Generate list of all palindromic numbers of 3 digits in python

Question

I want to generate list of all palindromic numbers of 3 digits in python. I can code this in a crude way but is there some intuitive way using list comprehension or itertools etc?

And also, How to do it if it is given the number is k digits instead of just 3?

jadkik94 · Accepted Answer · 2012-05-20T13:46:39.820

11

>>> L = [int("%d%d%d" % (x,y,x)) for x in range(1,10) for y in range(10)]
>>> L
[101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252,
 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414,
 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575,
 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737,
 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898,
 909, 919, 929, 939, 949, 959, 969, 979, 989, 999]

UPDATE: To be more memory and speed efficient, you can replace string formatting and int conversion by x+y*10+x*100. Thanks @larsmans.

UPDATE 2: And that's for k digits!

[int(''.join(map(str, (([x]+list(ys)+[z]+list(ys)[::-1]+[x]) if k%2
                  else ([x]+list(ys)+list(ys)[::-1]+[x])))))
            for x in range(1,10)
            for ys in itertools.permutations(range(10), k/2-1)
            for z in (range(10) if k%2 else (None,))]

And that's optimized to NOT use strings!

[sum([n*(10**i) for i,n in enumerate(([x]+list(ys)+[z]+list(ys)[::-1]+[x]) if k%2
                                else ([x]+list(ys)+list(ys)[::-1]+[x]))])
            for x in range(1,10)
            for ys in itertools.permutations(range(10), k/2-1)
            for z in (range(10) if k%2 else (None,))]

I used permutations, and a different loop for the first digit, which cannot be 0, and the last loop is to add all possible digits in the middle if k%2 == 1 (k is odd).

Suggestions to optimize this are welcome!

edited May 20 '12 at 13:46

answered May 20 '12 at 12:37

jadkik94

7,000
2
30
39

111 is also palindromic, or isn't it? – prongs May 20 '12 at 12:38
@jamylak that's because of the `x!=y` condition, not a biggie – prongs May 20 '12 at 12:40
@jamylak As OP said, I had explicitly omitted these numbers, I thought they were not palindromes, but I looked it up, and it seems they are. – jadkik94 May 20 '12 at 12:43
3

Nice one, though it would be harder to adapt this approach to any number of digits. – omz May 20 '12 at 12:43
2

This can be done an order of magnitude faster if you compute the numbers as `100*x+10*y+x` instead of through string interpolation. – Fred Foo May 20 '12 at 12:49
I'm pretty sure underlying string parsing, a mechanism like this is involved – prongs May 20 '12 at 12:50
@prongs: yes, but the allocation of two strings and the parsing done by `%` is costly. – Fred Foo May 20 '12 at 12:51
1

@larsmans True, thought of that too. And even if `int` does the same, there's the string conversion and format string parsing. It sure is faster. – jadkik94 May 20 '12 at 12:51
@prongs I'm still thinking about `k` digits... It is hard, indeed. – jadkik94 May 20 '12 at 12:53
@jadkik94: Yeah. I always say: "To generalize something you can either pay the `software engineering cost` or you can write a 500 line code" :P – prongs May 20 '12 at 12:55
@prongs Finally, here you have it with `k` digits! – jadkik94 May 20 '12 at 13:47
@jadkik94: Wow! you are quite fast. I also did some coding: [http://pastie.org/3940027](http://pastie.org/3940027) – prongs May 20 '12 at 14:05

score 2 · Answer 2 · answered May 20 '12 at 12:44

For the generalisation to k digits, the most obvious way is to do something like:

palindromes = [x for x in itertools.permutations(string.digits, k) if x == x[::-1]]

But it isn't very efficient - it generates every possible 3-digit number, and discards the ones that aren't palindromes. However, it is possible to generalise solutions like @jadkik94's - what you need to do is generate every combination of half the length (rounding down), and stick the mirror of it on the end:

palindromes = [x + x[::-1] for x in permutations(digits, k//2)]

Will work for all even k - for odd k, you add an extra loop to put all of 0-9 in between the mirrored halves.

score 0 · Answer 3 · answered May 05 '22 at 03:33

The simplest way would be converting the numbers to strings and checking if they are palindromes. To find the palindromes of the "k" number of digits, the list comprehension could be;

palindromes = [num for num in range(pow(10,k-1),pow(10,k)) if str(num) == str(num)[::-1]]

Generate list of all palindromic numbers of 3 digits in python

3 Answers3