# Consecutive identical integer members of a list

Can you help me to find and count sub-sequences of consecutive, identical integer members of a long list (at least 1000 members)? By sub-sequences I mean runs like 0, 0 or 5, 5, 5.

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• What is "at least"? The answers so far are sufficient for lists of a few thousand or so (unless the operation will be done repeatedly in some tight loop, then even that will add up). For large lists they can be significantly beaten.
– ciao
Oct 3, 2015 at 2:39
• @Magnesium ... so, could we help you or is there something left unclear? Maybe you also want to show what you have tried yourself?
– gwr
Oct 5, 2015 at 10:59

## 2 Answers

Modifying the approach by @bbgodfrey one might also use Tally to count all patterns:

list = RandomInteger[{0, 9}, 1000]; (* some integers *)

patternCount = Tally @ Split @ list (* returns a list of {{integer..},count} *)


Now we just take the ones that interest us (e.g. more than one integer):

patternCount2plus = Cases[
patternCount,
{ { Repeated[ _Integer, {2, Infinity} ] }, count_Integer }
]

(* { {{3,3}, 11}, {{4,4}, 8}, ... }*)


This might be sorted by the elements that are repeated and formatted more nicely:

patternCount2plusSorted = SortBy[ patternCount2plus, #[[1,1]]& ];

Grid[
patternCount2plusSorted,
Alignment -> Right,
Background -> { None, { {LightBlue, White} } }, (* alternate rows *)
Frame -> True
]


Giving something like this:

• Perhaps, your second line of code should be patternCount = Tally@Split@list, and your third patternCount2plus = Cases[patternCount, {{Repeated[_Integer, {2, Infinity}]}, count_Integer}]. Oct 2, 2015 at 17:38
• @bbgodfrey Yes, of course, thank you. That happens when one copies like the monks in medieval times. ;-)
– gwr
Oct 2, 2015 at 18:39

If the List is named lst, then

rept = Cases[Split[lst], z_ :> z /; Length[z] > 1]


finds all runs of repeated integers, and

Length[rept]


finds the number of them. Applied to

lst = {1, 2, 3, 3, 4, 5, 5, 5, 6, 4}


they give

(* {{3, 3}, {5, 5, 5}} *)
(* 2 *)


If only the number of repeated runs is desired, then

Count[Split[lst], z_ /; Length[z] > 1]


can be used. For instance,

SeedRandom[5];
Table[Count[Split[RandomInteger[{0, 9}, 10^i]], z_ /; Length[z] > 1], {i, 1, 6}]
(* {2, 9, 93, 890, 9063, 90270} *)