Discussion on: Daily Challenge #2 - String Diamond

View post

I would like to add:
1) Making str.center do some of the dirty work
2) Using ranges (and chain) to iterate up and down instead of keeping a state on num


from itertools import chain

def diamond(n): 
    if n <=0 or n % 2 == 0:
        raise ValueError('n must be odd and positive') 
    for i in chain(range(1, n + 1, 2), reversed(range(1, n, 2))): 
        print(('*' * i).center(n)) 

diamond(1)
diamond(3)
diamond(5)

It might be more readable to have two loops instead of using chain but I like chaining more because I don't need to repeat the print

Vicente G. Reyes • Sep 18 '19

Is this fast too?

orenovadia • Sep 18 '19

I wouldn't worry about it too much, since print requires IO, it is by far the bottleneck.

For the rest of the utilities that I used: range, chain, and str.center: they are all implemented in C if you are using the standard CPython (should be fast).

To avoid the IO, let's compare the two functions as generators of strings (I replaced the print with yield (('*' * i).center(n)) for both my implementation and Nicks:

In [39]: %timeit tuple(my_diamond(11))                                                                                                                                                                                 
3.59 µs ± 37.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [40]: %timeit tuple(nicks_diamond(11))                                                                                                                                                                           
4.66 µs ± 69.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Seems like mine is slower than Nicks.
However, on my machine, a single print statement takes about 4.17us, which is almost as long as diamond(11) takes without any prints!