reduce

I'm looking at this example of use of reduce() function. function add(runningTotal, currentValue) { return runningTotal + currentValue; } var nums = [1,2,3,4,5,6,7,8,9,10]; var sum = nums.reduce(add); print(sum); // displays 55 Could you give show me some other examples of using reduce() - I'm not sure I fully follow how it works. Thank you What reduces does is take an initialValue , a function with 2 essential parameters (can take more) and a list of values. If no initialValue is provided then it's assumed it's the first element of the list. The function is supposed to do something with the

Shuffle过程是MapReduce的核心，也被称为奇迹发生的地方。要想理解MapReduce， Shuffle是必须要了解的。我看过很多相关的资料，但每次看完都云里雾里的绕着，很难理清大致的逻辑，反而越搅越混。前段时间在做MapReduce job 性能调优的工作，需要深入代码研究MapReduce的运行机制，这才对Shuffle探了个究竟。考虑到之前我在看相关资料而看不懂时很恼火，所以在这里我尽最大的可能试着把Shuffle说清楚，让每一位想了解它原理的朋友都能有所收获。如果你对这篇文章有任何疑问或建议请留言到后面，谢谢！ Shuffle的正常意思是洗牌或弄乱，可能大家更熟悉的是Java API里的Collections.shuffle(List)方法，它会随机地打乱参数list里的元素顺序。如果你不知道MapReduce里Shuffle是什么，那么请看这张图： 这张是官方对Shuffle过程的描述。但我可以肯定的是，单从这张图你基本不可能明白Shuffle的过程，因为它与事实相差挺多，细节也是错乱的。后面我会具体描述Shuffle的事实情况，所以这里你只要清楚Shuffle的大致范围就成－怎样把map task的输出结果有效地传送到reduce端。也可以这样理解， Shuffle描述着数据从map task输出到reduce task输入的这段过程。

I have this code for a class where I'm supposed to use the reduce() method to find the min and max values in an array. However, we are required to use only a single call to reduce. The return array should be of size 2, but I know that the reduce() method always returns an array of size 1. I'm able to obtain the minimum value using the code below, however I don't know how to obtain the max value in that same call. I assume that once I do obtain the max value that I just push it to the array after the reduce() method finishes. /** * Takes an array of numbers and returns an array of size 2, *

最近在自学python,不知道怎么入手，就花了好几十大洋买了《python学习手册》来看，记忆力不好，怕以后忘记了，写下来当是笔记吧。也可以供那些python新手又没买这本书的人做参考哈~ 1.map把第一项传递给函数并收集结果 counters = [1,2,3,4] def inc(x):return x + 10 list(map(inc, counters)) list(map((lambda x:x+10), counters)) #另一种写法 得到的结果是：[11,12,13,14] map对每个列表中的元素都调用了inc函数，并将所有的返回值收集到一个新的列表中。 2.filter收集那些函数返回一个true值的项 list(filter((lambda x: x > 0), range(-5, 5))) #返回结果：[1,2,3,4] 3.reduce 通过对一个累加器和后续项应用函数来计算一个单个的值 from functools import reduce reduce((lambda x, y: x + y), [1,2,3,4]) #得到结果：10 reduce((lambda x, y: x * y), [1,2,3,4]) #得到结果：24 注意：reduce 在python 3.0的functools 模块中可用，而不是在内置作用域中可用。 来源：

I have a list of 2-tuples like this: l = [('a', 1), ('b', 2)] and I want to be able to map this onto a dictionary object, so that I can do something like l.a #=> 1 So I tried this, but why does it fail? d = reduce(lambda y,x : y.update({x[0]:x[1]}),l,{}) This gives the error: AttributeError: 'NoneType' object has no attribute 'update' What am I doing wrong? Andrey Sboev >>> l = [('a', 1), ('b', 2)] >>> d = dict(l) >>> d['a'] 1 Why not just do this: d = dict(l) Also, to answer your question, your solution is failing because y (which is a 2-tuple) has no method update, since it's not a dict.

When using the reduce() operation on a parallelstream the OCP exam book states that there are certain principles the reduce() arguments must adhere too. Those arguments are the following: The identity must be defined such that for all elements in the stream u, combiner.apply(identity, u) is equal to u. The accumulator operator op must be associative and stateless such that (a op b) op c is equal to a op (b op c) . The combiner operator must also be associative and stateless and compatible with the identity, such that for all of u and t combiner.apply(u, accumulator.apply(identity, t)) is equal

I have to save and load a cython class instance. My cython class is this plus several methods: import numpy as np cimport numpy as np cimport cython cdef class Perceptron_avg_my: cdef int wlen,freePos cdef np.ndarray w,wtot,wac,wtotc #np.ndarray[np.int32_t] cdef np.ndarray wmean #np.ndarray[np.float32_t] cdef public dict fpos def __cinit__(self,np.int64_t wlen=4*10**7): self.fpos= dict() self.freePos=1 self.wlen=wlen self.w=np.zeros(wlen,np.int32) self.wtot=np.zeros(wlen,np.int32) self.wac=np.zeros(wlen,np.int32) self.wtotc=np.zeros(wlen,np.int32) self.wmean=np.zeros(wlen,np.float32) cpdef