我有一个想法,使用相同的键组合不同行中的数据帧值(列表)。 该组合在不同行中必须具有相同或多个值,因此我不能仅使用 df.groupBy('id') 来获取结果。
示例如下:
+---------+--------------------+
|id |num_list |
+---------+--------------------+
|apple |[11, 12] |
|apple |[11, 13 ,14] |
|apple |[10, 22, 25] |
|banana |[15, 26] |
|banana |[15, 29] |
|banana |[15, 27] |
+---------+--------------------+
我们可以发现id=apple
有两条记录和两个列表,例如:[11, 12],[11, 13, 14]
,所以它们会被合并到新记录 id=apple,num_list=[11, 12, 13, 14]
但是 id=apple,num_list=[10, 22, 25]
不会合并。
这就是我想要的答案:
+---------+--------------------+
|id |num_list |
+---------+--------------------+
|apple |[11, 12, 13, 14] |
|apple |[10, 22, 25] |
|banana |[15, 26, 27, 29] |
+---------+--------------------+
编辑:
有一些规则我必须解释一下。
正如@Usernamenotfound 评论的那样,
假设 apple 有三个值 [11, 12, 14]
、[9, 13 ,14]
和 [12,13,27]
,答案将是 [9, 11, 12, 13 ,14, 27]
而不是 [9, 11, 12, 13 ,14]
和 >[12,13,27]
。
有一些新的例子:
+--------------------+------------+
| id|num_list |
+--------------------+------------+
|apple | [0]|
|apple | [0]|
|apple | [1]|
|apple | [1]|
|apple | [2]|
|apple | [3]|
|apple | [4]|
|apple | [5]|
|apple | [6]|
|apple | [6]|
|apple | [7]|
|apple | [7, 8]|
|apple | [9]|
|apple | [9]|
|apple | [9]|
|apple | [9, 10]|
|apple | [9, 17, 18]|
|apple | [10]|
|apple | [10]|
|apple | [10]|
+--------------------+------------+
如果我尝试使用@mayank 的代码将会得到错误的答案。
+--------------------------------+---------------------------------------------------------------------------------------+
| id|num_list |
+--------------------------------+---------------------------------------------------------------------------------------+
|apple |[0] |
|apple |[0] |
|apple |[1] |
|apple |[2] |
|apple |[3] |
|apple |[4] |
|apple |[5] |
|apple |[6] |
|apple |[8, 7] |
|apple |[9, 10, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 36, 37, 38]|
|apple |[12] |
|apple |[14] |
|apple |[24] |
|apple |[31] |
|apple |[32] |
|apple |[33, 34] |
|apple |[35] |
|apple |[39] |
+--------------------------------+---------------------------------------------------------------------------------------+
任何帮助将不胜感激。
最佳答案
也许不是最有效的解决方案,但它解决了您的问题。
import pyspark.sql.functions as F
from pyspark.sql.types import *
def get_combinations(lis):
final = []
for li in lis:
if [y for y in final if [z for z in li if z in y]]:
found = [y for y in final if [z for z in li if z in y]][0]
to_add = list(set(found + li))
final[final.index(found)] = to_add
else:
final.append(li)
return final
apply_udf = F.udf(lambda x:get_combinations(x),ArrayType(ArrayType(IntegerType())))
df = df.groupby('id').agg(F.collect_list('num_list').alias('num_list'))\
.select(['id', F.explode(apply_udf('num_list')).alias('num_list')])
关于python - pyspark 如果列具有相同的值,如何组合列值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47673889/