I took this line from online to make a random 4 digit number
number = str(random.randint(0,9999))
but sometimes it give me random 3 digit number why?
Is their a better way to make a random multiple digit number in python
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Did you read the [docs](https://docs.python.org/3/library/random.html#random.randint) of `randint`? It says: *Return a random integer N such that `a <= N <= b`. Alias for `randrange(a, b+1)`.* So in your code `0` is also an option. It is not 4 digits. Try `randint(1000,9999)` – Tomerikoo Jan 20 '21 at 07:47
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You should read the base knowledge about random.randint(a, b) - Return a random integer N such that a <= N <= b. Alias for randrange(a, b+1). Detail here: https://docs.python.org/3/library/random.html#random.randint – Vu Phan Jan 20 '21 at 07:49
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@Tomerikoo It's arguable that most of the answers in the question for which this is marked a duplicate of could be superseded by using `random.choices()`. – mhawke Jan 20 '21 at 08:08
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@mhawke and? I don't see your point. This is still an obvious duplicate. If you have another solution not existing there, feel free to post it there – Tomerikoo Jan 20 '21 at 08:34
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@Tomerikoo: fair enough. – mhawke Jan 20 '21 at 10:09
2 Answers
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Because you've asked for a number between 0 and 9999 inclusive. That includes single, double and triple digit numbers as well.
If you want only 4 digit numbers start at 1000:
number = str(random.randint(1000, 9999))
See the randint() documentation.
If you want 4 digits that could include leading zeroes then you use random.choices():
from string import digits
number = ''.join(random.choices(digits, k=4))
which picks 4 digits with replacement and joins them into a string.
mhawke
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`choices` could give something like `0123` which might not be an acceptable option... – Tomerikoo Jan 20 '21 at 08:33
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@Tomerikoo: or it might be acceptable, difficult to tell from the question. – mhawke Jan 20 '21 at 10:09
