Extract list of attributes from list of objects in python
attrs = [o.attr for o in objs] was the right code for making a list like the one you describe. Don’t try to subclass list for this. Is there something you did not like about that snippet?
attrs = [o.attr for o in objs] was the right code for making a list like the one you describe. Don’t try to subclass list for this. Is there something you did not like about that snippet?
In Java terms, Scala’s Seq would be Java’s List, and Scala’s List would be Java’s LinkedList. Note that Seq is a trait, which is similar to Java’s interface, but with the equivalent of up-and-coming defender methods. Scala’s List is an abstract class that is extended by Nil and ::, which are the concrete implementations of … Read more
List already has a slice method – it takes a subset of elements between a start and end index. What you’re looking for is repeated application of the span method: def span(p: (A) ⇒ Boolean): (List[A], List[A]) Which is documented as: Splits this list into a prefix/suffix pair according to a predicate. Note: c span … Read more
You can do this by resorting to manual recursion, but I like to believe Haskell is a more evolved language. Let’s see if we can develop a solution that uses existing recursion strategies. First some preliminaries. {-# LANGUAGE NoMonomorphismRestriction #-} — because who wants to write type signatures, amirite? import Data.List.Split — from package split … Read more
You can’t really use map to sum up a list, because map treats each list element independently from the others. You can use map for example to increment each value in a list like in map (+1) [1,2,3,4] — gives [2,3,4,5] Another way to implement your addm would be to use foldl: addm’ = foldl … Read more
DataFrame.from_records treats string as a character list. so it needs as many columns as length of string. You could simply use the DataFrame constructor. In [3]: pd.DataFrame(q_list, columns=[‘q_data’]) Out[3]: q_data 0 112354401 1 116115526 2 114909312 3 122425491 4 131957025 5 111373473
(note: the answer is at the bottom of this post) The 2nd function, (define (swap-ends x) ; swap [] = [] (if (or (equal (length x) 0) (equal (length x) 1)) ; swap [x] = [x] x ; swap (x:xs) (cons (first (swap-ends (rest x))) ; | (a:b) <- swap xs (swap-ends (cons (first x) … Read more
Also, not sure why you’re dead against cuts, so long as their removal would not change the declaritive meaning of the code, as per your link. For example: inter([], _, []). inter([H1|T1], L2, [H1|Res]) :- member(H1, L2), inter(T1, L2, Res). inter([_|T1], L2, Res) :- inter(T1, L2, Res). test(X):- inter([1,3,5,2,4], [6,1,2], X), !. test(X). X = … Read more
Beware of clbuttic mistakes. “Apple made the clbuttic mistake of forcing out their visionary – I mean, look at what NeXT has been up to!” Hmm. “clbuttic”. Google “clbuttic” – thousands of hits! There’s someone who call his car ‘clbuttic’. There are “Clbuttic Steam Engine” message boards. Webster’s dictionary – no help. Hmm. What can … Read more
The following is based on my previous answer to “Remove duplicates in list (Prolog)” and on this previous answer to the question “Prolog union for A U B U C“. list_item_subtracted_count0_count/5 is derived from list_item_subtracted/3. list_counts/2 is derived from list_setB/2, which were both defined here. list_item_subtracted_count0_count([], _, [], N,N). list_item_subtracted_count0_count([A|As], E, Bs1, N0,N) :- if_(A … Read more