reduce

问题 I am using a 0 reduce approach to my problem. I wish to preprocess data from one file and then to write it out as another file, but with no new lines and tab delimeters? How can I output my map job that has processed my data with the same file format it came in minus the preprocess. That is, I have something like this: Preprocess: <TITLE> Herp derp </Title> I am a major general Post Process: Herp Derp I am a major general What I want it to do is this: Herp Derp I am a major general I believe

问题 What are the priorities of the following 3 options for setting number of reduces? In other words, if all three are set, which one will be taken into account? Option1: setNumReduceTasks(2) within the application code Option2: -D mapreduce.job.reduces=2 as command line argument Option3: through $HADOOP_CONF_DIR/mapred-site.xml file <property> <name>mapreduce.job.reduces</name> <value>2</value> </property> 回答1: You have them racked in priority order - option 1 will override 2, and 2 will

问题 The signature of reduceLeft on some Seq[A] is def reduceLeft [B >: A] (f: (B, A) => B): B The type of A is known, but the lower bound >: tells us that B can be any supertype of A . Why is it like this? Why not def reduceLeft (f: (A, A) => A): A We already know that the head of the sequence is type A and so I can't think of how B could be anything other than equal to A . Can you provide an example where B is some super-type? 回答1: Let's say your class B has a method combine(other:B): B . Now

问题 This is probably straightforward, but I just can't get over it. I have a data structure that is a nested map, like this: (def m {:1 {:1 2 :2 5 :3 10} :2 {:1 2 :2 50 :3 25} :3 {:1 42 :2 23 :3 4}}) I need to set every m[i][i]=0 . This is simple in non-functional languages, but I cant make it work on Clojure. How is the idiomatic way to do so, considering that I do have a vector with every possible value? (let's call it v ) doing (map #(def m (assoc-in m [% %] 0)) v) will work, but using def

问题 I have 19 nested lists generated from a lapply and split operation. These lists are in the form: #list1 Var col1 col2 col3 A 2 3 4 B 3 4 5 #list2 Var col1 col2 col3 A 5 6 7 B 5 4 4 ...... #list19 Var col1 col2 col3 A 3 6 7 B 7 4 4 I have been able to merge the lists with merge.all <- function(x, y) merge(x, y, all=TRUE, by="Var") out <- Reduce(merge.all, DataList) I am however getting an error due to the similarity in the names of the other columns. How can I concatenate the name of the list

问题 In setting the number of processes, I'd be keen to see how many threads I can actually use on my machine - how do I find this? Is there a way to determine the number of threads available to me? 回答1: Do you want to know the CPU count? According to the docs, when you start a pool and don't sepecify the number of processes, the default number is the number of cpu's on the system: processes is the number of worker processes to use. If processes is None then the number returned by cpu_count() is

问题 Is there standard implementation of fold (reduce, aggregate etc) for one dimensional vector in Octave? If no, is there any way to express fold without using a loop statement? 回答1: The miscellaneous package provides the function reduce. For example, octave:6> reduce(@(x,y)(x*y), [1:5]) ans = 120 If you look at the source code for reduce , you'll see that it is a fairly simple Octave function that is implemented with a for loop, so it won't be more efficient than implementing the reduction with

问题 based on my question: https://stackoverflow.com/a/40661953/2392461, i open a new question with sample data. I want to merge/reduce this: var array = [{ 'key1': { 'key11': 0, 'key12': 1 }, 'key2': 0, 'key3': { 'key31': [1, 2], 'key32': { 'key321': 3, 'key322': [1, 2] } }, 'key4': 'test' }, { 'key1': { 'key11': 1, 'key12': 9 }, 'key2': 2, 'key3': { 'key31': [4, 3], 'key32': { 'key321': 6, 'key322': [8, 9] } }, 'key4': 'test' }, { 'key1': { 'key11': 3, 'key12': 4 }, 'key2': 7, 'key3': { 'key31':

问题 I'm trying to make a generic function that would reduce a list like so : func(['a','b','c'],str.join) # --> ['a','b','c','ab','ac','bc','abc'] func(['a','b','c'],lambda: a,b:a+'x'+b) # --> ['a','b','c','axb','axc','bxc','axbxc'] I don't really know how to do it. I did a few tries, but none was successful. I'm pretty sure there is a way to do it with reduce but i'm not very comfortable with the use of this function. Here are some attempts : reduce(lambda a,b:[a,b,str(a)+str(b)],['a','b','c'])

问题 I've achived these two things. Find all possible sublists of a list in given range (i ,j) . A = [ 44, 55, 66, 77, 88, 99, 11, 22, 33 ] Let, i = 2 and j = 4 Then, Possible sublists of the list "A" in the given range (2,4) is : [66], [66,77], [66,77,88], [77], [77,88], [88] And, minimum of the resultant product after multipying all the elements of the sublists: So, the resultant list after multiplying all the elements in the above sublists will become X = [66, 5082, 447216, 77, 6776, 88]` Now,