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Before, I asked that for search or filter values in two-segment list in Google Earth Engine at this link(Search or filter values in two-segment list in Google Earth Engine) and this question is answered by Daniel Wiell with very good explanation (https://gis.stackexchange.com/a/392774/112517).

I want to search for value in first column of the list, get finding result from second column and save to result to another new list. I want to search value from first column of lists and get second column value as result, for example, I want to search 101 and I get 12155 and save 12155 to new list.

var list = ee.List([
  [101, 12155],
  [123, 12333],
  [102, 12157],
  [102, 12222],
  [102, 12244],
  [103, 13333],
  [103, 12158], 
  ]) // This list contains thousands of paired values

If this list have repeated values such(102), I don't know how to achieve other paired values(12222, 12244). For example, if I search for 102, I get only 12157 as result.

How can I achieve other paired values(12222, 12244) and append results to new list?

1
  • 1
    I'm just wondering, what is the specific use case you have on Earth Engine with this list querying?
    – JonasV
    Apr 21 at 9:38
2

You can try following script. It finds indices of repeated values (in this case 102) in column 0 of list variable. Finally, these indices are used for printing repeated paired values in corresponding list.

var list = ee.List([
      [101, 12155],
      [123, 12333],
      [102, 12157],
      [102, 12222],
      [102, 12244],
      [103, 13333],
      [103, 12158], 
      ]); // This list contains thousands of paired values

print("original list", list);

var Column = 0;

var keyValues = list.map(function(inner) {
    return ee.List(inner).get(Column);
});

print("selecting column where 102 is", keyValues);

var num = ee.Number(102);

var seq = ee.List.sequence(0, keyValues.size().subtract(1));

var indexRep = seq.map(function (e) {

  var list = keyValues.map(function (ele) {
  
    return ee.Algorithms.If(ee.Number(ele).eq(num), ele, 0);
  
  });

  return ee.Algorithms.If(ee.Number(ee.List(list).get(e)).neq(0), e, -1);

}).removeAll([-1]);

print("index of 102 values", indexRep);

var pairedRep = indexRep.map(function (ele) {
  
  return list.get(ele);
  
});

print("paired values with 102", pairedRep);

Result of running above script in GEE console editor is as follows:

paired values with 102
[[102,12157],[102,12222],[102,12244]]

Editing Note:

Modified above script for printing only values (102 as test key value):

var list = ee.List([
  [101, 12155],
  [123, 12333],
  [102, 12157],
  [102, 12222],
  [102, 12244],
  [103, 13333],
  [103, 12158], 
  ]); // This list contains thousands of paired values

print("original list", list);

var Column = 0;

var keyValues = list.map(function(inner) {
    return ee.List(inner).get(Column);
});

print("selecting column where 102 is", keyValues);

var num = ee.Number(102);

var seq = ee.List.sequence(0, keyValues.size().subtract(1));

var indexRep = seq.map(function (e) {

  var list = keyValues.map(function (ele) {
  
    return ee.Algorithms.If(ee.Number(ele).eq(num), ele, 0);
  
  });

  return ee.Algorithms.If(ee.Number(ee.List(list).get(e)).neq(0), e, -1);

}).removeAll([-1]);

print("index of 102 values", indexRep);

var valuesPairedRep = indexRep.map(function (ele) {
  
  return ee.List(list.get(ele)).get(1);
  
});

print("values for repeated 102", valuesPairedRep);

Modified script by using 'iterate' method (101 as test key value):

var list = ee.List([
  [101, 12155],
  [123, 12333],
  [102, 12157],
  [102, 12222],
  [102, 12244],
  [103, 13333],
  [103, 12158], 
  ]); // This list contains thousands of paired values

print("original list", list);

var Column = 0;

var keyValues = list.map(function(inner) {
    return ee.List(inner).get(Column);
});

print("selecting column where 101 is", keyValues);

var num = ee.Number(101);

var new_seq = ee.List.sequence(0, 6);

var first = new_seq;

var appendValues = function(iter, previous) {
  
  return ee.List(previous)
    .set(iter, ee.Algorithms
    .If(ee.Number(ee.List(list.get(iter)).get(0)).eq(num), iter, -1));

};
  
var indexRep = ee.List(new_seq.iterate(appendValues, first))
  .removeAll([-1]);

print("indices for 101", indexRep);

var valuesPairedRep = indexRep.map(function (ele) {
  
  return ee.List(list.get(ele)).get(1);
  
});

print("values for 101", valuesPairedRep);
5
  • Thanks Xunilk, but it does not work for 101, and i only need the result value. I have thousands of paired values like above list, i must search value from first column and get result from second column. It must be iterate for all list values.
    – Mehmet C
    Apr 21 at 16:18
  • 1
    I fixed the first issue. I forgot that the list also has index 0. I will see other issues later because I have a shorter solution with iterate.
    – xunilk
    Apr 21 at 19:30
  • 1
    I put an Editing Note in my answer with another script by using 'iterate' list method.
    – xunilk
    Apr 21 at 21:45
  • Thanks Xunilk, this code works for one value. But, i can not complete this code to search all of the first values, find second column value and add all of them to a new list.
    – Mehmet C
    Apr 23 at 18:48
  • Dear Xunilk, how can i iterate all of first ID s and save secondary values to a new list? Your code run only for one search and save it to a list
    – Mehmet C
    May 14 at 7:46

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