Question

我有一个Spark DataFrame（使用PySpark 1.5.1）并想添加一个新列。

我尝试了以下内容但没有取得任何成功：

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"])

使用此错误也是错误的：

my_df_spark.withColumn("hours",  sc.parallelize(randomed_hours))

那么如何使用PySpark将新列（基于Python矢量）添加到现有的DataFrame中呢？

Answer 1

您无法在Spark中向DataFrame添加任意列。只能使用文字创建新列（其他文字类型在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|
## +---+---+-----+---+

转换现有列：

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...|
## +---+---+-----+---+--------------------+

包含使用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|
## +---+---+-----+---+--------------------+----+

或使用函数/ 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|
## +---+---+-----+---+--------------------+----+-------------------+

性能方面的内置函数（pyspark.sql.functions）映射到Catalyst表达式，通常比Python用户定义的函数更受欢迎。

如果您想将任意RDD的内容添加为列，您可以

添加row numbers to existing data frame
在RDD上调用zipWithIndex并将其转换为数据框
使用index作为连接键加入

Answer 2

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|
## +---+---+-----+---------+

Answer 3

Spark 2.0

# assumes schema has 'age' column 
df.select('*', (df.age + 10).alias('agePlusTen'))

Answer 4

我们可以通过多种方式在pySpark中添加新列。

首先创建一个简单的DataFrame。

date = [27, 28, 29, None, 30, 31]
df = spark.createDataFrame(date, IntegerType())

现在让我们尝试将列值加倍并将其存储在新列中。 PFB很少有几种方法可以达到相同的目的。

# Approach - 1 : using withColumn function
df.withColumn("double", df.value * 2).show()

# Approach - 2 : using select with alias function.
df.select("*", (df.value * 2).alias("double")).show()

# Approach - 3 : using selectExpr function with as clause.
df.selectExpr("*", "value * 2 as double").show()

# Approach - 4 : Using as clause in SQL statement.
df.createTempView("temp")
spark.sql("select *, value * 2 as double from temp").show()

有关Spark DataFrame函数的更多示例和说明，您可以访问我的blog。

我希望这会有所帮助。

Answer 5

添加列的最简单方法是使用“ withColumn”。由于数据框是使用sqlContext创建的，因此您必须指定架构或默认情况下可以在数据集中使用。如果指定了架构，则每次更改时工作量都会变得很乏味。

以下是您可以考虑的示例：

from pyspark.sql import SQLContext
from pyspark.sql.types import *
sqlContext = SQLContext(sc) # SparkContext will be sc by default 

# Read the dataset of your choice (Already loaded with schema)
Data = sqlContext.read.csv("/path", header = True/False, schema = "infer", sep = "delimiter")

# For instance the data has 30 columns from col1, col2, ... col30. If you want to add a 31st column, you can do so by the following:
Data = Data.withColumn("col31", "Code goes here")

# Check the change 
Data.printSchema()

Answer 6

我们可以通过以下步骤直接向DataFrame添加其他列：

from pyspark.sql.functions import when
df = spark.createDataFrame([["amit", 30], ["rohit", 45], ["sameer", 50]], ["name", "age"])
df = df.withColumn("profile", when(df.age >= 40, "Senior").otherwise("Executive"))
df.show()

Answer 7

添加带有一些自定义值或动态值计算的新列，这些值将根据现有列进行填充。

例如

|ColumnA | ColumnB |
|--------|---------|
| 10     | 15      |
| 10     | 20      |
| 10     | 30      |

和新的 ColumnC 作为 ColumnA+ColumnB

|ColumnA | ColumnB | ColumnC|
|--------|---------|--------|
| 10     | 15      | 25     |
| 10     | 20      | 30     |
| 10     | 30      | 40     |

使用

#to add new column
def customColumnVal(row):
rd=row.asDict()
rd["ColumnC"]=row["ColumnA"] + row["ColumnB"]

new_row=Row(**rd)
return new_row
----------------------------
#convert DF to RDD
df_rdd= input_dataframe.rdd

#apply new fucntion to rdd
output_dataframe=df_rdd.map(customColumnVal).toDF()

input_dataframe 是将被修改的数据框，customColumnVal 函数具有添加新列的代码。

Answer 8

添加udf时，您可以定义新的column_name：

u_f = F.udf(lambda :yourstring,StringType())
a.select(u_f().alias('column_name')

Answer 9

from pyspark.sql.functions import udf
from pyspark.sql.types import *
func_name = udf(
    lambda val: val, # do sth to val
    StringType()
)
df.withColumn('new_col', func_name(df.old_col))

如何向Spark DataFrame添加新列（使用PySpark）？

9 个答案: