Counting subsequences (i.e., patterns) within a list?

data = {1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1};
sub = {{1, 1}, {1, 2}, {2, 1}, {2, 2}};

Count[Partition[data, 2, 1], Alternatives @@ sub]

10

Or each separately:

Count[Partition[data, 2, 1], #] & /@ sub

{2, 3, 3, 2}

---

More general, for different length subsequences:

Length[ReplaceList[data, {___, ##, ___} :> {}]] & @@@ sub

{2, 3, 3, 2}

 Length[ReplaceList[data, {___, ##, ___} :> {}]] & @@@ {{1, 2, 1}, {1}}

{2, 6}

String processing is often useful for these problems if it can be applied.

data = {1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1};

pat = {{1, 1}, {1, 2}, {2, 1}, {2, 2}};

ts = ToString @ Row[#, ","] &;

StringCount[ts@data, ts /@ pat, Overlaps -> True]

10

Individually:

With[{ds = ts@data},
  StringCount[ds, ts@#, Overlaps -> True] & /@ pat
]

{2, 3, 3, 2}

Or perhaps better:

StringCases[ts@data, ts@# -> # & /@ pat, Overlaps -> True] // Tally

{{{1, 1}, 2}, {{1, 2}, 3}, {{2, 1}, 3}, {{2, 2}, 2}}

Another way with ListCorrelate used with proper arguments, for example :

data = {1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1};
sub = {{1, 1}, {1, 2}, {2, 1}, {2, 2}};

Total@Boole@
    ListCorrelate[#, data, {1, -1}, "paddingUselessHere", Equal, And] & /@ sub

{2, 3, 3, 2}

Another example

Let's say you're just looking for the longer pattern {1,2,1} :

Total@Boole@ListCorrelate[{1, 2, 1}, data, {1, -1}, "paddingUselessHere", Equal, And]

2

Explanation

It is easy to understand what ListCorrelate does with this example :

ListCorrelate[{p1, p2}, {a, b, c, d}, {1, -1}, "whatever", f, g]

{g[f[p1, a], f[p2, b]], g[f[p1, b], f[p2, c]], g[f[p1, c], f[p2, d]]}

(When the 4th argument of ListCorrelate is {1,-1} (default value), the 5th argument (padding) is useless, that's why i set it to "whatever".)