# Counting actual prolog lines loaded

**URL:** <https://swi-prolog.discourse.group/t/counting-actual-prolog-lines-loaded/4156>\
**Category:** Help!\
**Created:** [July 12, 2021, 6:51am UTC](https://swi-prolog.discourse.group/t/counting-actual-prolog-lines-loaded/4156 "2021-07-12T06:51:30Z")\
**Posts on this page:** 1\
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**Author:** ![Boris](https://yyz2.discourse-cdn.com/free1/user_avatar/swi-prolog.discourse.group/boris/32/7486_2.png) [@Boris](https://swi-prolog.discourse.group/u/Boris)\
**Post date:** [July 19, 2021, 1:16pm UTC](https://swi-prolog.discourse.group/t/counting-actual-prolog-lines-loaded/4156/4 "2021-07-19T13:16:39Z")

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(I hope I didn’t misunderstand the conversation)

The aggregate predicate family is useful but at the moment not generic enough. The obvious to me solution to single-pass aggregation (very large files, streams) is as follows:

1. Create a predicate that backtracks over the solutions;
2. Use a non-backtrackable data structure for aggregation.

For the second one, I have used `(nb)_rbtrees` with this tiny bit of code:

```prolog
nbdict_apply(X, Key, Pred, Init) :-
    ( nb_rb_get_node(X, Key, Node)
    -> nb_rb_node_value(Node, Val0),
        call(Pred, Val0, Val1),
        nb_rb_set_node_value(Node, Val1)
    ; nb_rb_insert(X, Key, Init)
    ).

```

This will either insert a default **Init** ial value or apply the **Pred** icate to the existing value associated with the **Key**. [This post by Jan W](https://swi-prolog.discourse.group/t/del-dict-4-behaviour-and-dict-iteration/4135/7) discusses the computational complexity. It also suggests an obvious and better way to count word frequency that _does not work_ if your input is big enough. I guess your question refers in part to that post?

If you do it like this, you can simply use a forall/2. If we assume you have defined `file_word/2` predicate that succeeds once for every word in a file, you would do:

```prolog
rb_empty(Freqs),
forall(file_word(File, Word),
    nbdict_apply(Freqs, Word, succ, 1)),
forall(rb_in(Key, Val, Freqs),
    format("~w ~w~n", [Key, Val]))

```

But this is still not optimal, you need to hand-roll both the backtracking predicate and the non-backtrackable accumulator for anything non-trivial. Do you have a better idea?

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