For an assignment in my CS class I have to print all powers of 3 (1, 3, 9, 27, etc.) less than 1000 using no more than 2 lines of code. I know I could just use
for x in range(7):
print(3**x)
because I know that 3^6 is the last power that gives a result under 1000 but I was wondering if there was a way to conditionally check that 3^x is under 1000 before printing still using only 2 lines of code at most. I may just be overthinking it but I want to make sure for my own information.
Usually in python we prefer however many lines can give the most readable code.
So, it's a strange requirement to limit the lines of code like that. At a guess, your instructor may have been looking for some mathematical insight on how a bound for iteration could be precomputed:
>>> for i in range(1 + int(math.log(1000,3))):
... print(3**i)
...
1
3
9
27
81
243
729
If you are allowed to use semicolons:
i=3;
while i<1000: print(i);i = i*3
>>> from itertools import takewhile, count
>>> print(*map(lambda x: 3**x, takewhile(lambda x: 3**x < 1000, count(0))), sep='\n')
1
3
9
27
81
243
729
You could have found that 6 is the lowest allowed exponent using logs. Here's a one-line solution:
import math
def printer(exp_num, target_num):
for i in range(int(math.log(target_num, exp_num))+1): print exp_num**i
printer(3, 1000)
itertools.takewhile is the correct way to express what you're asking, but it depends on a module which necessarily adds a line.
>>> import itertools
>>> print(list(itertools.takewhile(lambda x: x < 1000, (3**x for x in itertools.count()))))
[1, 3, 9, 27, 81, 243, 729]
Expressed in a more sane 3-liner format...
>>> import itertools
>>> for x in itertools.takewhile(lambda x: x < 1000, (3**x for x in itertools.count())):
... print(x)
...
1
3
9
27
81
243
729
Let's explain what's going on here, starting from the innermost. Each step builds on the previous.
itertools.count() is a generator that produces the numbers 1, 2, 3, 4, ... forever. It's how you express range(infinity).
(3**x for x in itertools.count()) is a generator that produces the numbers 1, 3, 9, 27, ... forever.
itertools.takewhile(lambda x: x < 1000, (3**x for x in itertools.count())) is a generator that produces the numbers 1, 3, 9, 27, ... forever while x < 1000 is true.
After that it's just a matter of printing the generated numbers.
The itertools module is a pretty important part of Python, I suggest learning it in general as it solves a lot of problems like this.
one liner with a list comprehension
print([3 ** x for x in range(7) if 3 ** x < 1000])
addressing kevin's comment and trying to stick to two lines of code
for i in range(1000):
print(3 ** i) if (3 ** i) < 1000 else i
i don't like the else statement
Probably, if you can work out how to use a ternary operator. This answer may help, but my best-guess (I don't work in Python) would be along the lines of...
for x in range(10):
print(3**x) if (3**x) < 1000 else print()
def power(x): # define function
for x in range(10):
if (3**x) < 1000:
print(3**x)
else:
print()
power(3) # call the function
