Best way to get the max value in a Spark dataframe column

>df1.show()
+-----+--------------------+--------+----------+-----------+
|floor|           timestamp|     uid|         x|          y|
+-----+--------------------+--------+----------+-----------+
|    1|2014-07-19T16:00:...|600dfbe2| 103.79211|71.50419418|
|    1|2014-07-19T16:00:...|5e7b40e1| 110.33613|100.6828393|
|    1|2014-07-19T16:00:...|285d22e4|110.066315|86.48873585|
|    1|2014-07-19T16:00:...|74d917a1| 103.78499|71.45633073|

>row1 = df1.agg({"x": "max"}).collect()[0]
>print row1
Row(max(x)=110.33613)
>print row1["max(x)"]
110.33613

The answer is almost the same as method3. but seems the "asDict()" in method3 can be removed

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Max value for a particular column of a dataframe can be achieved by using -

your_max_value = df.agg({"your-column": "max"}).collect()[0][0]

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Remark: Spark is intended to work on Big Data - distributed computing. The size of the example DataFrame is very small, so the order of real-life examples can be altered with respect to the small example.

Slowest: Method_1, because .describe("A") calculates min, max, mean, stddev, and count (5 calculations over the whole column).

Medium: Method_4, because, .rdd (DF to RDD transformation) slows down the process.

Faster: Method_3 ~ Method_2 ~ Method_5, because the logic is very similar, so Spark's catalyst optimizer follows very similar logic with minimal number of operations (get max of a particular column, collect a single-value dataframe; .asDict() adds a little extra-time comparing 2, 3 vs. 5)

import pandas as pd
import time

time_dict = {}

dfff = self.spark.createDataFrame([(1., 4.), (2., 5.), (3., 6.)], ["A", "B"])
#--  For bigger/realistic dataframe just uncomment the following 3 lines
#lst = list(np.random.normal(0.0, 100.0, 100000))
#pdf = pd.DataFrame({'A': lst, 'B': lst, 'C': lst, 'D': lst})
#dfff = self.sqlContext.createDataFrame(pdf)

tic1 = int(round(time.time() * 1000))
# Method 1: Use describe()
max_val = float(dfff.describe("A").filter("summary = 'max'").select("A").collect()[0].asDict()['A'])
tac1 = int(round(time.time() * 1000))
time_dict['m1']= tac1 - tic1
print (max_val)

tic2 = int(round(time.time() * 1000))
# Method 2: Use SQL
dfff.registerTempTable("df_table")
max_val = self.sqlContext.sql("SELECT MAX(A) as maxval FROM df_table").collect()[0].asDict()['maxval']
tac2 = int(round(time.time() * 1000))
time_dict['m2']= tac2 - tic2
print (max_val)

tic3 = int(round(time.time() * 1000))
# Method 3: Use groupby()
max_val = dfff.groupby().max('A').collect()[0].asDict()['max(A)']
tac3 = int(round(time.time() * 1000))
time_dict['m3']= tac3 - tic3
print (max_val)

tic4 = int(round(time.time() * 1000))
# Method 4: Convert to RDD
max_val = dfff.select("A").rdd.max()[0]
tac4 = int(round(time.time() * 1000))
time_dict['m4']= tac4 - tic4
print (max_val)

tic5 = int(round(time.time() * 1000))
# Method 5: Use agg()
max_val = dfff.agg({"A": "max"}).collect()[0][0]
tac5 = int(round(time.time() * 1000))
time_dict['m5']= tac5 - tic5
print (max_val)

print time_dict

Result on an edge-node of a cluster in milliseconds (ms):

small DF (ms): {'m1': 7096, 'm2': 205, 'm3': 165, 'm4': 211, 'm5': 180}

bigger DF (ms): {'m1': 10260, 'm2': 452, 'm3': 465, 'm4': 916, 'm5': 373}

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Another way of doing it:

df.select(f.max(f.col("A")).alias("MAX")).limit(1).collect()[0].MAX

On my data, I got this benchmarks:

df.select(f.max(f.col("A")).alias("MAX")).limit(1).collect()[0].MAX
CPU times: user 2.31 ms, sys: 3.31 ms, total: 5.62 ms
Wall time: 3.7 s

df.select("A").rdd.max()[0]
CPU times: user 23.2 ms, sys: 13.9 ms, total: 37.1 ms
Wall time: 10.3 s

df.agg({"A": "max"}).collect()[0][0]
CPU times: user 0 ns, sys: 4.77 ms, total: 4.77 ms
Wall time: 3.75 s

