Spark dataframe transform multiple rows to column

问题:

I am a novice to spark, and I want to transform below source dataframe (load from JSON file):

+--+-----+-----+
|A |count|major|
+--+-----+-----+
| a|    1|   m1|
| a|    1|   m2|
| a|    2|   m3|
| a|    3|   m4|
| b|    4|   m1|
| b|    1|   m2|
| b|    2|   m3|
| c|    3|   m1|
| c|    4|   m3|
| c|    5|   m4|
| d|    6|   m1|
| d|    1|   m2|
| d|    2|   m3|
| d|    3|   m4|
| d|    4|   m5|
| e|    4|   m1|
| e|    5|   m2|
| e|    1|   m3|
| e|    1|   m4|
| e|    1|   m5|
+--+-----+-----+

Into below result dataframe:

+--+--+--+--+--+--+
|A |m1|m2|m3|m4|m5|
+--+--+--+--+--+--+
| a| 1| 1| 2| 3| 0|
| b| 4| 2| 1| 0| 0|
| c| 3| 0| 4| 5| 0|
| d| 6| 1| 2| 3| 4|
| e| 4| 5| 1| 1| 1|
+--+--+--+--+--+--+

Here is the Transformation Rule:

1. The result dataframe is consisted with A + (n major columns) where the major columns names are specified by:

   sorted(src_df.map(lambda x: x[2]).distinct().collect())
   

2. The result dataframe contains m rows where the values for A column are provided by:

   sorted(src_df.map(lambda x: x[0]).distinct().collect())
   

3. The value for each major column in result dataframe is the value from source dataframe on the corresponding A and major (e.g. the count in Row 1 in source dataframe is mapped to the box where A is a and column m1)

4. The combinations of A and major in source dataframe has no duplication (please consider it a primary key on the two columns in SQL)

回答1:

Lets start with example data:

df = sqlContext.createDataFrame([
    ("a", 1, "m1"), ("a", 1, "m2"), ("a", 2, "m3"),
    ("a", 3, "m4"), ("b", 4, "m1"), ("b", 1, "m2"),
    ("b", 2, "m3"), ("c", 3, "m1"), ("c", 4, "m3"),
    ("c", 5, "m4"), ("d", 6, "m1"), ("d", 1, "m2"),
    ("d", 2, "m3"), ("d", 3, "m4"), ("d", 4, "m5"),
    ("e", 4, "m1"), ("e", 5, "m2"), ("e", 1, "m3"),
    ("e", 1, "m4"), ("e", 1, "m5")], 
    ("a", "cnt", "major"))

Please note that I've changed count to cnt. Count is a reserved keyword in most of the SQL dialects and it is not a good choice for a column name.

There are at least two ways to reshape this data:

• aggregating over DataFrame

  from pyspark.sql.functions import col, when, max
  
  majors = sorted(df.select("major")
      .distinct()
      .map(lambda row: row[0])
      .collect())
  
  cols = [when(col("major") == m, col("cnt")).otherwise(None).alias(m) 
      for m in  majors]
  maxs = [max(col(m)).alias(m) for m in majors]
  
  reshaped1 = (df
      .select(col("a"), *cols)
      .groupBy("a")
      .agg(*maxs)
      .na.fill(0))
  
  reshaped1.show()
  
  ## +---+---+---+---+---+---+
  ## |  a| m1| m2| m3| m4| m5|
  ## +---+---+---+---+---+---+
  ## |  a|  1|  1|  2|  3|  0|
  ## |  b|  4|  1|  2|  0|  0|
  ## |  c|  3|  0|  4|  5|  0|
  ## |  d|  6|  1|  2|  3|  4|
  ## |  e|  4|  5|  1|  1|  1|
  ## +---+---+---+---+---+---+
  

• groupBy over RDD

  from pyspark.sql import Row
  
  grouped = (df
      .map(lambda row: (row.a, (row.major, row.cnt)))
      .groupByKey())
  
  def make_row(kv):
      k, vs = kv
      tmp = dict(list(vs) + [("a", k)])
      return Row(**{k: tmp.get(k, 0) for k in ["a"] + majors})
  
  reshaped2 = sqlContext.createDataFrame(grouped.map(make_row))
  
  reshaped2.show()
  
  ## +---+---+---+---+---+---+
  ## |  a| m1| m2| m3| m4| m5|
  ## +---+---+---+---+---+---+
  ## |  a|  1|  1|  2|  3|  0|
  ## |  e|  4|  5|  1|  1|  1|
  ## |  c|  3|  0|  4|  5|  0|
  ## |  b|  4|  1|  2|  0|  0|
  ## |  d|  6|  1|  2|  3|  4|
  ## +---+---+---+---+---+---+
  

回答2:

Using zero323's dataframe,

df = sqlContext.createDataFrame([
("a", 1, "m1"), ("a", 1, "m2"), ("a", 2, "m3"),
("a", 3, "m4"), ("b", 4, "m1"), ("b", 1, "m2"),
("b", 2, "m3"), ("c", 3, "m1"), ("c", 4, "m3"),
("c", 5, "m4"), ("d", 6, "m1"), ("d", 1, "m2"),
("d", 2, "m3"), ("d", 3, "m4"), ("d", 4, "m5"),
("e", 4, "m1"), ("e", 5, "m2"), ("e", 1, "m3"),
("e", 1, "m4"), ("e", 1, "m5")], 
("a", "cnt", "major"))

you could also use

reshaped_df = df.groupby('a').pivot('major').max('cnt').fillna(0)