如何在 Spark 中对两个具有不同列数的 DataFrame 执行联合?

2025-03-14 08:57:00
admin
原创
57
摘要:问题描述:我有 2DataFrame个:我需要这样的联盟:由于列的数量和名称不同,该unionAll函数不起作用。我怎样才能做到这一点?解决方案 1:Spark 3.1+df = df1.unionByName(df2, allowMissingColumns=True) 测试结果:from pyspark....

问题描述:

我有 2DataFrame个:

源数据

我需要这样的联盟:

在此处输入图片描述

由于列的数量和名称不同,该unionAll函数不起作用。

我怎样才能做到这一点?


解决方案 1:

Spark 3.1+

df = df1.unionByName(df2, allowMissingColumns=True)

测试结果:

from pyspark.sql import SparkSession

spark = SparkSession.builder.getOrCreate()

data1=[
(1 , '2016-08-29', 1 , 2, 3),
(2 , '2016-08-29', 1 , 2, 3),
(3 , '2016-08-29', 1 , 2, 3)]
df1 = spark.createDataFrame(data1, ['code' , 'date' , 'A' , 'B', 'C'])
data2=[
(5 , '2016-08-29', 1, 2, 3, 4),
(6 , '2016-08-29', 1, 2, 3, 4),
(7 , '2016-08-29', 1, 2, 3, 4)]
df2 = spark.createDataFrame(data2, ['code' , 'date' , 'B', 'C', 'D', 'E'])

df = df1.unionByName(df2, allowMissingColumns=True)
df.show()
#     +----+----------+----+---+---+----+----+
#     |code|      date|   A|  B|  C|   D|   E|
#     +----+----------+----+---+---+----+----+
#     |   1|2016-08-29|   1|  2|  3|null|null|
#     |   2|2016-08-29|   1|  2|  3|null|null|
#     |   3|2016-08-29|   1|  2|  3|null|null|
#     |   5|2016-08-29|null|  1|  2|   3|   4|
#     |   6|2016-08-29|null|  1|  2|   3|   4|
#     |   7|2016-08-29|null|  1|  2|   3|   4|
#     +----+----------+----+---+---+----+----+

Spark 2.3+

diff1 = [c for c in df2.columns if c not in df1.columns]
diff2 = [c for c in df1.columns if c not in df2.columns]
df = df1.select('*', *[F.lit(None).alias(c) for c in diff1]) \n    .unionByName(df2.select('*', *[F.lit(None).alias(c) for c in diff2]))

测试结果:

from pyspark.sql import SparkSession, functions as F

spark = SparkSession.builder.getOrCreate()

data1=[
(1 , '2016-08-29', 1 , 2, 3),
(2 , '2016-08-29', 1 , 2, 3),
(3 , '2016-08-29', 1 , 2, 3)]
df1 = spark.createDataFrame(data1, ['code' , 'date' , 'A' , 'B', 'C'])
data2=[
(5 , '2016-08-29', 1, 2, 3, 4),
(6 , '2016-08-29', 1, 2, 3, 4),
(7 , '2016-08-29', 1, 2, 3, 4)]
df2 = spark.createDataFrame(data2, ['code' , 'date' , 'B', 'C', 'D', 'E'])

diff1 = [c for c in df2.columns if c not in df1.columns]
diff2 = [c for c in df1.columns if c not in df2.columns]
df = df1.select('*', *[F.lit(None).alias(c) for c in diff1]) \n    .unionByName(df2.select('*', *[F.lit(None).alias(c) for c in diff2]))
df.show()
#     +----+----------+----+---+---+----+----+
#     |code|      date|   A|  B|  C|   D|   E|
#     +----+----------+----+---+---+----+----+
#     |   1|2016-08-29|   1|  2|  3|null|null|
#     |   2|2016-08-29|   1|  2|  3|null|null|
#     |   3|2016-08-29|   1|  2|  3|null|null|
#     |   5|2016-08-29|null|  1|  2|   3|   4|
#     |   6|2016-08-29|null|  1|  2|   3|   4|
#     |   7|2016-08-29|null|  1|  2|   3|   4|
#     +----+----------+----+---+---+----+----+

解决方案 2:

在 Scala 中,您只需将所有缺失的列附加为nulls

import org.apache.spark.sql.functions._

// let df1 and df2 the Dataframes to merge
val df1 = sc.parallelize(List(
  (50, 2),
  (34, 4)
)).toDF("age", "children")

val df2 = sc.parallelize(List(
  (26, true, 60000.00),
  (32, false, 35000.00)
)).toDF("age", "education", "income")

val cols1 = df1.columns.toSet
val cols2 = df2.columns.toSet
val total = cols1 ++ cols2 // union

def expr(myCols: Set[String], allCols: Set[String]) = {
  allCols.toList.map(x => x match {
    case x if myCols.contains(x) => col(x)
    case _ => lit(null).as(x)
  })
}

df1.select(expr(cols1, total):_*).unionAll(df2.select(expr(cols2, total):_*)).show()

