减少对 - python

Question

如果我必须减少一对值，我该如何编写lambda表达式。

testr = [('r1', (1, 1)), ('r1', (1, 5)),('r2', (1, 1)),('r3', (1, 1))] 

所需的输出是

('r1', (2, 6)),('r2', (1, 1)),('r3', (1, 1)) 

Answer 1

通过重点减少碳排放量

.reduceByKey(lambda a, b: (a[0]+b[0], a[1]+b[1])) 

你可以把它与拉链任意长度元组更通用的：

.reduceByKey(lambda a, b: tuple(x+y for x,y in zip(a,b))) 

Answer 2

it is not clear for me how reduce can use to reduce with lambda to reduce list tuples with different keys. My solution is can reduce list of tuples, but it uses function, which is perhaps too troublesome to do in pure lambda, if not impossible. 

def reduce_tuple_list(tl): 

    import operator as op 
    import functools as fun 
    import itertools as it 

    # sort the list for groupby 
    tl = sorted(tl,key=op.itemgetter(0)) 
    # this function with reduce lists with the same key 
    def reduce_with_same_key(tl): 
     def add_tuple(t1,t2): 
      k1, tl1 = t1 
      k2, tl2 = t2 
      if k1 == k2: 
       l1,r1 = tl1 
       l2,r2 = tl2 
       l = l1+l2 
       r = r1+r2 
       return k1,(l,r) 
      else: 
       return t1,t2 
     return tuple(fun.reduce(add_tuple, tl)) 

    # group by keys 
    groups = [] 
    for k, g in it.groupby(tl, key=op.itemgetter(0)): 
     groups.append(list(g)) 

    new_list = [] 
    # we need to add only lists whose length is greater than one 
    for el in groups: 
     if len(el) > 1: # reduce 
      new_list.append(reduce_with_same_key(el)) 
     else: # single tuple without another one with the same key 
      new_list.append(el[0]) 
    return new_list 


    testr = [('r1', (1, 1)), ('r3', (11, 71)), ('r1', (1, 5)),('r2', (1, 1)),('r3', (1, 1))] 

    >>> reduce_tuple_list(testr) 

    [('r1', (2, 6)), ('r2', (1, 1)), ('r3', (12, 72))] 

Answer 3

可以使用combineByKey方法

testr = sc.parallelize((('r1', (1, 1)), ('r1', (1, 5)),('r2', (1, 1)),('r3', (1, 1)))) 

testr.combineByKey(lambda x:x,lambda x,y:(x[0]+y[0],x[1]+y[1]),lambda x,y:(x[0]+x[1],y[0]+y[1])).collect()