图解Spark Transformation算子

时间：2021-12-06 20:48:32

0. 写在前面

Spark总共有两类算子，分别是Transformation算子和Action算子。Transformation算子变换不触发提交作业，而Action算子会触发SparkContext提交Job作业，下面主要使用pySpark API来作为事例，图解Spark的Transformation算子。

1. countByKey

图解Spark Transformation算子

# countByKey
x = sc.parallelize([('B',1),('B',2),('A',3),('A',4),('A',5)])
y = x.countByKey()
print(x.collect())
print(y)

[('B', 1), ('B', 2), ('A', 3), ('A', 4), ('A', 5)]
defaultdict(<type 'int'>, {'A': 3, 'B': 2})
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2. join

图解Spark Transformation算子

# join
x = sc.parallelize([('C',4),('B',3),('A',2),('A',1)])
y = sc.parallelize([('A',8),('B',7),('A',6),('D',5)])
z = x.join(y)
print(x.collect())
print(y.collect())
print(z.collect())

[('C', 4), ('B', 3), ('A', 2), ('A', 1)]
[('A', 8), ('B', 7), ('A', 6), ('D', 5)]
[('A', (2, 8)), ('A', (2, 6)), ('A', (1, 8)), ('A', (1, 6)), ('B', (3, 7))]
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3. leftOuterJoin

图解Spark Transformation算子

# leftOuterJoin
x = sc.parallelize([('C',4),('B',3),('A',2),('A',1)])
y = sc.parallelize([('A',8),('B',7),('A',6),('D',5)])
z = x.leftOuterJoin(y)
print(x.collect())
print(y.collect())
print(z.collect())

[('C', 4), ('B', 3), ('A', 2), ('A', 1)]
[('A', 8), ('B', 7), ('A', 6), ('D', 5)]
[('A', (2, 8)), ('A', (2, 6)), ('A', (1, 8)), ('A', (1, 6)), ('C', (4, None)), ('B', (3, 7))]
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4. rightOuterJoin

图解Spark Transformation算子

# rightOuterJoin
x = sc.parallelize([('C',4),('B',3),('A',2),('A',1)])
y = sc.parallelize([('A',8),('B',7),('A',6),('D',5)])
z = x.rightOuterJoin(y)
print(x.collect())
print(y.collect())
print(z.collect())

[('C', 4), ('B', 3), ('A', 2), ('A', 1)]
[('A', 8), ('B', 7), ('A', 6), ('D', 5)]
[('A', (2, 8)), ('A', (2, 6)), ('A', (1, 8)), ('A', (1, 6)), ('B', (3, 7)), ('D', (None, 5))]
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5. partitionBy

图解Spark Transformation算子

# partitionBy
x = sc.parallelize([(0,1),(1,2),(2,3)],2)
y = x.partitionBy(numPartitions = 3, partitionFunc = lambda x: x)  # only key is passed to paritionFunc
print(x.glom().collect())
print(y.glom().collect())

[[(0, 1)], [(1, 2), (2, 3)]]
[[(0, 1)], [(1, 2)], [(2, 3)]]
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6. combineByKey

图解Spark Transformation算子

# combineByKey
x = sc.parallelize([('B',1),('B',2),('A',3),('A',4),('A',5)])
createCombiner = (lambda el: [(el,el**2)]) 
mergeVal = (lambda aggregated, el: aggregated + [(el,el**2)]) # append to aggregated
mergeComb = (lambda agg1,agg2: agg1 + agg2 )  # append agg1 with agg2
y = x.combineByKey(createCombiner,mergeVal,mergeComb)
print(x.collect())
print(y.collect())

[('B', 1), ('B', 2), ('A', 3), ('A', 4), ('A', 5)]
[('A', [(3, 9), (4, 16), (5, 25)]), ('B', [(1, 1), (2, 4)])]
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7. aggregateByKey

图解Spark Transformation算子

# aggregateByKey
x = sc.parallelize([('B',1),('B',2),('A',3),('A',4),('A',5)])
zeroValue = [] # empty list is 'zero value' for append operation
mergeVal = (lambda aggregated, el: aggregated + [(el,el**2)])
mergeComb = (lambda agg1,agg2: agg1 + agg2 )
y = x.aggregateByKey(zeroValue,mergeVal,mergeComb)
print(x.collect())
print(y.collect())

[('B', 1), ('B', 2), ('A', 3), ('A', 4), ('A', 5)]
[('A', [(3, 9), (4, 16), (5, 25)]), ('B', [(1, 1), (2, 4)])]
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8. foldByKey

图解Spark Transformation算子

# foldByKey
x = sc.parallelize([('B',1),('B',2),('A',3),('A',4),('A',5)])
zeroValue = 1 # one is 'zero value' for multiplication
y = x.foldByKey(zeroValue,lambda agg,x: agg*x )  # computes cumulative product within each key
print(x.collect())
print(y.collect())

[('B', 1), ('B', 2), ('A', 3), ('A', 4), ('A', 5)]
[('A', 60), ('B', 2)]
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9. groupByKey

图解Spark Transformation算子

# groupByKey
x = sc.parallelize([('B',5),('B',4),('A',3),('A',2),('A',1)])
y = x.groupByKey()
print(x.collect())
print([(j[0],[i for i in j[1]]) for j in y.collect()])

[('B', 5), ('B', 4), ('A', 3), ('A', 2), ('A', 1)]
[('A', [3, 2, 1]), ('B', [5, 4])]
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10. flatMapValues