+---+--------+---------+-------+
|age|children|education| income|
+---+--------+---------+-------+
| 50|       2|     null|   null|
| 34|       4|     null|   null|
| 26|    null|     true|60000.0|
| 32|    null|    false|35000.0|
+---+--------+---------+-------+

更新

两个时间DataFrames都会有相同的列顺序,因为我们total在两种情况下都进行了映射。

df1.select(expr(cols1, total):_*).show()
df2.select(expr(cols2, total):_*).show()

+---+--------+---------+------+
|age|children|education|income|
+---+--------+---------+------+
| 50|       2|     null|  null|
| 34|       4|     null|  null|
+---+--------+---------+------+

+---+--------+---------+-------+
|age|children|education| income|
+---+--------+---------+-------+
| 26|    null|     true|60000.0|
| 32|    null|    false|35000.0|
+---+--------+---------+-------+

解决方案 3:

这是我的 Python 版本:

from pyspark.sql import SparkSession, HiveContext
from pyspark.sql.functions import lit
from pyspark.sql import Row

def customUnion(df1, df2):
    cols1 = df1.columns
    cols2 = df2.columns
    total_cols = sorted(cols1 + list(set(cols2) - set(cols1)))
    def expr(mycols, allcols):
        def processCols(colname):
            if colname in mycols:
                return colname
            else:
                return lit(None).alias(colname)
        cols = map(processCols, allcols)
        return list(cols)
    appended = df1.select(expr(cols1, total_cols)).union(df2.select(expr(cols2, total_cols)))
    return appended

以下是示例用法:

data = [
    Row(zip_code=58542, dma='MIN'),
    Row(zip_code=58701, dma='MIN'),
    Row(zip_code=57632, dma='MIN'),
    Row(zip_code=58734, dma='MIN')
]

firstDF = spark.createDataFrame(data)

data = [
    Row(zip_code='534', name='MIN'),
    Row(zip_code='353', name='MIN'),
    Row(zip_code='134', name='MIN'),
    Row(zip_code='245', name='MIN')
]

secondDF = spark.createDataFrame(data)

customUnion(firstDF,secondDF).show()

解决方案 4:

以下是使用 pyspark 的 Python 3.0 代码:

from pyspark.sql.functions import lit


def __order_df_and_add_missing_cols(df, columns_order_list, df_missing_fields):
    """ return ordered dataFrame by the columns order list with null in missing columns """
    if not df_missing_fields:  # no missing fields for the df
        return df.select(columns_order_list)
    else:
        columns = []
        for colName in columns_order_list:
            if colName not in df_missing_fields:
                columns.append(colName)
            else:
                columns.append(lit(None).alias(colName))
        return df.select(columns)


def __add_missing_columns(df, missing_column_names):
    """ Add missing columns as null in the end of the columns list """
    list_missing_columns = []
    for col in missing_column_names:
        list_missing_columns.append(lit(None).alias(col))

    return df.select(df.schema.names + list_missing_columns)


def __order_and_union_d_fs(left_df, right_df, left_list_miss_cols, right_list_miss_cols):
    """ return union of data frames with ordered columns by left_df. """
    left_df_all_cols = __add_missing_columns(left_df, left_list_miss_cols)
    right_df_all_cols = __order_df_and_add_missing_cols(right_df, left_df_all_cols.schema.names,
                                                        right_list_miss_cols)
    return left_df_all_cols.union(right_df_all_cols)


def union_d_fs(left_df, right_df):
    """ Union between two dataFrames, if there is a gap of column fields,
     it will append all missing columns as nulls """
    # Check for None input
    if left_df is None:
        raise ValueError('left_df parameter should not be None')
    if right_df is None:
        raise ValueError('right_df parameter should not be None')
        # For data frames with equal columns and order- regular union
    if left_df.schema.names == right_df.schema.names:
        return left_df.union(right_df)
    else:  # Different columns
        # Save dataFrame columns name list as set
        left_df_col_list = set(left_df.schema.names)
        right_df_col_list = set(right_df.schema.names)
        # Diff columns between left_df and right_df
        right_list_miss_cols = list(left_df_col_list - right_df_col_list)
        left_list_miss_cols = list(right_df_col_list - left_df_col_list)
        return __order_and_union_d_fs(left_df, right_df, left_list_miss_cols, right_list_miss_cols)

