Most efficient way to replace/ update the values multiple rows and columns of a Dataset

Question

Suppose we wish to replace the cols "b" and "c" of rows 3 to 5 in the following dataset to b and c respectively:

dataset = Dataset[{
   <|"a" -> 1, "b" -> "x", "c" -> {1}|>,
   <|"a" -> 2, "b" -> "y", "c" -> {2, 3}|>,
   <|"a" -> 3, "b" -> "z", "c" -> {3}|>,
   <|"a" -> 4, "b" -> "x", "c" -> {4, 5}|>,
   <|"a" -> 5, "b" -> "y", "c" -> {5, 6, 7}|>,
   <|"a" -> 6, "b" -> "z", "c" -> {}|>}]

1. What is the most efficient way to do it at scale, i.e. assuming the replacements are going to be across multiple/ all rows in large datasets.
2. Is is possible to do it inline, i.e., in the current dataset, without having to recreate a new dataset?
3. It would be preferrable if the updates are computed (i.e., not constants as shown in this toy example)

Answer 1

You can use ReplacePart directly on the dataset but it seems a tedious/inefficient way of doing this. It is scalable as long as you can use patterns or Alternative to select the positions you want to replace. I expect someone else will be able to provide a more efficient method that works directly on the dataset without first converting to a list of associations.

dataset=ReplacePart[dataset, {{3 | 4 | 5, "b"} -> b, {3 | 4 | 5, "c"} -> c}]

Answer 2

I mentioned it in the comments but basically I like to keep it pretty simple and while I'm not sure if it is the fastest way to do it it is nice and clean and the memory usage is not as bad as it would seem due to Mathematica's good choice of persistent data structures to build Associations on.

In[1]:= MemoryInUse[] / 1000000 // N
Out[1]= 96.5472
In[2]:= data = ResourceData["US State Income"]
In[3]:= MemoryInUse[] / 1000000 // N
Out[3]= 117.604
(* Updating a field *)
In[4]:= data2 = Dataset[<|"State"->#State, "Income"->#Income/2.0|>&/@Normal@data]
In[5]:= MemoryInUse[] / 1000000 // N
Out[5]= 117.885
(* Appending a field *)
In[6]:= data3 = Dataset[<|#, "IncomeSquared"->#Income^2|>&/@Normal@data]
In[7]:= MemoryInUse[] / 1000000 // N
Out[7]= 118.574

Answer 3

Actually, I don't now how to do that with a Dataset but with nested Lists and Association, in-place modification is possible:

assoc = Normal[dataset];
assoc[[3 ;; 5, {"b", "c"}]] = ConstantArray[{b, c}, 3];
assoc // Dataset

So, if you don't need any specific properties of Dataset, I'd suggest to stick with nested Associations.