How to count sequential NaN values

Hello,

I have a matrix

x = [1,NaN,3;4,NaN,6;7,NaN,NaN;1,2,NaN;3,NaN,3;NaN,9,1]

creates

     1   NaN     3
     4   NaN     6
     7   NaN   NaN
     1     2   NaN
     3   NaN     3
   NaN     9     1

I want to be able to write to a variable every time a series of NaN (1 or more) occur in a column, and how many.

For example for the above code I want it to calculate

1 occurance for column 1 for 1 instance

1 occurance for column 2 for 3 instances

1 occurance for column 2 for 1 instance

1 occurance for column 3 for 2 instances

Do I put in a for loop for each column, search for NaN, then use "Diff" to find a difference of zero, then record how many gaps there are?

Or, do I delete every NaN value, then search for gaps in the data and use "Diff" like below?

threshold = 5;

gap = diff(data1);

idx = find(gap>threshold);

Thank you.

Answer 1

[dpb]

Presuming I get your intent,

>> sum(diff([zeros(1,3);isnan(x)])==1)
ans =
   1     2     1
>>

ADDENDUM

Locations...

>> [i j]=find(diff([zeros(1,3);isnan(x)])==1);
>> [i j]
ans =
   6     1
   1     2
   5     2
   3     3
>>

[Steven Crisp]

Yes kind of. I want the output for every sequence of nan's to be

col 1, 1 instance, row 6

col 2, 3 instance, row 1

col 2, 1 instance, row 5

col 3, 2 instance, row 3

Where I just want the sequential NaN's from columns, with no relation between columns. A new column is a whole new assortment.

Does that make sense?