图解Spark Transformation算子

# flatMapValues
x = sc.parallelize([('A',(1,2,3)),('B',(4,5))])
y = x.flatMapValues(lambda x: [i**2 for i in x]) # function is applied to entire value, then result is flattened print(x.collect()) print(y.collect()) [('A', (1, 2, 3)), ('B', (4, 5))] [('A', 1), ('A', 4), ('A', 9), ('B', 16), ('B', 25)]
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11. mapValues

图解Spark Transformation算子

# mapValues
x = sc.parallelize([('A',(1,2,3)),('B',(4,5))])
y = x.mapValues(lambda x: [i**2 for i in x]) # function is applied to entire value print(x.collect()) print(y.collect()) [('A', (1, 2, 3)), ('B', (4, 5))] [('A', [1, 4, 9]), ('B', [16, 25])]
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12. groupWith

图解Spark Transformation算子

# groupWith
x = sc.parallelize([('C',4),('B',(3,3)),('A',2),('A',(1,1))])
y = sc.parallelize([('B',(7,7)),('A',6),('D',(5,5))])
z = sc.parallelize([('D',9),('B',(8,8))])
a = x.groupWith(y,z)
print(x.collect())
print(y.collect())
print(z.collect())
print("Result:")
for key,val in list(a.collect()): 
    print(key, [list(i) for i in val])

[('C', 4), ('B', (3, 3)), ('A', 2), ('A', (1, 1))]
[('B', (7, 7)), ('A', 6), ('D', (5, 5))]
[('D', 9), ('B', (8, 8))]
Result:
D [[], [(5, 5)], [9]]
C [[4], [], []]
B [[(3, 3)], [(7, 7)], [(8, 8)]]
A [[2, (1, 1)], [6], []]
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13. cogroup

图解Spark Transformation算子

# cogroup
x = sc.parallelize([('C',4),('B',(3,3)),('A',2),('A',(1,1))])
y = sc.parallelize([('A',8),('B',7),('A',6),('D',(5,5))])
z = x.cogroup(y)
print(x.collect())
print(y.collect())
for key,val in list(z.collect()):
    print(key, [list(i) for i in val])

[('C', 4), ('B', (3, 3)), ('A', 2), ('A', (1, 1))]
[('A', 8), ('B', 7), ('A', 6), ('D', (5, 5))]
A [[2, (1, 1)], [8, 6]]
C [[4], []]
B [[(3, 3)], [7]]
D [[], [(5, 5)]]
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14. sampleByKey

图解Spark Transformation算子

# sampleByKey
x = sc.parallelize([('A',1),('B',2),('C',3),('B',4),('A',5)])
y = x.sampleByKey(withReplacement=False, fractions={'A':0.5, 'B':1, 'C':0.2})
print(x.collect())
print(y.collect())

[('A', 1), ('B', 2), ('C', 3), ('B', 4), ('A', 5)]
[('B', 2), ('C', 3), ('B', 4)]
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15. subtractByKey

图解Spark Transformation算子

# subtractByKey
x = sc.parallelize([('C',1),('B',2),('A',3),('A',4)])
y = sc.parallelize([('A',5),('D',6),('A',7),('D',8)])
z = x.subtractByKey(y)
print(x.collect())
print(y.collect())
print(z.collect())

[('C', 1), ('B', 2), ('A', 3), ('A', 4)]
[('A', 5), ('D', 6), ('A', 7), ('D', 8)]
[('C', 1), ('B', 2)]
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16. subtract

图解Spark Transformation算子

# subtract
x = sc.parallelize([('C',4),('B',3),('A',2),('A',1)])
y = sc.parallelize([('C',8),('A',2),('D',1)])
z = x.subtract(y)
print(x.collect())
print(y.collect())
print(z.collect())

[('C', 4), ('B', 3), ('A', 2), ('A', 1)]
[('C', 8), ('A', 2), ('D', 1)]
[('A', 1), ('C', 4), ('B', 3)]
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17. keyBy

图解Spark Transformation算子

# keyBy
x = sc.parallelize([1,2,3])
y = x.keyBy(lambda x: x**2)
print(x.collect()) print(y.collect()) [1, 2, 3] [(1, 1), (4, 2), (9, 3)]
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18. repartition

图解Spark Transformation算子

# repartition
x = sc.parallelize([1,2,3,4,5],2)
y = x.repartition(numPartitions=3)
print(x.glom().collect())
print(y.glom().collect())

[[1, 2], [3, 4, 5]]
[[], [1, 2, 3, 4], [5]]
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19. coalesce

图解Spark Transformation算子

# coalesce
x = sc.parallelize([1,2,3,4,5],2)
y = x.coalesce(numPartitions=1)
print(x.glom().collect())
print(y.glom().collect())

[[1, 2], [3, 4, 5]]
[[1, 2, 3, 4, 5]]
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20. zip

图解Spark Transformation算子

# zip
x = sc.parallelize(['B','A','A'])
# zip expects x and y to have same #partitions and #elements/partition
y = x.map(lambda x: ord(x))  
z = x.zip(y)
print(x.collect())
print(y.collect())
print(z.collect())

['B', 'A', 'A']
[66, 65, 65]
[('B', 66), ('A', 65), ('A', 65)]
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21. zipWithIndex

图解Spark Transformation算子

# zipWithIndex
x = sc.parallelize(['B','A','A'],2)
y = x.zipWithIndex()
print(x.glom().collect())
print(y.collect())

[['B'], ['A', 'A']]
[('B', 0), ('A', 1), ('A', 2)]
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21. zipWithUniqueId

图解Spark Transformation算子

# zipWithUniqueId
x = sc.parallelize(['B','A','A'],2)
y = x.zipWithUniqueId()
print(x.glom().collect())
print(y.collect())

[['B'], ['A', 'A']]
[('B', 0), ('A', 1), ('A', 3)]
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