Gavin’s answer is simple and elegant. But if there are many columns, a much faster solution would be: lapply(seq_len(ncol(x)), function(i) x[,i]) The speed difference is 6x in the example below: > x <- matrix(1:1e6, 10) > system.time( as.list(data.frame(x)) ) user system elapsed 1.24 0.00 1.22 > system.time( lapply(seq_len(ncol(x)), function(i) x[,i]) ) user system elapsed 0.2 … Read more
Operators like <= in Python are generally not overriden to mean something significantly different than “less than or equal to”. It’s unusual for the standard library does this–it smells like legacy API to me. Use the equivalent and more clearly-named method, set.issubset. Note that you don’t need to convert the argument to a set; it’ll … Read more
The definition of flatten2/2 you’ve given is busted; it actually behaves like this: ?- flatten2([a, [b,c], [[d],[],[e]]], R). R = [a, b, c] ; false. So, given the case where you’ve already bound R to [a,b,c,d,e], the failure isn’t surprising. Your definition is throwing away the tail of lists (ListTail) in the 3rd clause – … Read more
PowerShell does some default formatting when it presents data/objects to the user. Usually objects are displayed in tabular form when they have up to 4 properties, and in list form when they have more than 4 properties. If you output several things in a row, PowerShell applies the format (list/table) from the first object to … Read more
You need to use agg. Example: from pyspark import SparkContext from pyspark.sql import HiveContext from pyspark.sql import functions as F sc = SparkContext(“local”) sqlContext = HiveContext(sc) df = sqlContext.createDataFrame([ (“a”, None, None), (“a”, “code1”, None), (“a”, “code2”, “name2”), ], [“id”, “code”, “name”]) df.show() +—+—–+—–+ | id| code| name| +—+—–+—–+ | a| null| null| | a|code1| … Read more
Mark, this is already answered in your previous topic. But OK, here it is again: Suppose ${list} points to a List<Object>, then the following <c:forEach items=”${list}” var=”item”> ${item}<br> </c:forEach> does basically the same as as following in “normal Java”: for (Object item : list) { System.out.println(item); } If you have a List<Map<K, V>> instead, then … Read more
Here’s how you can find the “powers of two” in logically-pure way! Using sicstus-prolog 4.3.5, library(reif) and library(clpz): :- use_module([library(reif), library(clpz)]). power_of_two_t(I, T) :- L #= min(I,1), M #= I /\ (I-1), call((L = 1, M = 0), T). % using (=)/3 and (‘,’)/3 of library(reif) Sample query1 using meta-predicate tfilter/3 in combination with power_of_two_t/2: … Read more
Prolog rules are read independently of each other, so you need one rule for the case where the element is unique and one where it is not. Provided the order of the elements is not relevant, you might use: ?- remUniqueVals([A,B,C], [1,1]). A = B, B = 1, dif(C, 1) ; A = C, C … Read more
You can use indexing, if you index a number beyond the size of the object it returns NA. This works for any arbitrary number of rows defined with foo: nm <- list(1:8,3:8,1:5) foo <- 8 sapply(nm, ‘[‘, 1:foo) EDIT: Or in one line using the largest vector as number of rows: sapply(nm, ‘[‘, seq(max(sapply(nm,length)))) From … Read more
This answer shows a logically pure way to do it. The following is based on clpfd. :- use_module(library(clpfd)). We define meta-predicate tcount/3 similarly to tfilter/3! :- meta_predicate tcount(2,?,?). tcount(P_2,Xs,N) :- N #>= 0, list_pred_tcount_(Xs,P_2,0,N). :- meta_predicate list_pred_tcount_(?,2,?,?). list_pred_tcount_([] , _ ,N ,N). list_pred_tcount_([X|Xs],P_2,N0,N) :- if_(call(P_2,X), (N1 is N0+1, N1 #=< N), N1 = N0), list_pred_tcount_(Xs,P_2,N1,N). … Read more