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lib.lists.commonPrefix: init
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commit
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2 changed files with 60 additions and 1 deletions
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@ -3,7 +3,7 @@
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{ lib }:
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{ lib }:
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let
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let
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inherit (lib.strings) toInt;
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inherit (lib.strings) toInt;
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inherit (lib.trivial) compare min;
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inherit (lib.trivial) compare min id;
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inherit (lib.attrsets) mapAttrs;
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inherit (lib.attrsets) mapAttrs;
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in
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in
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rec {
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rec {
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@ -663,6 +663,32 @@ rec {
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else if start + count > len then len - start
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else if start + count > len then len - start
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else count);
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else count);
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/* The common prefix of two lists.
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Type: commonPrefix :: [a] -> [a] -> [a]
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Example:
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commonPrefix [ 1 2 3 4 5 6 ] [ 1 2 4 8 ]
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=> [ 1 2 ]
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commonPrefix [ 1 2 3 ] [ 1 2 3 4 5 ]
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=> [ 1 2 3 ]
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commonPrefix [ 1 2 3 ] [ 4 5 6 ]
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=> [ ]
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*/
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commonPrefix =
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list1:
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list2:
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let
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# Zip the lists together into a list of booleans whether each element matches
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matchings = zipListsWith (fst: snd: fst != snd) list1 list2;
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# Find the first index where the elements don't match,
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# which will then also be the length of the common prefix.
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# If all elements match, we fall back to the length of the zipped list,
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# which is the same as the length of the smaller list.
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commonPrefixLength = findFirstIndex id (length matchings) matchings;
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in
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take commonPrefixLength list1;
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/* Return the last element of a list.
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/* Return the last element of a list.
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This function throws an error if the list is empty.
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This function throws an error if the list is empty.
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@ -488,6 +488,39 @@ runTests {
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expected = { a = [ 2 3 ]; b = [7]; c = [8];};
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expected = { a = [ 2 3 ]; b = [7]; c = [8];};
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};
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};
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testListCommonPrefixExample1 = {
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expr = lists.commonPrefix [ 1 2 3 4 5 6 ] [ 1 2 4 8 ];
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expected = [ 1 2 ];
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};
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testListCommonPrefixExample2 = {
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expr = lists.commonPrefix [ 1 2 3 ] [ 1 2 3 4 5 ];
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expected = [ 1 2 3 ];
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};
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testListCommonPrefixExample3 = {
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expr = lists.commonPrefix [ 1 2 3 ] [ 4 5 6 ];
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expected = [ ];
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};
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testListCommonPrefixEmpty = {
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expr = lists.commonPrefix [ ] [ 1 2 3 ];
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expected = [ ];
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};
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testListCommonPrefixSame = {
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expr = lists.commonPrefix [ 1 2 3 ] [ 1 2 3 ];
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expected = [ 1 2 3 ];
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};
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testListCommonPrefixLazy = {
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expr = lists.commonPrefix [ 1 ] [ 1 (abort "lib.lists.commonPrefix shouldn't evaluate this")];
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expected = [ 1 ];
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};
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# This would stack overflow if `commonPrefix` were implemented using recursion
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testListCommonPrefixLong =
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let
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longList = genList (n: n) 100000;
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in {
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expr = lists.commonPrefix longList longList;
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expected = longList;
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};
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testSort = {
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testSort = {
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expr = sort builtins.lessThan [ 40 2 30 42 ];
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expr = sort builtins.lessThan [ 40 2 30 42 ];
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expected = [2 30 40 42];
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expected = [2 30 40 42];
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