3

i've been stuck on a question for some time now:

I'm looking to create a python function that consumes a string and a positive integer. The function will print the string n times, for n lines. I cannot use loops, i must only use recursion

e.g.

>>> repeat("hello", 3)
hellohellohello
hellohellohello
hellohellohello

whenever i try to make a function that does this, the function decreases the length of the string, progressively:

e.g.

>>> repeat("hello", 3)
hellohellohello
hellohello
hello

here's what my code looks like:

def repeat(a, n):
    if n == 0:
        print(a*n)
    else:
        print(a*n)
        repeat(a, n-1)

What is wrong with this attempt? How can I fix it?

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3
  • So, you are confused what the n-1 is doing? Commented Oct 12, 2016 at 21:05
  • yes, i want to keep the number of times the string is printed in one line as a fixed sort of parameter, but i want the function to repeat the line n times. Commented Oct 12, 2016 at 21:06
  • Hint: you can modify the first argument to repeat the same way you modify the second. Instead of passing n on the recursive call, you pass n-1. So instead of passing a, maybe pass something else? Commented Oct 12, 2016 at 21:09

5 Answers 5

Reset to default
7

One liner

def repeat(a,n):
    print((((a*n)+'\n')*n)[:-1])

Let's split this up

  1. a*n repeats string n times, which is what you want in one line
  2. +'\n' adds a new line to the string so that you can go to the next line
  3. *n because you need to repeat it n times
  4. the [:-1] is to remove the last \n as print puts a new-line by default.
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3 Comments

Doesn't use recursion, though
I assumed "use recursion if needed"
@zyxue I explained the steps. You can always have temporary variables for the steps.
5

Try this

def f(string, n, c=0):
    if c < n:
        print(string * n)
        f(string, n, c=c + 1)

f('abc', 3)
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2 Comments

I like the trick of counting up so you can use a default 0 for the 3rd parameter.
this is great and easy to understand! thank you! the count up is a great way to approach it
4

So you just need extra argument that will tell you how many times you already ran that function, and it should have default value, because in first place function must take two arguments(str and positive number).

def repeat(a, n, already_ran=0):
    if n == 0:
        print(a*(n+already_ran))
    else:
        print(a*(n+already_ran))
        repeat(a, n-1, already_ran+1)
repeat('help', 3)

Output

helphelphelp
helphelphelp
helphelphelp
helphelphelp
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Comments

2

You should (optionally) pass a 3rd parameter that handles the decrementing of how many lines are left:

def repeat(string, times, lines_left=None):
    print(string * times)

    if(lines_left is None):
        lines_left = times
    lines_left = lines_left - 1

    if(lines_left > 0):
        repeat(string, times, lines_left)
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3 Comments

This could be reduced, but I wanted to be verbose for its self-explanatory nature.
camelCase is not very pythonic.
Ugh, thank you @zyxue! It's been a while since I've been in Python...forgot about that one.
2

You were really close.

def repeat(a, n):
    def rep(a, c):
        if c > 0:
            print(a)
            rep(a, c - 1)
    return rep(a * n, n)
print(repeat('ala', 2))
alaala
alaala

A function with closure would do the job.

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4 Comments

super close, i'm looking to print (a*n) n times
No idea why this is upvoted when it literally does the same thing as the OP's method.
Usually, it's better to avoid nested function unless you have a very strong argument for it. It's also harder to test. In this case, I can see the motivation, an advantage of this solution over a 3-argument function is that it eliminates the possibility of mistakenly passing 3 arguments. You can make rep a bit cleaner by changing the condition to if c > 0, then you don't need to write a single-line return statement
@zyxue yes, it might be better idea to change the statement as you said. Edited.

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