Converting seconds in date form Numpy Python - arrays

Question

The delta_s function calculates the difference of time between 2 dates in the dates with seconds. Then the average median and max values for the differences is calculated. I am trying to convert the times in seconds for average median and max values but it does not work. i want to convert it in the form of x days x hours x minutes x seconds.
Code:
import numpy as np
dates= np.array(['2017-09-15 07:11:00' ,'2017-09-15 11:25:30', '2017-09-15 12:11:10', '2021-04-07 22:43:12', '2021-04-08 00:49:18'],
dtype="datetime64[ns]")
delta_s = np.diff(dates) // 1e9 # nanoseconds to seconds
delta_s = delta_s.astype(np.float64)
delta_avg = np.average(delta_s)
delta_median= np.median(delta_s)
delta_max = np.max(delta_s)
delta_max_index= np.argmax(delta_s)

Answer 1

The line delta_s = np.diff(dates) // 1e9 does not actually convert nanoseconds to seconds. It simply divides 1e9 to the timedelta object but the time unit is preserved timedelta64[ns].
>>> np.diff(dates)
array([ 15270000000000, 2740000000000, 112357922000000000,
7566000000000], dtype='timedelta64[ns]')
>>> np.diff(dates) // 1e9
array([ 15270, 2740, 112357922, 7566],
dtype='timedelta64[ns]')
This may mess up with any calculations you're doing.
Use
delta_s = np.array([np.timedelta64(td, 's') for td in np.diff(dates) ])
Currently there are no inbuilt functions to format timedelta to strings. However you can use something of this sort.
# Function to convert seconds to Human readable Timedelta string
def seconds_to_tdstring(total_seconds):
days, remainder = divmod(total_seconds, 60 * 60 * 24)
hours, remainder = divmod(remainder, 60 * 60)
minutes, seconds = divmod(remainder, 60)
return '{:02} Days {:02} Hours {:02} Minutes {:02} Seconds'.format(int(days), int(hours), int(minutes), int(seconds))
print(seconds_to_tdstring(delta_avg))
Output:
325 Days 04 Hours 24 Minutes 34 Seconds
I have modified the answer from a similar question .