解决方案 5:

一个非常简单的方法来做到这一点 -select数据框中的列按相同的顺序使用unionAll

df1.select('code', 'date', 'A', 'B', 'C', lit(None).alias('D'), lit(None).alias('E'))\n   .unionAll(df2.select('code', 'date', lit(None).alias('A'), 'B', 'C', 'D', 'E'))

解决方案 6:

这是一个 pyspark 解决方案。

它假设如果 中的字段df1缺失df2,则将该缺失字段添加到df2并使用空值。但是它还假设如果该字段存在于两个数据框中,但该字段的类型或可空性不同,则两个数据框会发生冲突并且无法组合。在这种情况下,我提出了TypeError

from pyspark.sql.functions import lit

def harmonize_schemas_and_combine(df_left, df_right):
    left_types = {f.name: f.dataType for f in df_left.schema}
    right_types = {f.name: f.dataType for f in df_right.schema}
    left_fields = set((f.name, f.dataType, f.nullable) for f in df_left.schema)
    right_fields = set((f.name, f.dataType, f.nullable) for f in df_right.schema)

    # First go over left-unique fields
    for l_name, l_type, l_nullable in left_fields.difference(right_fields):
        if l_name in right_types:
            r_type = right_types[l_name]
            if l_type != r_type:
                raise TypeError, "Union failed. Type conflict on field %s. left type %s, right type %s" % (l_name, l_type, r_type)
            else:
                raise TypeError, "Union failed. Nullability conflict on field %s. left nullable %s, right nullable %s"  % (l_name, l_nullable, not(l_nullable))
        df_right = df_right.withColumn(l_name, lit(None).cast(l_type))

    # Now go over right-unique fields
    for r_name, r_type, r_nullable in right_fields.difference(left_fields):
        if r_name in left_types:
            l_type = left_types[r_name]
            if r_type != l_type:
                raise TypeError, "Union failed. Type conflict on field %s. right type %s, left type %s" % (r_name, r_type, l_type)
            else:
                raise TypeError, "Union failed. Nullability conflict on field %s. right nullable %s, left nullable %s" % (r_name, r_nullable, not(r_nullable))
        df_left = df_left.withColumn(r_name, lit(None).cast(r_type))    

    # Make sure columns are in the same order
    df_left = df_left.select(df_right.columns)

    return df_left.union(df_right)

解决方案 7:

lit(None)如果你只是使用简单的解决方法(这也是我知道的唯一方法),我发现这里的大多数 python 答案写得有点太笨拙了。作为替代方案,这可能会有用:

# df1 and df2 are assumed to be the given dataFrames from the question

# Get the lacking columns for each dataframe and set them to null in the respective dataFrame.
# First do so for df1...
for column in [column for column in df1.columns if column not in df2.columns]:
    df1 = df1.withColumn(column, lit(None))

# ... and then for df2
for column in [column for column in df2.columns if column not in df1.columns]:
    df2 = df2.withColumn(column, lit(None))

之后只需执行union()您想执行的操作。

注意:如果您的列顺序df1df2使用不同unionByName()

result = df1.unionByName(df2)

解决方案 8:

修改了 Alberto Bonsanto 的版本以保留原始列顺序(OP 暗示顺序应与原始表格一致)。此外,该match部分引发了 Intellij 警告。

这是我的版本:

def unionDifferentTables(df1: DataFrame, df2: DataFrame): DataFrame = {

  val cols1 = df1.columns.toSet
  val cols2 = df2.columns.toSet
  val total = cols1 ++ cols2 // union

  val order = df1.columns ++  df2.columns
  val sorted = total.toList.sortWith((a,b)=> order.indexOf(a) < order.indexOf(b))

  def expr(myCols: Set[String], allCols: List[String]) = {
      allCols.map( {
        case x if myCols.contains(x) => col(x)
        case y => lit(null).as(y)
      })
  }

  df1.select(expr(cols1, sorted): _*).unionAll(df2.select(expr(cols2, sorted): _*))
}

解决方案 9:

在pyspark中:

df = df1.join(df2, ['each', 'shared', 'col'], how='full')

解决方案 10:

我遇到了同样的问题,使用 join 而不是 union 解决了我的问题。因此,例如使用python,而不是这行代码:
result = left.union(right),这将无法针对不同数量的列执行,您应该使用这个:

result = left.join(right, left.columns if (len(left.columns) < len(right.columns)) else right.columns, "outer")

请注意,第二个参数包含两个 DataFrame 之间的公共列。如果不使用它,结果将具有重复的列,其中一个为空,另一个不为空。希望对您有所帮助。

解决方案 11:

