He is looking for median, not mean/avg.
You have to implement the median logic by yourself, as there is no directly
implementation from Spark. You can use RDD API, if you are using 1.6.x, or
dataset if 2.x
The following example gives you an idea how to calculate the median using
dataset API. You can even change the code to add additional logic to calculate
the diff of every value with the median.
scala> spark.version
res31: String = 2.1.0
scala> val ds = Seq((100,0.43),(100,0.33),(100,0.73),(101,0.29),(101,0.96),
(101,0.42),(101,0.01)).toDF("id","value").as[(Int, Double)]
ds: org.apache.spark.sql.Dataset[(Int, Double)] = [id: int, value: double]
scala> ds.show
+---+-----+
| id|value|
+---+-----+
|100| 0.43|
|100| 0.33|
|100| 0.73|
|101| 0.29|
|101| 0.96|
|101| 0.42|
|101| 0.01|
+---+-----+
scala> def median(seq: Seq[Double]) = {
| val size = seq.size
| val sorted = seq.sorted
| size match {
| case even if size % 2 == 0 => (sorted((size-2)/2) + sorted(size/2))
/ 2
| case odd => sorted((size-1)/2)
| }
| }
median: (seq: Seq[Double])Double
scala> ds.groupByKey(_._1).mapGroups((id, iter) => (id,
median(iter.map(_._2).toSeq))).show
+---+-----+
| _1| _2|
+---+-----+
|101|0.355|
|100| 0.43|
+---+-----+
Yong
________________________________
From: ayan guha <guha.a...@gmail.com>
Sent: Wednesday, March 22, 2017 7:23 PM
To: Craig Ching
Cc: Yong Zhang; user@spark.apache.org
Subject: Re: calculate diff of value and median in a group
I would suggest use window function with partitioning.
select group1,group2,name,value, avg(value) over (partition group1,group2 order
by name) m
from t
On Thu, Mar 23, 2017 at 9:58 AM, Craig Ching
<craigch...@gmail.com<mailto:craigch...@gmail.com>> wrote:
Are the elements count big per group? If not, you can group them and use the
code to calculate the median and diff.
They're not big, no. Any pointers on how I might do that? The part I'm having
trouble with is the grouping, I can't seem to see how to do the median per
group. For mean, we have the agg feature, but not for median (and I understand
the reasons for that).
Yong
________________________________
From: Craig Ching <craigch...@gmail.com<mailto:craigch...@gmail.com>>
Sent: Wednesday, March 22, 2017 3:17 PM
To: user@spark.apache.org<mailto:user@spark.apache.org>
Subject: calculate diff of value and median in a group
Hi,
When using pyspark, I'd like to be able to calculate the difference between
grouped values and their median for the group. Is this possible? Here is some
code I hacked up that does what I want except that it calculates the grouped
diff from mean. Also, please feel free to comment on how I could make this
better if you feel like being helpful :)
from pyspark import SparkContext
from pyspark.sql import SparkSession
from pyspark.sql.types import (
StringType,
LongType,
DoubleType,
StructField,
StructType
)
from pyspark.sql import functions as F
sc = SparkContext(appName='myapp')
spark = SparkSession(sc)
file_name = 'data.csv'
fields = [
StructField(
'group2',
LongType(),
True),
StructField(
'name',
StringType(),
True),
StructField(
'value',
DoubleType(),
True),
StructField(
'group1',
LongType(),
True)
]
schema = StructType(fields)
df = spark.read.csv(
file_name, header=False, mode="DROPMALFORMED", schema=schema
)
df.show()
means = df.select([
'group1',
'group2',
'name',
'value']).groupBy([
'group1',
'group2'
]).agg(
F.mean('value').alias('mean_value')
).orderBy('group1', 'group2')
cond = [df.group1 == means.group1, df.group2 == means.group2]
means.show()
df = df.select([
'group1',
'group2',
'name',
'value']).join(
means,
cond
).drop(
df.group1
).drop(
df.group2
).select('group1',
'group2',
'name',
'value',
'mean_value')
final = df.withColumn(
'diff',
F.abs(df.value - df.mean_value))
final.show()
sc.stop()
And here is an example dataset I'm playing with:
100,name1,0.43,0
100,name2,0.33,0
100,name3,0.73,0
101,name1,0.29,0
101,name2,0.96,0
101,name3,0.42,0
102,name1,0.01,0
102,name2,0.42,0
102,name3,0.51,0
103,name1,0.55,0
103,name2,0.45,0
103,name3,0.02,0
104,name1,0.93,0
104,name2,0.16,0
104,name3,0.74,0
105,name1,0.41,0
105,name2,0.65,0
105,name3,0.29,0
100,name1,0.51,1
100,name2,0.51,1
100,name3,0.43,1
101,name1,0.59,1
101,name2,0.55,1
101,name3,0.84,1
102,name1,0.01,1
102,name2,0.98,1
102,name3,0.44,1
103,name1,0.47,1
103,name2,0.16,1
103,name3,0.02,1
104,name1,0.83,1
104,name2,0.89,1
104,name3,0.31,1
105,name1,0.59,1
105,name2,0.77,1
105,name3,0.45,1
and here is what I'm trying to produce:
group1,group2,name,value,median,diff
0,100,name1,0.43,0.43,0.0
0,100,name2,0.33,0.43,0.10
0,100,name3,0.73,0.43,0.30
0,101,name1,0.29,0.42,0.13
0,101,name2,0.96,0.42,0.54
0,101,name3,0.42,0.42,0.0
0,102,name1,0.01,0.42,0.41
0,102,name2,0.42,0.42,0.0
0,102,name3,0.51,0.42,0.09
0,103,name1,0.55,0.45,0.10
0,103,name2,0.45,0.45,0.0
0,103,name3,0.02,0.45,0.43
0,104,name1,0.93,0.74,0.19
0,104,name2,0.16,0.74,0.58
0,104,name3,0.74,0.74,0.0
0,105,name1,0.41,0.41,0.0
0,105,name2,0.65,0.41,0.24
0,105,name3,0.29,0.41,0.24
1,100,name1,0.51,0.51,0.0
1,100,name2,0.51,0.51,0.0
1,100,name3,0.43,0.51,0.08
1,101,name1,0.59,0.59,0.0
1,101,name2,0.55,0.59,0.04
1,101,name3,0.84,0.59,0.25
1,102,name1,0.01,0.44,0.43
1,102,name2,0.98,0.44,0.54
1,102,name3,0.44,0.44,0.0
1,103,name1,0.47,0.16,0.31
1,103,name2,0.16,0.16,0.0
1,103,name3,0.02,0.16,0.14
1,104,name1,0.83,0.83,0.0
1,104,name2,0.89,0.83,0.06
1,104,name3,0.31,0.83,0.52
1,105,name1,0.59,0.59,0.0
1,105,name2,0.77,0.59,0.18
1,105,name3,0.45,0.59,0.14
Thanks for any help!
Cheers,
Craig
--
Best Regards,
Ayan Guha
