Alternating Lists
February 1, 2019
Today’s exercise is a simple task in list manipulation:
Write a program that takes one or more lists and returns a single list containing the elements of the input lists, taken alternately from the lists. If the lists are of different lengths, continue taking items from the lists that remain.
For instance, given the inputs (1 2 3 4 5), (a b c) and (w x y z), the desired output is (1 a w 2 b x 3 c y 4 z 5).
Your task is to write a program to take items alternately from multiple lists. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
In Ruby. Assumes elements are in the lists are not nil.
Mumps version
LISTS(LISTS) ; S MAX=0 F I=1:1:$L(LISTS,"|") S LEN=$L($P(LISTS,"|",I),";"),MAX=$S(LEN>MAX:LEN,1:MAX) S LIST="" F J=1:1:MAX F I=1:1:$L(LISTS,"|") S CHAR=$P($P(LISTS,"|",I),";",J) S:CHAR'="" LIST=LIST_CHAR_" " Q $E(LIST,1,$L(LIST)-1)1 a w 2 b x 3 c y 4 z 5
— Here’s some Haskell, I’ll leave it to enthusiasts to convert to pointfree style.
-- Interleave elements from a list of lists. -- Corecursive so works with infinite lists. interleave s = aux (filter (not . null) s) where aux [] = [] aux s = map head s ++ interleave (map tail s) -- Specialize to Int to avoid problems with [] iinterleave = interleave :: [[Int]] -> [Int] main = print (iinterleave [[],[]]) >> print (iinterleave [[1,2,3],[]]) >> print (iinterleave [[],[1,2,3]]) >> print (iinterleave [[1,2,3],[4,5,6]]) >> print (iinterleave [[1,2,3],[4,5,6],[7,8,9]]) >> print (take 10 (iinterleave [[1,4..],[2,5..],[3,6..]]))In Python.
from itertools import zip_longest sentinel = object() def alt_lists(*lists): return [xi for x in zip_longest(*lists, fillvalue=sentinel) for xi in x if xi is not sentinel] print(alt_lists([1,2,3,4,5], ["a", "b", "c"], ["w", "x", "y", "z"]))I think my solution is a bit simpler:
(define (alternate x . y)
(cond
((null? y) x)
((null? x)
(apply alternate y))
(else
(cons
(car x)
(apply
alternate
(append y (list (cdr x))))))))
Here’s another Haskell solution. (In pointfree style! :-)
{-# OPTIONS_GHC -fno-warn-type-defaults #-} import Data.List (transpose) alternate :: [[a]] -> [a] alternate = concat . transpose main :: IO () main = do -- From the original problem. print $ alternate [['1', '2', '3', '4', '5'], ['a', 'b', 'c'], ['w', 'x', 'y', 'z']] -- From matthew's solution. print $ alternate [[], [] :: [()]] print $ alternate [[1, 2, 3], []] print $ alternate [[], [1, 2, 3]] print $ alternate [[1, 2, 3], [4, 5, 6]] print $ alternate [[1, 2, 3], [4, 5, 6], [7, 8, 9]] print $ take 10 $ alternate [[1, 4..], [2, 5..], [3, 6..]]@Globules: very nice – it would be good to have a pointfree definition of transpose as well!
Here’s another Haskell version, this one is a simplification of a modification of the Haskell library definition of transpose. I’ve not seen that list comprehension with partial matches idiom before:
And here’s a pointfree Haskell solution that doesn’t seem too incomprehensible:
Here’s an simple-minded scheme solution with filter and map:
(define (not-null? x) (not (null? x)))
(define (splice . lists)
(let ((flists (filter not-null? lists)))
(if (not-null? flists)
(append (map car flists)
(apply splice (map cdr flists)))
'())))
Here’s a solution in C that uses linked lists.
#include <ctype.h> #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct node { char* val; struct node* next; } node_t; void print_list(node_t* list) { while (list != NULL) { printf("%s", list->val); if (list->next != NULL) printf("->"); list = list->next; } printf("\n"); } void free_list(node_t* list) { while (list != NULL) { node_t* next = list->next; free(list); list = next; } } void alternate(node_t** lists, int n, node_t** output) { node_t* list = NULL; int done = 0; while (!done) { done = 1; for (int i = 0; i < n; ++i) { if (lists[i] == NULL) continue; done = 0; node_t* node = malloc(sizeof(node_t)); node->val = lists[i]->val; node->next = list; list = node; lists[i] = lists[i]->next; } } node_t* prev = NULL; node_t* node = NULL; node_t* next = list; while (next != NULL) { prev = node; node = next; next = next->next; node->next = prev; } list = node; *output = list; } int main(int argc, char* argv[]) { int n = 0; int capacity = 1; node_t** lists = malloc(sizeof(node_t*) * capacity); node_t* list = NULL; char* sep = argv[1]; for (int i = argc - 1; i >= 1; --i) { char* arg = argv[i]; if (!strcmp(sep, arg)) { if (list == NULL) continue; ++n; if (n > capacity) { capacity *= 2; lists = realloc(lists, sizeof(node_t*) * capacity); } lists[n - 1] = list; list = NULL; continue; } node_t* node = malloc(sizeof(node_t)); node->next = list; node->val = arg; list = node; } for (int i = 0; i < n / 2; ++i) { node_t* tmp = lists[i]; lists[i] = lists[n - i - 1]; lists[n - i - 1] = tmp; } node_t* alternating = NULL; alternate(lists, n, &alternating); print_list(alternating); for (int i = 0; i < n; ++i) { free_list(lists[i]); } free_list(alternating); free(lists); return EXIT_SUCCESS; }Example:
Here’s a more raw, packageless python implementation I thought of:
def alternator(lists):
max_len = 0
if name == “main“:
lists = [[1,2,3,4,5],[‘a’,’b’,’c’],[‘w’,’x’,’y’,’z’]]
alternator(lists)
Here’s a solution in Common Lisp.
