Best way to get the max value in a Spark dataframe column
I'm trying to figure out the best way to get the largest value in a Spark dataframe column.
Consider the following example:
df = spark.createDataFrame([(1., 4.), (2., 5.), (3., 6.)], ["A", "B"])
df.show()
Which creates:
+---+---+
| A| B|
+---+---+
|1.0|4.0|
|2.0|5.0|
|3.0|6.0|
+---+---+
My goal is to find the largest value in column A (by inspection, this is 3.0). Using PySpark, here are four approaches I can think of:
# Method 1: Use describe()
float(df.describe("A").filter("summary = 'max'").select("A").first().asDict()['A'])
# Method 2: Use SQL
df.registerTempTable("df_table")
spark.sql("SELECT MAX(A) as maxval FROM df_table").first().asDict()['maxval']
# Method 3: Use groupby()
df.groupby().max('A').first().asDict()['max(A)']
# Method 4: Convert to RDD
df.select("A").rdd.max()[0]
Each of the above gives the right answer, but in the absence of a Spark profiling tool I can't tell which is best.
Any ideas from either intuition or empiricism on which of the above methods is most efficient in terms of Spark runtime or resource usage, or whether there is a more direct method than the ones above?
Solution 1:
>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
Solution 2:
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]
Solution 3:
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}
Solution 4:
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
Solution 5:
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|
+------+------+------+