question already has answers here:
How do I Pandas group-by to get sum?
(11 answers)
Closed 4 years ago.
There is a dataframe, which includes one column of time and another column of bill. As can be seen from the table, there can have multiple records for a single day. The order of time can be random
time bill
2006-1-18 10.11
2006-1-18 9.02
2006-1-19 12.34
2006-1-20 6.86
2006-1-12 10.05
Based on these information, I would like to generate a time series dataframe, which has two columns Time and total bill
The time column will save the date in order, the total bill will save the sum of multiple bill records belonging to one day.
newdf = pd.DataFrame(df.groupby('time').bill.sum())
newdf.rename(columns={'time':'Time', 'bill': 'total bill'}, inplace = True)
newdf
output:
Time total_bill
0 2006-1-18 10.11
1 2006-1-18 9.02
2 2006-1-19 12.34
3 2006-1-20 6.86
4 2006-1-12 10.05