What is the best way to choose data with the earliest timestamp for each key in an RDD?

I am working with an RDD that contains two variables ID and time. The time variable is in the format of datetime.datetime. Here is a snapshot of the first few rows of the RDD data:

 [[41186, datetime.datetime(2014, 3, 1, 20, 48, 5, 630000)],
 [32036, datetime.datetime(2014, 3, 2, 0, 25, 41, 950000)],
 [41186, datetime.datetime(2014, 3, 2, 0, 31, 29, 380000)],
 [40479, datetime.datetime(2014, 3, 2, 0, 39, 6, 800000)],
 [4598, datetime.datetime(2014, 3, 2, 1, 48, 47, 430000)]]

In this dataset, the same ID can appear multiple times with different date times. I am interested in selecting each ID with the latest time only.

For example, from the provided sample data, I need to extract the following records:

 [[41186, datetime.datetime(2014, 3, 1, 20, 48, 5, 630000)],
 [32036, datetime.datetime(2014, 3, 2, 0, 25, 41, 950000)],
 [40479, datetime.datetime(2014, 3, 2, 0, 39, 6, 800000)],
 [4598, datetime.datetime(2014, 3, 2, 1, 48, 47, 430000)]]

Could someone please guide me on how to write a query to achieve this output? Thank you.

pythonpysparkrdd

Answer №1

Utilize the groupByKey method and then implement the min function:

print(rdd.groupByKey().mapValues(min).collect())
#[(15792, datetime.datetime(2017, 5, 15, 10, 12, 29, 210000)),
# (45270, datetime.datetime(2017, 5, 15, 9, 24, 15, 440000)),
# (3984, datetime.datetime(2017, 5, 15, 8, 45, 51, 360000)),
# (9014, datetime.datetime(2017, 5, 15, 11, 59, 32, 710000))]

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