有许多简洁的方法可以解决这个问题,但需要适度牺牲性能。

def unionWithDifferentSchema(a: DataFrame, b: DataFrame): DataFrame = {
    sparkSession.read.json(a.toJSON.union(b.toJSON).rdd)
}

这是可以解决问题的函数。对每个数据框使用 toJSON 会生成一个 json Union。这样可以保留顺序和数据类型。

唯一的缺点是 toJSON 相对昂贵(不过不多,速度可能会减慢 10-15%)。不过这可以保持代码整洁。

解决方案 12:

我的 Java 版本:

    private static Dataset<Row> unionDatasets(Dataset<Row> one, Dataset<Row> another) {
        StructType firstSchema = one.schema();
        List<String> anotherFields = Arrays.asList(another.schema().fieldNames());
        another = balanceDataset(another, firstSchema, anotherFields);
        StructType secondSchema = another.schema();
        List<String> oneFields = Arrays.asList(one.schema().fieldNames());
        one = balanceDataset(one, secondSchema, oneFields);
        return another.unionByName(one);
    }

    private static Dataset<Row> balanceDataset(Dataset<Row> dataset, StructType schema, List<String> fields) {
        for (StructField e : schema.fields()) {
            if (!fields.contains(e.name())) {
                dataset = dataset
                        .withColumn(e.name(),
                                lit(null));
                dataset = dataset.withColumn(e.name(),
                        dataset.col(e.name()).cast(Optional.ofNullable(e.dataType()).orElse(StringType)));
            }
        }
        return dataset;
    }

解决方案 13:

此函数接收两个具有不同架构的数据框(df1 和 df2)并将它们合并。首先,我们需要通过将所有(缺失的)列从 df1 添加到 df2 并反之亦然,将它们带到同一个架构。要向 df 添加新的空列,我们需要指定数据类型。

import pyspark.sql.functions as F
    
def union_different_schemas(df1, df2):
   # Get a list of all column names in both dfs
   columns_df1 = df1.columns
   columns_df2 = df2.columns
   # Get a list of datatypes of the columns
   data_types_df1 = [i.dataType for i in df1.schema.fields]
   data_types_df2 = [i.dataType for i in df2.schema.fields]
   # We go through all columns in df1 and if they are not in df2, we add 
   # them (and specify the correct datatype too)
   for col, typ in zip(columns_df1, data_types_df1):
      if col not in df2.columns:
         df2 = df2\n            .withColumn(col, F.lit(None).cast(typ))
   # Now df2 has all missing columns from df1, let's do the same for df1
   for col, typ in zip(columns_df2, data_types_df2):
      if col not in df1.columns:
         df1 = df1\n            .withColumn(col, F.lit(None).cast(typ))
   # Now df1 and df2 have the same columns, not necessarily in the same 
   # order, therefore we use unionByName
   combined_df = df1\n      .unionByName(df2)

   return combined_df

解决方案 14:

这是 Scala 中的版本,也在这里回答了,还有一个 Pyspark 版本..(Spark - 将具有不同模式(列名和序列)的 DataFrame 合并/合并到具有主通用模式的 DataFrame) -

它需要合并数据框列表。所有数据框中相同名称的列应该具有相同的数据类型。

def unionPro(DFList: List[DataFrame], spark: org.apache.spark.sql.SparkSession): DataFrame = {

    /**
     * This Function Accepts DataFrame with same or Different Schema/Column Order.With some or none common columns
     * Creates a Unioned DataFrame
     */

    import spark.implicits._

    val MasterColList: Array[String] = DFList.map(_.columns).reduce((x, y) => (x.union(y))).distinct

    def unionExpr(myCols: Seq[String], allCols: Seq[String]): Seq[org.apache.spark.sql.Column] = {
      allCols.toList.map(x => x match {
        case x if myCols.contains(x) => col(x)
        case _                       => lit(null).as(x)
      })
    }

    // Create EmptyDF , ignoring different Datatype in StructField and treating them same based on Name ignoring cases

    val masterSchema = StructType(DFList.map(_.schema.fields).reduce((x, y) => (x.union(y))).groupBy(_.name.toUpperCase).map(_._2.head).toArray)

    val masterEmptyDF = spark.createDataFrame(spark.sparkContext.emptyRDD[Row], masterSchema).select(MasterColList.head, MasterColList.tail: _*)

    DFList.map(df => df.select(unionExpr(df.columns, MasterColList): _*)).foldLeft(masterEmptyDF)((x, y) => x.union(y))

