Forum Discussion
List.Generate how it updates
- 1 year ago
In the second example, by substituting, we can see that
z = [y] + y = alist{[x]} + alist{x} = alist{[x]} + alist{[x]+1}Thus z is always the sum of the trailing two alist items.
In the first example, y adds onto it's prior accumulated value rather than resetting as alist{x}. Let each set of [brackets] denote the value from the previous iteration. Then
y = [y] + alist{x} = [y] + alist([x]+1} = [[y]] + alist{[x]} + alist{[x]+1} = [[[y]]] + alist{[[x]]} + alist{[x]} + alist{[x]+1} = ... = alist{0} + ... + alist{[x]} + alist{[x]+1}This is a cumulative sum rather than just the trailing two items.
The key difference is a self-referential update step. z doesn't accumulate because it never refers to its value from prior iterations.
Here's what the iterations look like as a table: - 1 year ago
Hi Dicken ,
AlexisOlson has already provided a great explanation of why your two List.Generate approaches behave differently. If you’re interested in even more flexibility or want to experiment with more advanced accumulations, here are some extra tips:
- You can manage multiple accumulators by extending your record in List.Generate (e.g., [x=..., y=..., z=...]) and updating each one based on previous iteration values.
- For more complex rolling calculations (like moving averages or conditional accumulations), add fields to your state record and update them step by step—this lets you track any state across iterations.
- If you want to “debug” or visualize how your state evolves, output the full record at each step (not just one variable). This helps you see exactly how values are carried forward or reset.
- Beyond List.Generate, consider List.Accumulate for cases where you want to process a list with an accumulator function. Sometimes it’s a bit more straightforward for cumulative calculations.
If you have a specific accumulation or algorithm in mind beyond what’s already solved, feel free to share it—happy to dive deeper or optimize it together!
Happy coding!
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In the second example, by substituting, we can see that
z = [y] + y
= alist{[x]} + alist{x}
= alist{[x]} + alist{[x]+1}
Thus z is always the sum of the trailing two alist items.
In the first example, y adds onto it's prior accumulated value rather than resetting as alist{x}. Let each set of [brackets] denote the value from the previous iteration. Then
y = [y] + alist{x}
= [y] + alist([x]+1}
= [[y]] + alist{[x]} + alist{[x]+1}
= [[[y]]] + alist{[[x]]} + alist{[x]} + alist{[x]+1}
= ...
= alist{0} + ... + alist{[x]} + alist{[x]+1}
This is a cumulative sum rather than just the trailing two items.
The key difference is a self-referential update step. z doesn't accumulate because it never refers to its value from prior iterations.
Here's what the iterations look like as a table: