Can you suggest an approach to select January Landsat 5 data between 2000 to 2010 and calculate its mean?

In other words, the mean of the all January images from 2000 to 2010.


You nee to use ee.Filter.calendarRange(start,end,field) to filter by all Januaries from 2000 to 2010.

field could be:

  • field (String, default: "day_of_year"): The calendar field to filter over. Options are: 'year', 'month', 'hour', 'minute', 'day_of_year', 'day_of_month', and 'day_of_week'.

So, the code is:

var imageCollection = ee.ImageCollection("LANDSAT/LT05/C01/T1_TOA");

// Use ee.Filter.calendarRange to filter by year and month

var img = imageCollection.filter(ee.Filter.calendarRange(2000,2010,'year'))

// reduce image collection with mean()
var mean = img.mean();

// Set visualization parameters
var vizParams = {
  bands: ['B3', 'B2', 'B1'],
  min: 0,
  max: 0.5,
  gamma: [0.95, 1.1, 1]

// add layer

This code process specifically your task description, doesn't consider cloud mask or a cloud coverage filter.

  • Thank You for your valuable answer. Thanks a lot, but i have a question here, var img = imageCollection.filter(ee.Filter.calendarRange(2000,2010,'year')) .filter(ee.Filter.calendarRange(1,1,'month')); It is taking only January 2000 and January 2010, – P_P Sep 23 '17 at 9:11
  • var img = imageCollection.filter(ee.Filter.calendarRange(2000,2010,'year')) .filter(ee.Filter.calendarRange(1,1,'month')); It is filtering all images from January 2000 to January2010. I need only january from every year. Thank you – P_P Sep 23 '17 at 9:34
  • @P_P weird, I test it and takes every year. Try with .filterDate(“2000-01-01” , “2010-01-31”) instead of first filter. – aldo_tapia Sep 23 '17 at 12:02
  • Thank You for your kind reply. I started with this .filterDate(“2000-01-01” , “2010-01-31”) but its is taking all images between these dates. – P_P Sep 24 '17 at 5:24
  • @P_P yes, but I said as remplace of first filter (year filter), you need to keep the second filter (month filter). – aldo_tapia Sep 24 '17 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.