Векторизация трех вложенных циклов, которые вычисляют среднесуточное значение почасовых данных ⇐ Python

[Anonymous]

Есть ли способ векторизовать следующий трехвложенный цикл, вычисляющий среднее дневное значение почасовых данных? Функция ниже сначала выполняет цикл по году, затем по месяцам и, наконец, по дням. Он также проверяет последний месяц и день, чтобы гарантировать, что цикл не выходит за пределы последнего месяца или дня данных.
def hourly2daily(my_var,my_periods):

import pandas as pd
import sys

print('######### Daily2monthly function ##################')
Frs_year =my_periods[0].year
Frs_month =my_periods[0].month
Frs_day =my_periods[0].day
Frs_hour =my_periods[0].hour

Last_year =my_periods[-1].year
Last_month =my_periods[-1].month
Last_day =my_periods[-1].day
Last_hour =my_periods[-1].hour

print('First year is '+str(Frs_year) +'\n'+\
'First months is '+str(Frs_month)+'\n'+\
'First day is '+str(Frs_day)+'\n'+\
'First hour is '+str(Frs_hour))
print(' ')

print('Last year is '+str(Last_year)+'\n'+\
'Last months is '+str(Last_month)+'\n'+\
'Last day is '+str(Last_day)+'\n'+\
'Last hour is '+str(Last_hour))

#### Trick to be used for pd.data_range function only #######
# The following if condition was written for "data_range" function
# to understand why we did that, try
# pd.date_range('01/2000','12/2000',freq='M')
# you will find that there is no Dec, and the only way to do that is
# to do the following if condition tricks.

# at 12 data range know that now it should look at the next year.
Last_year_ = Last_year
Last_month_= Last_month

if (Last_month == 12):
Last_year_ = Last_year+1
Last_month_ = 1

Frs = str(Frs_year)+'/'+str(Frs_month)+'/'+str(Frs_day)+' '+str(Frs_hour)+":00"

Lst = str(Last_year_)+'/'+str(Last_month_)+'/'+str(Last_day)+' '+str(Last_hour)+":00"

my_daily_time=pd.date_range(Frs,Lst,freq='D')

## END of the data_range tricks ###########

nt_days=len(my_daily_time)
nd=np.ndim(my_var)

if (nd == 1): # only time series
var_mean=np.full((nt_days),np.nan)

if (nd == 2): # e.g., time, lat or lon or lev
n1=np.shape(my_var)[1]
var_mean=np.full((nt_days,n1),np.nan)

if (nd == 3): # e.g., time, lat, lon
n1=np.shape(my_var)[1]
n2=np.shape(my_var)[2]
var_mean=np.full((nt_days,n1,n2),np.nan)

if (nd == 4): # e.g., time, lat , lon, lev
n1=np.shape(my_var)[1]
n2=np.shape(my_var)[2]
n3=np.shape(my_var)[3]
var_mean=np.full((nt_days,n1,n2,n3),np.nan)

end_mm=12
k=0
# First loop over Years
#######################
for yy in np.arange(Frs_year,Last_year+1):
print('working on Year '+str(yy))
# in case the last month is NOT 12
if (yy == Last_year):
end_mm=Last_month
print('The last month is '+str(end_mm))
# Second loop over Months
#########################
for mm in np.arange(1,end_mm+1):
end_day=pd.Period(str(yy)+'-'+str(mm)).days_in_month
# in case the last day is not at the end of the month.
if ((yy == Last_year) & (mm == Last_month)):
end_day=Last_day
# Third loop over Days
#######################
for dd in np.arange(1,end_day+1):
print(str(yy)+'-'+str(mm)+'-'+str(dd))
#list all days of the month and year.
I=np.where((my_periods.year == yy) &\
(my_periods.month == mm) &\
(my_periods.day == dd ))[0]

print(I)
# if there is a discontinuity in time.
if len(I) == 0 :
print('Warning time shift here >>')
print('Check the continuity of your time sequence')
sys.exit()

var_mean[k,...]=np.nanmean(my_var[I,...],0)
k=k+1
return var_mean,my_daily_time



Подробнее здесь: https://stackoverflow.com/questions/793 ... ourly-data