Question

Python 3.4 Human readable time intervals problem

Write the function readable-interval that takes a string consisting of a human-readable time interval Homework #3 CSE 101 Spring 2017 Page 3 and converts it into a list of six integers, given in this order: of years, of months, t of days, #of hours, of minutes, of seconds] The input values stored within the string can be quite scrambled and consist of multiple, repeated time components (days, months, etc.), as shown in the examples below. If a time component appears more than once, the values are be added. The time units might be given in singular or plural form (e.g., "month" and "months" are both valid). There will always be at least (but not necessarily exactly) one space between the tokens (i.e., components) of the string. Examples: Function Call: readable-interval 1 year 2 months 3 seconds') Return Value: 1, 2, 0, 0, 0, 31 Function Call: readable-interval 1 second 2 years 3 seconds 7 months 6 days' Return Value: [2, 7, 6, 0, 0, 4 Function Call 4 hours 17 minutes') readable interval 1 days 4 days 3 years Return Value: 13, 0, 5, 4, 17, 01 Function Call readable interval 6 hours 20 minutes 421 seconds Return Value: [0, 0, 0, 6, 20 421] Function Call readable interval 0 days 0 months 7 months 18 years') Return Value: 18 7, 0, 0, 0, 01

I really need help with this problem. I don't know how to start and what method will be most helpful and easy.

Answer 1

Program: def readable interval(intervalstring): value = interval-string Split(" ") interval= [0, 0, 0, 0, 0, 0] I for i in range(len(value)) ifi%21=0 if value[i]"years" or value[i"year": interval[0] +int(valueli-1]) elif valuei"month" or valuei"months": interval [1]+int(value[i-1]) elífvalue [i] == "day" or value [i] == "days": interval[2] += int(value[i-1]) elif valuei"hour" or value[i"hours" erval3] int(value[i-1]) elif value[i"minute" or value[il"minutes": interval [4] += int(value [i-1) elifvalueli] == "second" or value [i]--"seconds": interval[5] int(value[i - 1]) return interval

print readable interval'1 year 2 months 3 seconds')) print adable_interval('1 second 2 years 3 seconds 7 months 6 days') print(readable intervall'1 days 4 days 3 years 4 hours print readable interval('6 hours 20 minutes 421 seconds')) print(readable interval('O days O months 7 months 18 years')) 17 minutes ')

Sample output: [GCC 4.8.2] on linux [2, 7, 6, e, e, 4] [3, θ, 5, 4, 17, θ] [a, θ, θ, 6, 28, 421] [18, 7, θ, θ, θ, θ]

Editable code:

def readable_interval(interval_string):
  
value = interval_string.split(" ")
interval = [0, 0, 0, 0, 0, 0]   
for i in range(len(value)):
if i % 2 != 0:
if value[i] == "years" or value[i] == "year":
interval[0] += int(value[i - 1])
elif value[i] == "month" or value[i] == "months":
interval[1] += int(value[i - 1])
elif value[i] == "day" or value[i] == "days":
interval[2] += int(value[i - 1])
elif value[i] == "hour" or value[i] == "hours":
interval[3] += int(value[i - 1])
elif value[i] == "minute" or value[i] == "minutes":
interval[4] += int(value[i - 1])
elif value[i] == "second" or value[i] == "seconds":
interval[5] += int(value[i - 1])
return interval
  
print(readable_interval('1 year 2 months 3 seconds'))
print(readable_interval('1 second 2 years 3 seconds 7 months 6 days'))
print(readable_interval('1 days 4 days 3 years 4 hours 17 minutes '))
print(readable_interval('6 hours 20 minutes 421 seconds'))
print(readable_interval('0 days 0 months 7 months 18 years'))