Calculate SFC per year

[SQPaul]

Hello,

First of everything thanks for the package, it's very helpful. My question is if there is an option to get the streamflow characteristics per year instead of all the years?

Cheers!

Paul

[ThibHlln]

Hi @SQPaul,

I am really glad you find this package helpful! 🙂

No, I am afraid there is no straightforward way to output the SFCs per year with eflowcalc.

I was about to suggest you do a numpy.reshape(flows, (365, -1)) to benefit from the fact that you can compute SFCs on multiple streamflow time series at once (i.e. computing on multiple years instead of multiple time series), but this won't work because of leap years. And you may want to do that on multiple streamflow time series at once anyway, so you would end up with a 3D array, which is problematic too.

eflowcalc was designed with the assumption that the number of streamflow time series at hand would be much greater than the number of hydrological years (given that hydrological records are typically several decades long at best), so that a loop on (<100) hydrological years was deemed computationally affordable. So if you do have a lot of streamflow time series to compute at once (or even if you don't), I would suggest you stick to this principle and use eflowcalc.calculator on each hydrological year separately in a loop, as I don't see a worthwhile gain in trying fancy alternatives (e.g. to work with ragged arrays or similar) to bypass the leap year problem.

For example, something like this should work:

from datetime import datetime, timedelta
import numpy as np
import eflowcalc as efc


times = np.array([datetime(2010, 10, 1) + timedelta(days=d) for d in range(3653)])  # 10-year period
flows = np.random.uniform(3, 50, (3653, 100))  # 100 made up streamflow time series
area = 120.7  # km2

hydro_year = '01/10'

sfcs = (efc.ma41, efc.dh4, efc.ra7)

per_year = np.zeros((len(sfcs), 10, 100))

for i, y in enumerate(range(2010, 2020)):

    s = datetime.strptime(f'{hydro_year}/{y}', '%d/%m/%Y')
    e = datetime.strptime(f'{hydro_year}/{y+1}', '%d/%m/%Y')

    mask = (times >= s) & (times < e)

    per_year[:, i, :] = efc.calculator(sfcs, times[mask], flows[mask], area, hydro_year=hydro_year)

Hope this helps,
T