我有一個Spark DataFrame(使用PySpark 1.5.1)，並想添加一個新的列。
我已經嘗試了以下方法，但沒有任何成功的：
type(randomed_hours) # => list
# Create in Python and transform to RDD
new_col = pd.DataFrame(randomed_hours, columns=['new_col'])
spark_new_col = sqlContext.createDataFrame(new_col)
my_df_spark.withColumn("hours", spark_new_col["new_col"])
還有一個錯誤使用這個：
my_df_spark.withColumn("hours", sc.parallelize(randomed_hours))
那麽如何使用PySpark將新的列(基於Python向量)添加到現有的DataFrame？
最佳解決方法
您不能將任意列添加到Spark中的DataFrame。新列隻能使用literal創建(其他literal類型在How to add a constant column in a Spark DataFrame?中描述)
from pyspark.sql.functions import lit
df = sqlContext.createDataFrame(
[(1, "a", 23.0), (3, "B", -23.0)], ("x1", "x2", "x3"))
df_with_x4 = df.withColumn("x4", lit(0))
df_with_x4.show()
## +---+---+-----+---+
## | x1| x2| x3| x4|
## +---+---+-----+---+
## | 1| a| 23.0| 0|
## | 3| B|-23.0| 0|
## +---+---+-----+---+
轉換現有的列：
from pyspark.sql.functions import exp
df_with_x5 = df_with_x4.withColumn("x5", exp("x3"))
df_with_x5.show()
## +---+---+-----+---+--------------------+
## | x1| x2| x3| x4| x5|
## +---+---+-----+---+--------------------+
## | 1| a| 23.0| 0| 9.744803446248903E9|
## | 3| B|-23.0| 0|1.026187963170189...|
## +---+---+-----+---+--------------------+
包括使用join：
from pyspark.sql.functions import exp
lookup = sqlContext.createDataFrame([(1, "foo"), (2, "bar")], ("k", "v"))
df_with_x6 = (df_with_x5
.join(lookup, col("x1") == col("k"), "leftouter")
.drop("k")
.withColumnRenamed("v", "x6"))
## +---+---+-----+---+--------------------+----+
## | x1| x2| x3| x4| x5| x6|
## +---+---+-----+---+--------------------+----+
## | 1| a| 23.0| 0| 9.744803446248903E9| foo|
## | 3| B|-23.0| 0|1.026187963170189...|null|
## +---+---+-----+---+--------------------+----+
或使用function /udf生成：
from pyspark.sql.functions import rand
df_with_x7 = df_with_x6.withColumn("x7", rand())
df_with_x7.show()
## +---+---+-----+---+--------------------+----+-------------------+
## | x1| x2| x3| x4| x5| x6| x7|
## +---+---+-----+---+--------------------+----+-------------------+
## | 1| a| 23.0| 0| 9.744803446248903E9| foo|0.41930610446846617|
## | 3| B|-23.0| 0|1.026187963170189...|null|0.37801881545497873|
## +---+---+-----+---+--------------------+----+-------------------+
映射到Catalyst表達式的性能優先、內置函數(pyspark.sql.functions)通常優於Python用戶定義的函數。
如果你想添加一個任意RDD的內容作為一個列，你可以
-
在RDD上調用
zipWithIndex並將其轉換為數據幀 -
加入這兩個使用索引作為連接鍵
次佳解決方法
使用UDF添加列：
df = sqlContext.createDataFrame(
[(1, "a", 23.0), (3, "B", -23.0)], ("x1", "x2", "x3"))
from pyspark.sql.functions import udf
from pyspark.sql.types import *
def valueToCategory(value):
if value == 1: return 'cat1'
elif value == 2: return 'cat2'
...
else: return 'n/a'
# NOTE: it seems that calls to udf() must be after SparkContext() is called
udfValueToCategory = udf(valueToCategory, StringType())
df_with_cat = df.withColumn("category", udfValueToCategory("x1"))
df_with_cat.show()
## +---+---+-----+---------+
## | x1| x2| x3| category|
## +---+---+-----+---------+
## | 1| a| 23.0| cat1|
## | 3| B|-23.0| n/a|
## +---+---+-----+---------+
第三種解決方法
# assumes schema has 'age' column
df.select('*', (df.age + 10).alias('agePlusTen'))
