How do I convert numpy array to days, hours, mins?

Question

Running with this series

X = number_of_logons_all.values
split = round(len(X) / 2)
X1, X2 = X[0:split], X[split:]
mean1, mean2 = X1.mean(), X2.mean()
var1, var2 = X1.var(), X2.var()
print('mean1=%f, mean2=%f' % (mean1, mean2))
print('variance1=%f, variance2=%f' % (var1, var2))

I get:

mean1=60785.792548, mean2=61291.266868
variance1=7483553053.651829, variance2=7603208729.348722

But I wanted something like this in my PyCharm console (pulled from another result):

>>> -103 days +04:37:13.802435724...

Tried to place the np.array in a pd.Dataframe() to get the expected value by adding

.apply(pd.to_timedelta, unit='s')

...this didn't work, so I tried

new = pd.DataFrame([mean1]).to_numpy(dtype='timedelta64[ns]')

...and (still) got something like this:

>>>> [[63394]]

Anyone out there who could assist me converting to an easily comprehended datetime result from my means calculation above?

Thx, in advance for your kind support.

Answer 1

You can use f-strings:

mean1, mean2 = 60785.792548, 61291.266868
variance1, variance2=7603208729.348722,7483553053.651829

print(f'mean1={pd.Timedelta(mean1, unit="s")}, mean2={pd.Timedelta(mean2, unit="s")}')
print(f'variance1={pd.Timedelta(variance1, unit="s")}, variance2={pd.Timedelta(variance2, unit="s")}')
mean1=0 days 16:53:05.792548, mean2=0 days 17:01:31.266868
variance1=88000 days 02:25:29.348722458, variance2=86615 days 04:44:13.651828766