Python: Print structure time from given number of seconds
Python Operating System Services: Exercise-25 with Solution
Write a Python program that takes a given number of seconds and pass since epoch as an argument. Print structure time in local time.
Sample Solution:
Python Code :
import time
result = time.localtime(414538698)
print("\nResult:", result)
print("\nYear:", result.tm_year)
Sample Output:
Result: time.struct_time(tm_year=1983, tm_mon=2, tm_mday=19, tm_hour=21, tm_min=38, tm_sec=18, tm_wday=5, tm_yday=50, tm_isdst=0) Year: 1983
Python Code Editor:
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Previous: Write a Python program to print simple format of time, full names and the representation format and preferred date time format.
Next: Write a Python program that takes a tuple containing 9 elements corresponding to structure time as an argument and returns a string representing it.
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Python: Tips of the Day
Python: Cache results with decorators
There is a great way to cache functions with decorators in Python. Caching will help save time and precious resources when there is an expensive function at hand.
Implementation is easy, just import lru_cache from functools library and decorate your function using @lru_cache.
from functools import lru_cache @lru_cache(maxsize=None) def fibo(a): if a <= 1: return a else: return fibo(a-1) + fibo(a-2) for i in range(20): print(fibo(i), end="|") print("\n\n", fibo.cache_info())
Output:
0|1|1|2|3|5|8|13|21|34|55|89|144|233|377|610|987|1597|2584|4181| CacheInfo(hits=36, misses=20, maxsize=None, currsize=20)
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