Dataframe 如何获取spark行的值_计数？

我有一个spark数据框，3列存储3种不同的预测。我想知道每个输出值的计数，以便选择最大次数获得的值作为最终输出

在pandas中，我可以通过为每一行调用lambda函数来轻松实现这一点，以获得如下所示的值_计数。我已经在这里将spark df转换为pandas df，但我需要能够直接在spark df上执行类似的操作

r=[Row(run_1=1, run_2=2, run_3=1, name='test run', id=1)]
df1=spark.createDataFrame(r)
df1.show()
df2=df1.toPandas()
r=df2.iloc[0]
val_counts=r[['run_1','run_2','run_3']].value_counts()
print(val_counts)
top_val=val_counts.index[0] 
top_val_cnt=val_counts.values[0]
print('Majority output = %s, occured %s out of 3 times'%(top_val,top_val_cnt))

输出告诉我值1出现的次数最多，在本例中是两次

+---+--------+-----+-----+-----+
| id|    name|run_1|run_2|run_3|
+---+--------+-----+-----+-----+
|  1|test run|    1|    2|    1|
+---+--------+-----+-----+-----+

1    2
2    1
Name: 0, dtype: int64

Majority output = 1, occured 2 out of 3 times

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我正在尝试编写一个udf函数，它可以获取每个df1行并获得top_val和top_val。有没有一种方法可以使用spark df实现这一点？

python的代码应该类似，也许它会对您有所帮助

  val df1 = Seq((1, 1, 1, 2), (1, 2, 3, 3), (2, 2, 2, 2)).toDF()
  df1.show()
  df1.select(array('*)).map(s=>{
    val list = s.getList(0)
    (list.toString(),list.toArray.groupBy(i => i).mapValues(_.size).toList.toString())
  }).show(false)

输出：

+---+---+---+---+
| _1| _2| _3| _4|
+---+---+---+---+
|  1|  1|  1|  2|
|  1|  2|  3|  3|
|  2|  2|  2|  2|
+---+---+---+---+

+------------+-------------------------+
|_1          |_2                       |
+------------+-------------------------+
|[1, 1, 1, 2]|List((2,1), (1,3))       |
|[1, 2, 3, 3]|List((2,1), (1,1), (3,2))|
|[2, 2, 2, 2]|List((2,4))              |
+------------+-------------------------+

---

让我们有一个类似于您的测试数据帧

 list = [(1,'test run',1,2,1),(2,'test run',3,2,3),(3,'test run',4,4,4)]
df=spark.createDataFrame(list, ['id', 'name','run_1','run_2','run_3'])

newdf = df.rdd.map(lambda x : (x[0],x[1],x[2:])) \
.map(lambda x : (x[0],x[1],x[2][0],x[2][1],x[2][2],[max(set(x[2]),key=x[2].count )])) \
.toDF(['id','test','run_1','run_2','run_3','most_frequent'])


>>> newdf.show()
+---+--------+-----+-----+-----+-------------+
| id|    test|run_1|run_2|run_3|most_frequent|
+---+--------+-----+-----+-----+-------------+
|  1|test run|    1|    2|    1|          [1]|
|  2|test run|    3|    2|    3|          [3]|
|  3|test run|    4|    4|    4|          [4]|
+---+--------+-----+-----+-----+-------------+

或者，当列表中的每个项目都不同时，您需要处理一个案例。i、 e返回空值

list = [(1,'test run',1,2,1),(2,'test run',3,2,3),(3,'test run',4,4,4),(4,'test run',1,2,3)]
df=spark.createDataFrame(list, ['id', 'name','run_1','run_2','run_3'])

from pyspark.sql.functions import udf

@udf
def most_frequent(*mylist): 
    counter = 1
    num = mylist[0] 

    for i in mylist: 
        curr_frequency = mylist.count(i) 
        if(curr_frequency> counter): 
            counter = curr_frequency 
            num = i 

            return num
    else:
            return None

将计数器初始化为“1”，并仅当其大于“1”时返回计数

df.withColumn('most_frequent', most_frequent('run_1', 'run_2', 'run_3')).show()

+---+--------+-----+-----+-----+-------------+
| id|    name|run_1|run_2|run_3|most_frequent|
+---+--------+-----+-----+-----+-------------+
|  1|test run|    1|    2|    1|            1|
|  2|test run|    3|    2|    3|            3|
|  3|test run|    4|    4|    4|            4|
|  4|test run|    1|    2|    3|         null|
+---+--------+-----+-----+-----+-------------+
+---+--------+-----+-----+-----+----+

---

谢谢@AndrzejS。我只是在找这样的东西：-）