问题描述
我的问题类似于给定的here,但我还有一个我想从中获得总和的字段,即我的RDD如下(我将其显示为数据框)
+----------+----------------+----------------+ | c1 | c2 | val | +----------+----------------+----------------+ | t1| [a,b] | [11,12]| | t2| [a,b,c ] | [13,14,15]| | t3| [a,c,d] |[16,17,18,19]| +----------+----------------+----------------+
@H_502_5@我想得到这样的东西:
+----------+----------------+----------------+ | c1 | c2 | sum(val) | +----------+----------------+----------------+ | t1| [a,b] | 23 | | t2| [a,b] | 27 | | t2| [a,c] | 28 | | t2| [b,d] | 29 | | t3| [a,b] | 33 | | t3| [a,c] | 34 | | t3| [a,d] | 35 | | t3| [b,c] | 35 | | t3| [b,d] | 36 | | t3| [c,d] | 37 | +----------+----------------+----------------+
@H_502_5@使用以下代码,我得到前两列
def combinations(row): l = row[1] k = row[0] m = row[2] return [(k,v) for v in itertools.combinations(l,2)] a.map(combinations).flatMap(lambda x: x).take(5)
@H_502_5@def combinations(row): l = row[1] k = row[0] m = row[2] return [(k,v,x) for v in itertools.combinations(l,2) for x in map(sum,itertools.combinations(m,2)) ] a.map(combinations).flatMap(lambda x: x).take(5)
@H_502_5@感谢您的帮助。
解决方法
尝试以下方法:
,a = sc.parallelize([ (1,[1,2,3,4],[11,12,13,14]),(2,[3,4,5,6],[15,16,17,18]),(3,[-1,[19,20,21,22]) ]) def combinations(row): l = row[1] k = row[0] m = row[2] return [(k,v,x) for v in itertools.combinations(l,2) for x in map(sum,itertools.combinations(m,2))] a.map(combinations).flatMap(lambda x: x).take(5)
如下解决
def combinations(row): l = row[1] k = row[0] m = row[2] return [(k,m[l.index(v[0])]+m[l.index(v[1])]) for v in itertools.combinations(l,2)] a.map(combinations).flatMap(lambda x: x).take(5)
由于第二列和第三列中的元素数量相同,因此我提取了元素并将其添加。感谢Lavesh的回答,他帮助我找到了解决方案。