To the collection of possible approaches here comes one more:

data = {0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0};
ComponentMeasurements[Image[1 - {data}], "Count"]

Really a nice solution. I learned quite a few things from all reactions.

Best regards, Chiel Geeraert (Netherlands)

Here is my workaround :

In[20]:= Test = {1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0,
    0, 0, 0, 0, 1, 0, 0};
(* The desired output is the number of zero's between the ones so
{4,2,5,7} *)

In[15]:= Test = Flatten[Position[Test, 1]]

Out[15]= {1, 6, 9, 15, 23}

In[17]:= Result = 
 Table[Test[[k + 1]] - Test[[k]] - 1, {k, 1, Length[Test] - 1}]

Out[17]= {4, 2, 5, 7}

Thank you very much. In the meantime I have found a simple , not very sophisticated, workaround solution but it works.

I will try your suggestion and try to understand it.

Best regards, Chiel Geeraert

Can be done even shorter:

SequenceCases[{1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0}, x : {0 ..} :> Length[x]]

You may try SequenceSplit:

l = {1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0};
SequenceSplit[l, {1 ..}]
Map[Length, %]

This counts also the trailing three zeros. You can correct that problem if you really need it.