Recursive one liner function to print numbers from N to 1

Question

def recur(n):
    print(n)
    if n>1:
        recur(n-1)

How can this be converted to just line. I can not find a way to use the logic as recur = lambda x: print(x) if.......

I can not even use

func = lambda x: [print(i) for i in range(x,0,-1)]

Does this answer your question? [Can a lambda function call itself recursively in Python?](https://stackoverflow.com/questions/481692/can-a-lambda-function-call-itself-recursively-in-python) — dangee1705, May 08 '20 at 14:51

Alain T. · Accepted Answer · 2020-05-08T14:59:33.030

2

If you want the numbers printed on the same line, you can use the end= parameter of the print function:

printDown = lambda n: print(n,end=" ") or printDown(n-1) if n>1 else print(n)

printDown(10)

10 9 8 7 6 5 4 3 2 1

If you want them on separate lines:

printDown = lambda n: print(n) or (printDown(n-1) if n>1 else None)

printDown(10)

10
9
8
7
6
5
4
3
2
1

edited May 08 '20 at 14:59

answered May 08 '20 at 14:51

Alain T.

score 2 · Answer 2 · answered May 08 '20 at 14:54

2

One possible recursive lambda:

recur = lambda n: [print(n), recur if n>1 else lambda x: None][1](n-1)

recur(10)

Prints:

answered May 08 '20 at 14:54

Andrej Kesely

user2390182 · Answer 3 · 2020-05-08T15:03:36.960

2

In Python >= 3.8, you can do it in one line entirely:

print(*(x := lambda n: [n]+x(n-1) if n else [])(10), sep='\n')
10
9
8
7
6
5
4
3
2
1

Note that this also does not abuse a comprehension or expression for side effects. x can still be used to create a common list now:

x(5)
# [5, 4, 3, 2, 1]

edited May 08 '20 at 15:03

answered May 08 '20 at 14:57

user2390182

3 Answers3