Accumulator for outer nested loop Python
I'm learning python on my own from a book and solving problems. In one problem, the user inputs amount of rain for each month in one year across a period of years. I need to find the average of rain in each year (sum(monthly rain)/numb_months, and also the average of rain in that period, e.g. in two years. In the following code, I can get the average for each year (I used 3 months only instead of 12 months to avoid tedious input now), but I don't know where I need to put an accumulator for total rain in that period and then average it. I appreciate your help.
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
python loops nested accumulator
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I'm learning python on my own from a book and solving problems. In one problem, the user inputs amount of rain for each month in one year across a period of years. I need to find the average of rain in each year (sum(monthly rain)/numb_months, and also the average of rain in that period, e.g. in two years. In the following code, I can get the average for each year (I used 3 months only instead of 12 months to avoid tedious input now), but I don't know where I need to put an accumulator for total rain in that period and then average it. I appreciate your help.
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
python loops nested accumulator
add a comment |
I'm learning python on my own from a book and solving problems. In one problem, the user inputs amount of rain for each month in one year across a period of years. I need to find the average of rain in each year (sum(monthly rain)/numb_months, and also the average of rain in that period, e.g. in two years. In the following code, I can get the average for each year (I used 3 months only instead of 12 months to avoid tedious input now), but I don't know where I need to put an accumulator for total rain in that period and then average it. I appreciate your help.
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
python loops nested accumulator
I'm learning python on my own from a book and solving problems. In one problem, the user inputs amount of rain for each month in one year across a period of years. I need to find the average of rain in each year (sum(monthly rain)/numb_months, and also the average of rain in that period, e.g. in two years. In the following code, I can get the average for each year (I used 3 months only instead of 12 months to avoid tedious input now), but I don't know where I need to put an accumulator for total rain in that period and then average it. I appreciate your help.
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
python loops nested accumulator
python loops nested accumulator
asked Nov 14 '18 at 18:15
abenolabenol
173
173
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1 Answer
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If I understood what you want (to calculate the absolute average of the rain amount during the period), this should do the trick:
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
total_rain = 0
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
total_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
total_months = years_in_period * number_of_months
print("Absolute average of rain/month was", total_rain/total_months)
print("Absolute average of rain/year was", total_rain/years_in_period)
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If I understood what you want (to calculate the absolute average of the rain amount during the period), this should do the trick:
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
total_rain = 0
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
total_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
total_months = years_in_period * number_of_months
print("Absolute average of rain/month was", total_rain/total_months)
print("Absolute average of rain/year was", total_rain/years_in_period)
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
add a comment |
If I understood what you want (to calculate the absolute average of the rain amount during the period), this should do the trick:
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
total_rain = 0
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
total_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
total_months = years_in_period * number_of_months
print("Absolute average of rain/month was", total_rain/total_months)
print("Absolute average of rain/year was", total_rain/years_in_period)
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
add a comment |
If I understood what you want (to calculate the absolute average of the rain amount during the period), this should do the trick:
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
total_rain = 0
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
total_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
total_months = years_in_period * number_of_months
print("Absolute average of rain/month was", total_rain/total_months)
print("Absolute average of rain/year was", total_rain/years_in_period)
If I understood what you want (to calculate the absolute average of the rain amount during the period), this should do the trick:
number_of_months = 3
years_in_period = int(input("Please enter the number of years in the period. n"))
total_rain = 0
for year in range(years_in_period):
yearly_rain = 0
print('Year', year+1)
print('−−−−−−−−−−−−−−−−−')
for month in range(number_of_months):
print('Month', month+1, end='')
monthly_rain = float(input("Please enter rainfall for this month: n"))
yearly_rain += monthly_rain
total_rain += monthly_rain
average_yearly_rainfall = yearly_rain / number_of_months
print("Average yearly rainfall of year ", year+1, " is ", average_yearly_rainfall)
print("Year total rain is", yearly_rain)
print()
total_months = years_in_period * number_of_months
print("Absolute average of rain/month was", total_rain/total_months)
print("Absolute average of rain/year was", total_rain/years_in_period)
answered Nov 14 '18 at 18:28
Helena MartinsHelena Martins
923118
923118
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
add a comment |
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
That's what I was looking for! Thank you so much, Pedro Martins for quick help!
– abenol
Nov 14 '18 at 18:50
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
No problem! Just let me know if you need something else ;)
– Helena Martins
Nov 14 '18 at 19:02
add a comment |
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