Question

Regroup list elements to generate new lists of 2 pairs

  • 15 April 2019
  • 4 replies
  • 2 views
samisnunu
Badge +2

I have the following list as an example:

Original list:

mylist{0}.attr1 = a1

mylist{0}.attr2 = a2

mylist{0}.attr3 = a3

mylist{1}.attr1 = b1

mylist{1}.attr2 = b2

mylist{1}.attr3 = b3

mylist{2}.attr1 = c1

mylist{2}.attr2 = c2

mylist{2}.attr3 = c3

mylist{3}.attr1 = d1

mylist{3}.attr2 = d2

mylist{3}.attr3 = d3

mylist{4}.attr1 = e1

mylist{4}.attr2 = e2

mylist{4}.attr3 = e3

 

I want to regroup this list elements by recreating multiple lists by grouping the elements into 2 pairs while taking into account the previous element index as follows:

 

New regrouping of elements:

lstGroup1{0}.attr1 = a1

lstGroup1{0}.attr2 = a2

lstGroup1{0}.attr3 = a3

lstGroup1{1}.attr1 = b1

lstGroup1{1}.attr2 = b2

lstGroup1{1}.attr3 = b3

 

lstGroup2{0}.attr1 = b1

lstGroup2{0}.attr2 = b2

lstGroup2{0}.attr3 = b3

lstGroup2{1}.attr1 = c1

lstGroup2{1}.attr2 = c2

lstGroup2{1}.attr3 = c3

 

lstGroup3{0}.attr1 = c1

lstGroup3{0}.attr2 = c2

lstGroup3{0}.attr3 = c3

lstGroup3{1}.attr1 = d1

lstGroup3{1}.attr2 = d2

lstGroup3{1}.attr3 = d3

 

lstGroup4{0}.attr1 = d1

lstGroup4{0}.attr2 = d2

lstGroup4{0}.attr3 = d3

lstGroup4{1}.attr1 = e1

lstGroup4{1}.attr2 = e2

lstGroup4{1}.attr3 = e3

I tried creating a custom looping transformer using a Sampler, but not succeeding, may be a Python code can be easier to do the task..please help.

Note the challenge is the original list length can be varied (dynamic) could be 2, 3, 4, ..50+ etc

4 replies

ebygomm
Rank
Userlevel 1
Badge +10

Do you actually want this output?

lstGroup2{1}.attr1 = b1

lstGroup2{1}.attr2 = b2

lstGroup2{1}.attr3 = b3

lstGroup2{2}.attr1 = c1

lstGroup2{2}.attr2 = c2

lstGroup2{2}.attr3 = c3

 

or this?

lstGroup2{0}.attr1 = b1

lstGroup2{0}.attr2 = b2

lstGroup2{0}.attr3 = b3

lstGroup2{1}.attr1 = c1

lstGroup2{1}.attr2 = c2

lstGroup2{1}.attr3 = c3

samisnunu
Badge +2

oh thanks for the hint @egomm..you're right as the index for the new list will start at 0 , I'll edit the output example..

ebygomm
Rank
Userlevel 1
Badge +10

What is the intention afterwards? It's fairly straightforward to build the list you want in python but you would have to type in all the lstGroup names to expose them. 

import fme
import fmeobjects

def processFeature(feature):
        attr1 = feature.getAttribute('mylist{}.attr1')
        attr2 = feature.getAttribute('mylist{}.attr2')
        attr3 = feature.getAttribute('mylist{}.attr3')
        for i, val in enumerate(attr1):
            if i < len(attr1)-1:
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr1',attr1[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr1',attr1[i+1])
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr2',attr2[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr2',attr2[i+1])
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr3',attr3[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr3',attr3[i+1])

 

samisnunu
Badge +2

What is the intention afterwards? It's fairly straightforward to build the list you want in python but you would have to type in all the lstGroup names to expose them. 

import fme
import fmeobjects

def processFeature(feature):
        attr1 = feature.getAttribute('mylist{}.attr1')
        attr2 = feature.getAttribute('mylist{}.attr2')
        attr3 = feature.getAttribute('mylist{}.attr3')
        for i, val in enumerate(attr1):
            if i < len(attr1)-1:
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr1',attr1[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr1',attr1[i+1])
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr2',attr2[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr2',attr2[i+1])
                feature.setAttribute('lstGroup'+str(i+1)+'{0}.attr3',attr3[i])
                feature.setAttribute('lstGroup'+str(i+1)+'{1}.attr3',attr3[i+1])

 

many thanks!..the big challenge is I found the original list can be varied in length.. so it could be 4 elements or 50+ in some cases (dynamic list length), so the grouping should accommodate the list length.

Reply


Community Stats

28,670
Posts
109,365
Replies
36,531
Members
Terms & ConditionsCookie settings
A product of
Safe Software respectfully acknowledges that we live, learn and work on the traditional and unceded territories of the Kwantlen, Katzie, and Semiahmoo First Nations.