(defun alternate (&rest lists) (labels ((rec (result lists) (let ((lists (remove-if #'null lists))) (if (null lists) result (rec (reduce #'(lambda (a b) (cons b a)) (mapcar #'car lists) :initial-value result) (mapcar #'cdr lists)))))) (reverse (rec (list) lists))))Example.
Here’s a solution in C that uses arrays.
#include <stdlib.h> #include <stdio.h> #include <string.h> #define INIT_CAPACITY 1 typedef struct array { int n; char** data; } array_t; void print_array(array_t* array) { printf("["); for (int i = 0; i < array->n; ++i) { if (i > 0) printf(" "); printf("%s", array->data[i]); } printf("]\n"); } void alternate(array_t* arrays, int n, array_t* output) { output->n = 0; for (int i = 0; i < n; ++i) output->n += arrays[i].n; output->data = malloc(sizeof(char*) * output->n); int col = 0; // index in input data (column of jagged array) int idx = 0; // index in output data while (idx < output->n) { for (int i = 0; i < n; ++i) { if (col >= arrays[i].n) continue; output->data[idx++] = arrays[i].data[col]; } ++col; } } int main(int argc, char* argv[]) { char* sep = argv[1]; int n = 0; for (int i = 1; i < argc; ++i) if (!strcmp(sep, argv[i])) ++n; array_t* arrays = malloc(sizeof(array_t) * n); int capacity = INIT_CAPACITY; for (int i = 0; i < n; ++i) { arrays[i].n = 0; arrays[i].data = malloc(sizeof(char*) * capacity); } int pos = 0; for (int i = 2; i < argc; ++i) { if (!strcmp(sep, argv[i])) { ++pos; capacity = INIT_CAPACITY; continue; } ++(arrays[pos].n); if (arrays[pos].n > capacity) { capacity *= 2; arrays[pos].data = realloc(arrays[pos].data, sizeof(char*) * capacity); } arrays[pos].data[arrays[pos].n - 1] = argv[i]; } array_t alternating; alternate(arrays, n, &alternating); print_array(&alternating); for (int i = 0; i < n; ++i) free(arrays[i].data); free(arrays); free(alternating.data); return EXIT_SUCCESS; }Example:
Here’s a solution in Python.
def alternate(l): l = [list(reversed(x)) for x in l] output = [] while l: l = [x for x in l if x] for x in l: output.append(x.pop()) return output l = [[ 1 , 2 , 3 , 4, 5], ['a', 'b', 'c'], ['w', 'x', 'y', 'z']] print(alternate(l))Output:
Klong version (Does not handle empty list for list entry) Code: f::{[r l2]; r::x; l2::[]; {r::r,*x; l2::l2,(,(1_x))}'y; (,r),(,l2)} g2::{[t]; :[0=+/#'y;x;g2((t::f(x;y))@0; t@1)]} g::{g2([]; x)} Output: {.p(x," --> ",g(x))}'[[[1 2 3 4 5] ["a" "b" "c"] ["W" "X" "Y" "Z"]] [[] []] [[1 2 3] []] [[] [1 2 3]] [[1 2 3] [4 5 6]] [[1 2 3] [4 5 6] [7 8 9]] [[1 4 5 6 7 8 9 10 11 12] [2 5 6 7 8 9 10 11 12] [3 6 7 8 9 10 11 12 13 14]]] [[1 2 3 4 5] [a b c] [W X Y Z] --> 1 a W 2 b X 3 c Y 4 Z 5] [[] [] --> ] [[1 2 3] [] --> 1 2 3] [[] [1 2 3] --> 1 2 3] [[1 2 3] [4 5 6] --> 1 4 2 5 3 6] [[1 2 3] [4 5 6] [7 8 9] --> 1 4 7 2 5 8 3 6 9] [[1 4 5 6 7 8 9 10 11 12] [2 5 6 7 8 9 10 11 12] [3 6 7 8 9 10 11 12 13 14] --> 1 2 3 4 5 6 5 6 7 6 7 8 7 8 9 8 9 10 9 10 11 10 11 12 11 12 13 12 14]