reduce

The standard R expression outer(X, Y, f) evaluates to a matrix whose (i, j)-th entry has the value f(X[i], Y[j]) . I would like to implement the function multi.outer , an n-dimensional generalization of outer : multi.outer(f, X_1, ..., X_n) , where f is some n-ary function, would produce a (length(X_1) * ... * length(X_n)) array whose (i_1,...,i_n)-th entry has the value f(X_1[i_1], ..., X_n[i_n]) for all valid index sets (i_1,...,i_n). Clearly, for each i in {1, ..., n}, all the elements of X_i in multi.outer(f, X_1,...,X_i,..., X_n) must be allowable i-th arguments for the function f . For

I've a silly question involving fold and reduce in PySpark . I understand the difference between these two methods, but, if both need that the applied function is a commutative monoid, I cannot figure out an example in which fold cannot be substituted by reduce`. Besides, in the PySpark implementation of fold it is used acc = op(obj, acc) , why this operation order is used instead of acc = op(acc, obj) ? (this second order sounds more closed to a leftFold to me) Cheers Tomas Empty RDD It cannot be substituted when RDD is empty: val rdd = sc.emptyRDD[Int] rdd.reduce(_ + _) // java.lang

I am trying to understand how fold and foldLeft and the respective reduce and reduceLeft work. I used fold and foldLeft as my example scala> val r = List((ArrayBuffer(1, 2, 3, 4),10)) scala> r.foldLeft(ArrayBuffer(1,2,4,5))((x,y) => x -- y._1) scala> res28: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(5) scala> r.fold(ArrayBuffer(1,2,4,5))((x,y) => x -- y._1) <console>:11: error: value _1 is not a member of Serializable with Equals r.fold(ArrayBuffer(1,2,4,5))((x,y) => x -- y._1) Why fold didn't work as foldLeft ? What is Serializable with Equals ? I understand fold and foldLeft has

There are many ways to write a Python program that computes a histogram. By histogram, I mean a function that counts the occurrence of objects in an iterable and outputs the counts in a dictionary. For example: >>> L = 'abracadabra' >>> histogram(L) {'a': 5, 'b': 2, 'c': 1, 'd': 1, 'r': 2} One way to write this function is: def histogram(L): d = {} for x in L: if x in d: d[x] += 1 else: d[x] = 1 return d Are there more concise ways of writing this function? If we had dictionary comprehensions in Python, we could write: >>> { x: L.count(x) for x in set(L) } but since Python 2.6 doesn't have