All of them give the same answer

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The below example shows how to get the max value in a Spark dataframe column.

from pyspark.sql.functions import max

df = sql_context.createDataFrame([(1., 4.), (2., 5.), (3., 6.)], ["A", "B"])
df.show()
+---+---+
|  A|  B|
+---+---+
|1.0|4.0|
|2.0|5.0|
|3.0|6.0|
+---+---+

result = df.select([max("A")]).show()
result.show()
+------+
|max(A)|
+------+
|   3.0|
+------+

print result.collect()[0]['max(A)']
3.0

Similarly min, mean, etc. can be calculated as shown below:

from pyspark.sql.functions import mean, min, max

result = df.select([mean("A"), min("A"), max("A")])
result.show()
+------+------+------+
|avg(A)|min(A)|max(A)|
+------+------+------+
|   2.0|   1.0|   3.0|
+------+------+------+

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First add the import line:

from pyspark.sql.functions import min, max

To find the min value of age in the dataframe:

df.agg(min("age")).show()

+--------+
|min(age)|
+--------+
|      29|
+--------+

To find the max value of age in the dataframe:

df.agg(max("age")).show()

+--------+
|max(age)|
+--------+
|      77|
+--------+

---

I used another solution (by @satprem rath) already present in this chain.

To find the min value of age in the dataframe:

df.agg(min("age")).show()

+--------+
|min(age)|
+--------+
|      29|
+--------+

edit: to add more context.

While the above method printed the result, I faced issues when assigning the result to a variable to reuse later.

Hence, to get only the int value assigned to a variable:

from pyspark.sql.functions import max, min  

maxValueA = df.agg(max("A")).collect()[0][0]
maxValueB = df.agg(max("B")).collect()[0][0]

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In case some wonders how to do it using Scala (using Spark 2.0.+), here you go:

scala> df.createOrReplaceTempView("TEMP_DF")
scala> val myMax = spark.sql("SELECT MAX(x) as maxval FROM TEMP_DF").
    collect()(0).getInt(0)
scala> print(myMax)
117

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I believe the best solution will be using head()

Considering your example:

+---+---+
|  A|  B|
+---+---+
|1.0|4.0|
|2.0|5.0|
|3.0|6.0|
+---+---+

Using agg and max method of python we can get the value as following :

from pyspark.sql.functions import max df.agg(max(df.A)).head()[0]

This will return: 3.0

Make sure you have the correct import:
from pyspark.sql.functions import max The max function we use here is the pySPark sql library function, not the default max function of python.

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To just get the value use any of these

df1.agg({"x": "max"}).collect()[0][0] df1.agg({"x": "max"}).head()[0] df1.agg({"x": "max"}).first()[0]

Alternatively we could do these for 'min'

from pyspark.sql.functions import min, max
df1.agg(min("id")).collect()[0][0]
df1.agg(min("id")).head()[0]
df1.agg(min("id")).first()[0]

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in pyspark you can do this:

max(df.select('ColumnName').rdd.flatMap(lambda x: x).collect())

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Here is a lazy way of doing this, by just doing compute Statistics:

df.write.mode("overwrite").saveAsTable("sampleStats")
Query = "ANALYZE TABLE sampleStats COMPUTE STATISTICS FOR COLUMNS " + ','.join(df.columns)
spark.sql(Query)

df.describe('ColName')

or

spark.sql("Select * from sampleStats").describe('ColName')

or you can open a hive shell and

describe formatted table sampleStats;

You will see the statistics in the properties - min, max, distinct, nulls, etc.

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import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.functions._

val testDataFrame = Seq(
  (1.0, 4.0), (2.0, 5.0), (3.0, 6.0)
).toDF("A", "B")

val (maxA, maxB) = testDataFrame.select(max("A"), max("B"))
  .as[(Double, Double)]
  .first()
println(maxA, maxB)

And the result is (3.0,6.0), which is the same to the testDataFrame.agg(max($"A"), max($"B")).collect()(0).However, testDataFrame.agg(max($"A"), max($"B")).collect()(0) returns a List, [3.0,6.0]

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