I am trying to extract values from an ImageCollection for a set of coordinates in Google Earth Engine using the Python API.

First, I set up my EE environment:

import ee
import pprint

Then I load the image collection and points. Pts is intended to be read into a table with latitude, longitude, and siteID as columns.

img = ee.ImageCollection('UMT/NTSG/v2/MODIS/NPP')
pts = ee.FeatureCollection('users/name/sites_unique.csv');

The image collection contains many images, each representing a year. I would like to iterate through each year and extract the associated NPP (band: ‘annual_npp’) for each point in pts. I anticipate this looking like (pseudocode):

For year in [2010, 2011, 2012, 2013]:
    img_selection = img.select(year = yr)
    output = img_selection.sample(pts, 1000).get('annual_npp').getInfo()

Does anyone know a generalizable way to extract this information from an ImageCollection?

Edit I get something close to the intended outcome in Earth Engine using the following:

var dataset = ee.ImageCollection('UMT/NTSG/v2/MODIS/NPP')
                  .filter(ee.Filter.date('2010-01-01', '2013-01-01'));
var npp = dataset.select('annualNPP').mean();
var out = npp.select(['annualNPP']).sampleRegions({
  collection: pts,
  properties: ['SiteID'],
  geometries: true,
  scale: 1000

1 Answer 1


Maybe something like this?

def sample_point(point, image):
    return image \
        .select(['annualNPP']) \
        ) \
        .first() \
        .set('year', image.date().get('year')) \
        .set('point', point.geometry().coordinates())

def sample_image(image):
    return pts.map(lambda point: sample_point(point, image))

img = ee.ImageCollection('UMT/NTSG/v2/MODIS/NPP')
pts = ee.FeatureCollection([
    ee.Feature(ee.Geometry.Point([-100, 35])),
    ee.Feature(ee.Geometry.Point([-100, 40]))

dataset = ee.ImageCollection('UMT/NTSG/v2/MODIS/NPP') \
    .filter(ee.Filter.date('2010-01-01', '2013-01-01'))

nppFeatures = dataset \
    .map(sample_image) \

npp = [feature.get('properties') for feature in nppFeatures.getInfo().get('features')]

pp = pprint.PrettyPrinter()
  • Perfect, thank you Daniel!
    – Ben
    Nov 29, 2021 at 23:33

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