Zhigang Xu

Ah, the explanation to the above puzzle (i.e., NaN vs Double) lies in that whenever NaN is concatenated with other classes, it will be converted to zero first. Thus,

 uint8(nan)

ans =

    0

 int32(nan)

ans =

           0

[nan uint16([1 2])]

ans =

      0 1 2

whereas
[nan double([1 2])]

ans =

   NaN 1 2

A lesson to remember is that NaN is only a not-a-number within double class. It will become zero when it is converted to other classes.

Matt,

Thanks for your comments and rating. Your suggestion can evidently make the codes more concise. However, using logical indexing is usually faster than FIND, as the Matlab Editor would advise. For this reason, I am hesitating to implement it for now. I tested your suggestion with a large input array (x=randi(999, 7e6,1)), and did not find any significant improvement in speed, although it did not go slower either. I am wondering if you can supply me your test script to show the significant difference in speed. I will be very happy to implement your suggestion after I understand why it can do so. Thanks!

Zhigang

I tested fstr2num with a large string (>1,000,000) of numbers such as: 11 44 44 11 33 13 33 22 11 44. Separated by a single space. str2num takes 3-4 seconds to convert them to numeric. fstr2num took less than a second, but it produced many more numbers than actually are, probably breaking some of the numbers to single numbers (1 2 3 4).

In short, it doesn't work well. hope you can fix it because the world needs it!

Ah, the explanation to the above puzzle (i.e., NaN vs Double) lies in that whenever NaN is concatenated with other classes, it will be converted to zero first. Thus,

 uint8(nan)

ans =

    0

 int32(nan)

ans =

           0

[nan uint16([1 2])]

ans =

      0 1 2

whereas
[nan double([1 2])]

ans =

   NaN 1 2

A lesson to remember is that NaN is only a not-a-number within double class. It will become zero when it is converted to other classes.

To make the function work for other integer class, I would suggest to modify the two lines:

d=diff([NaN; x]);
d=[true; d~=0];

to:

d=diff(x);
d=[true; d~=0];

Otherwise unexpected result is obtained with
[u p]=numunique(uint8([0 1]))