| Re: Meaning of the assignment [message #93674 is a reply to message #93650] |
Tue, 27 September 2016 18:40  |
laura.hike
Messages: 87
Registered: September 2013
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Member |
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I suggest you use the online documentation.
On Wednesday, September 21, 2016 at 2:06:41 AM UTC-7, Sanu wrote:
> On Wednesday, September 21, 2016 at 12:39:12 PM UTC+5:30, Helder wrote:
>> On Wednesday, September 21, 2016 at 8:12:54 AM UTC+2, Sanu wrote:
>>> On Wednesday, September 21, 2016 at 11:26:48 AM UTC+5:30, Sanu wrote:
>>>> exp = 'BYTE(BYTE(b1 LE 650 AND b2 GT 200)*1 + 0)'
>>>>
>>>> For this b< 650 and b2> 200 will be assigned as class 1.
>>>> What is the meaning of 0 which is added?????
>>>> Kindly help.
>>>
>>> Similar thing i found on thid
>>>
>>> exp = 'BYTE(BYTE(b1 GE 0.4)*21 OR BYTE((b2/b3) GE 4.0)*21)'
>>> What is the meaning of *21
>>
>> Well, I also don't understand what you have. But you should maybe provide some more information: Normally a string is a string and will not be computed unless you use execute(). Is this what you are doing? Where is this exp being used?
>>
>> If the assignment were not a string, then you can figure out things as they were described in the post from yesterday.
>> Take things apart, a bit like an onion... layer by layer.
>> Let's consider this expression:
>> BYTE(BYTE(b1 GE 0.4)*21 OR BYTE((b2/b3) GE 4.0)*21)
>> You have at the most inner parts expressions like: b1 GE 0.4
>> IDL evaluates if b1 is greater or equal to 0.4. If it is greater or equal, it will "substitute" in the expression this part with a 1. If not, it will put a 0. IDL returns from such an operation a byte value, so the conversion to byte using the function BYTE() is useless. The expression you gave can be simplified to:
>> BYTE((b1 GE 0.4)*21 OR ((b2/b3) GE 4.0)*21)
>> Now it gets a bit tricky.
>> on the left side of the OR you have
>> (b1 GE 0.4)*21
>> on the right side
>> ((b2/b3) GE 4.0)*21
>>
>> You have to evaluate these two separately. The first will give you:
>> 1 * 21 ---> if b1 is greater or equal than 0.4
>> 0 * 21 ---> if b1 is smaller than 0.4
>> The second will give you
>> 1 * 21 ---> if b2/b3 is greater or equal than 4.0
>> 0 * 21 ---> if b2/b3 is smaller than 4.0
>>
>> The result of each operation is an integer. Either 0 or 21.
>>
>> The OR operation will return either 0 or 21 depending on the input:
>> 0 OR 0 ---> will give 0
>> 21 OR 0 ---> will give 21
>> 0 OR 21 ---> will give 21
>> 21 OR 21 ---> will give 21
>>
>> The outer shell of the onion is a byte function that converts the previous result to a byte value.
>> So depending on the last operation, you will either get a 0 or a 21 as a byte instead of an integer.
>>
>> You should be able to figure out the rest alone. If not, I seriously recommend some basic programming reference book.
>>
>> Helder
>
>
> Thank You Helder.
> Do you have some e-book which I can refer for ENVI IDL programming.
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