﻿ Python: Print structure time from given number of seconds - w3resource

# 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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## 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)```