[Solved] Python 3: How to get a random 4 digit number?

Question

If you want to extend your solution to make sure the leading digit isn’t 0, there’s probably no super-concise way to write it; just be explicit:

first = random.choice(range(1, 10))
leftover = set(range(10)) - {first}
rest = random.sample(leftover, 3)
digits = [first] + rest

Notice that I’m taking advantage of the fact that sample takes a sequence or set. In older versions of Python, it required a sequence. According to the 2.7 docs, this includes 2.7—but CPython and PyPy 2.7 both take a set anyway. Anyway, if this is a problem, you can just do leftover = [val for val in range(10) if val != first].

One more thing to consider, if performance matters. A nondeterminstic solution like noskio’s answer obviously has unbounded worst-case time, but it could be significantly faster in almost every case. There’s a 90% chance you’ll get your answer the first time, and a 99.9999999% within the first ten loops, and it’s at least plausible that a single call to sample could be about an order of magnitude faster than multiple set operations, etc.

Accepted Answer

If you want to extend your solution to make sure the leading digit isn’t 0, there’s probably no super-concise way to write it; just be explicit:

first = random.choice(range(1, 10))
leftover = set(range(10)) - {first}
rest = random.sample(leftover, 3)
digits = [first] + rest

Notice that I’m taking advantage of the fact that sample takes a sequence or set. In older versions of Python, it required a sequence. According to the 2.7 docs, this includes 2.7—but CPython and PyPy 2.7 both take a set anyway. Anyway, if this is a problem, you can just do leftover = [val for val in range(10) if val != first].

One more thing to consider, if performance matters. A nondeterminstic solution like noskio’s answer obviously has unbounded worst-case time, but it could be significantly faster in almost every case. There’s a 90% chance you’ll get your answer the first time, and a 99.9999999% within the first ten loops, and it’s at least plausible that a single call to sample could be about an order of magnitude faster than multiple set operations, etc.