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Tile Error in Google Earth Engine?

Geographic Information Systems Asked on December 28, 2020

The below code has written for ASTER radiometric correction in Google Earth Engine but in visualization step, returns this error:
enter image description here

code link:https://code.earthengine.google.com/3ccaaaa0e9ce68bdf80b8056be5d057d

Map.centerObject(table);
Map.addLayer(table)
 
var ast = ee.ImageCollection("ASTER/AST_L1T_003")
.filterDate('2000-01-01','2020-01-01')
.filterBounds(table)
.filter(ee.Filter.calendarRange(7,9,'month'))
.select('B0[4-9]')
.map(function(img){
  
  var gain4 = ee.Image(ee.Number(img.select('B04').get('GAIN_COEFFICIENT_B04')));
  var gain5 = ee.Image(ee.Number(img.select('B05').get('GAIN_COEFFICIENT_B05')));
  var gain6 = ee.Image(ee.Number(img.select('B06').get('GAIN_COEFFICIENT_B06')));
  var gain7 = ee.Image(ee.Number(img.select('B07').get('GAIN_COEFFICIENT_B07')));
  var gain8 = ee.Image(ee.Number(img.select('B08').get('GAIN_COEFFICIENT_B08')));
  var gain9 = ee.Image(ee.Number(img.select('B08').get('GAIN_COEFFICIENT_B09')));
  
  
  var b4 = img.select('B04').multiply(gain4).clip(table);
  var b5 = img.select('B05').multiply(gain5).clip(table);
  var b6 = img.select('B06').multiply(gain6).clip(table);
  var b7 = img.select('B07').multiply(gain7).clip(table);
  var b8 = img.select('B08').multiply(gain8).clip(table);
  var b9 = img.select('B09').multiply(gain9).clip(table);
  
  
  var stack = ee.Image.cat([b4,b5,b6,b7,b8,b9]);
  return stack
}).median();


print(ast);

Map.addLayer(ast)

One Answer

Not every image in the images you've selected has the properties GAIN_COEFFICIENT_B04 and so on. For example, ASTER/AST_L1T_003/20100709070749 has only

GAIN_COEFFICIENT_B01: 0.676
GAIN_COEFFICIENT_B02: 0.708
GAIN_COEFFICIENT_B10: 0.006822
GAIN_COEFFICIENT_B11: 0.00678
GAIN_COEFFICIENT_B12: 0.00659
GAIN_COEFFICIENT_B13: 0.005693
GAIN_COEFFICIENT_B14: 0.005225
GAIN_COEFFICIENT_B3N: 0.862

Thus, .get('GAIN_COEFFICIENT_B04') returns null, and your image calculation fails because null is not a valid value for a constant image.

You can add a filter on the image collection to ensure those properties are present:

.filter(ee.Filter.notNull([
  'GAIN_COEFFICIENT_B04',
  'GAIN_COEFFICIENT_B05',
  'GAIN_COEFFICIENT_B06',
  'GAIN_COEFFICIENT_B07',
  'GAIN_COEFFICIENT_B08',
  'GAIN_COEFFICIENT_B09',
]))

I noticed some other improvements that could be made to your code.

  var gain4 = ee.Image(ee.Number(img.select('B04').get('GAIN_COEFFICIENT_B04')));

Here, the .select('B04') does nothing (since you're getting a property, and properties are independent of bands). Also, there are variants of .get() that eliminate needing to write ee.Number:

  var gain4 = ee.Image(img.getNumber('GAIN_COEFFICIENT_B04'));

But this can be even simpler: it isn't necessary to write ee.Image(constant) in many cases; you can write a number and it will be converted to an image.

  var b4 = img.select('B04').multiply(img.getNumber('GAIN_COEFFICIENT_B04')).clip(table);

Finally, it is unnecessarily inefficient to .clip(table) the individual images rather than the result, and when you are clipping to a table/collection you should use the more efficient .clipToCollection(table) (though this doesn't matter since you only have one feature).

.map(function(img){
  ...
}).median().clipToCollection(table);

Putting it all together:

var ast = ee.ImageCollection("ASTER/AST_L1T_003")
  .filterDate('2000-01-01','2020-01-01')
  .filterBounds(table)
  .filter(ee.Filter.calendarRange(7,9,'month'))
  .select('B0[4-9]')
  .filter(ee.Filter.notNull([
    'GAIN_COEFFICIENT_B04',
    'GAIN_COEFFICIENT_B05',
    'GAIN_COEFFICIENT_B06',
    'GAIN_COEFFICIENT_B07',
    'GAIN_COEFFICIENT_B08',
    'GAIN_COEFFICIENT_B09',
  ]))
  .map(function(img){
    return ee.Image.cat([
      img.select('B04').multiply(img.getNumber('GAIN_COEFFICIENT_B04')),
      img.select('B05').multiply(img.getNumber('GAIN_COEFFICIENT_B05')),
      img.select('B06').multiply(img.getNumber('GAIN_COEFFICIENT_B06')),
      img.select('B07').multiply(img.getNumber('GAIN_COEFFICIENT_B07')),
      img.select('B08').multiply(img.getNumber('GAIN_COEFFICIENT_B08')),
      img.select('B09').multiply(img.getNumber('GAIN_COEFFICIENT_B09')),
    ]);
  })
  .median()
  .clipToCollection(table);

https://code.earthengine.google.com/3ba97a7c2009def6043c614c5aeae4f7

Correct answer by Kevin Reid on December 28, 2020

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