  }

以下是示例测试 -


    val aDF = Seq(("A", 1), ("B", 2)).toDF("Name", "ID")
    val bDF = Seq(("C", 1, "D1"), ("D", 2, "D2")).toDF("Name", "Sal", "Deptt")
    unionPro(List(aDF, bDF), spark).show

输出结果为 -

+----+----+----+-----+
|Name|  ID| Sal|Deptt|
+----+----+----+-----+
|   A|   1|null| null|
|   B|   2|null| null|
|   C|null|   1|   D1|
|   D|null|   2|   D2|
+----+----+----+-----+

解决方案 15:

派斯帕克

Alberto 的 Scala 版本运行良好。但是,如果你想要进行 for 循环或一些变量的动态赋值,你可能会遇到一些问题。解决方案来自 Pyspark - 干净的代码:

from pyspark.sql.functions import *

#defining dataframes
df1 = spark.createDataFrame(
    [
        (1, 'foo','ok'), 
        (2, 'pro','ok')
    ],
    ['id', 'txt','check']
)

df2 = spark.createDataFrame(
    [
        (3, 'yep',13,'mo'), 
        (4, 'bro',11,'re')
        
    ],
    ['id', 'txt','value','more'] 
) 

#retrieving columns
cols1 = df1.columns
cols2 = df2.columns

#getting columns from df1 and df2
total = list(set(cols2) | set(cols1)) 

#defining function for adding nulls (None in case of pyspark)
def addnulls(yourDF): 
  for x in total:
    if not x in yourDF.columns:
      yourDF = yourDF.withColumn(x,lit(None))
  return yourDF

df1 = addnulls(df1)
df2 = addnulls(df2)


#additional sorting for correct unionAll (it concatenates DFs by column number)
df1.select(sorted(df1.columns)).unionAll(df2.select(sorted(df2.columns))).show()

+-----+---+----+---+-----+
|check| id|more|txt|value|
+-----+---+----+---+-----+
|   ok|  1|null|foo| null|
|   ok|  2|null|pro| null|
| null|  3|  mo|yep|   13|
| null|  4|  re|bro|   11|
+-----+---+----+---+-----+

解决方案 16:

from functools import reduce
from pyspark.sql import DataFrame
import pyspark.sql.functions as F

def unionAll(*dfs, fill_by=None):
    clmns = {clm.name.lower(): (clm.dataType, clm.name) for df in dfs for clm in df.schema.fields}
    
    dfs = list(dfs)
    for i, df in enumerate(dfs):
        df_clmns = [clm.lower() for clm in df.columns]
        for clm, (dataType, name) in clmns.items():
            if clm not in df_clmns:
                # Add the missing column
                dfs[i] = dfs[i].withColumn(name, F.lit(fill_by).cast(dataType))
    return reduce(DataFrame.unionByName, dfs)
unionAll(df1, df2).show()
  1. 不区分大小写的列

  2. 将返回实际的列大小写

  3. 支持现有的数据类型

  4. 默认值可自定义

  5. 一次传递多个数据框(例如 unionAll(df1, df2, df3, ..., df10))

解决方案 17:

这是另一个:

def unite(df1: DataFrame, df2: DataFrame): DataFrame = {
    val cols1 = df1.columns.toSet
    val cols2 = df2.columns.toSet
    val total = (cols1 ++ cols2).toSeq.sorted
    val expr1 = total.map(c => {
      if (cols1.contains(c)) c else "NULL as " + c
    })
    val expr2 = total.map(c => {
      if (cols2.contains(c)) c else "NULL as " + c
    })
    df1.selectExpr(expr1:_*).union(
      df2.selectExpr(expr2:_*)
    )
}

解决方案 18:

Pyspark DataFrame 连接的联合和外联合。这适用于具有不同列的多个数据框。

def union_all(*dfs):
    return reduce(ps.sql.DataFrame.unionAll, dfs)

def outer_union_all(*dfs):

    all_cols = set([])
    for df in dfs:
        all_cols |= set(df.columns) 
    all_cols = list(all_cols)
    print(all_cols)

    def expr(cols, all_cols):

        def append_cols(col):
            if col in cols:
                return col
            else:
                return sqlfunc.lit(None).alias(col)

        cols_ = map(append_cols, all_cols)
        return list(cols_)

    union_df = union_all(*[df.select(expr(df.columns, all_cols)) for df in dfs])
    return union_df

解决方案 19:

还有一个用于合并列表的通用方法DataFrame

def unionFrames(dfs: Seq[DataFrame]): DataFrame = {
    dfs match {
      case Nil => session.emptyDataFrame // or throw an exception?
      case x :: Nil => x
      case _ =>
        //Preserving Column order from left to right DF's column order
        val allColumns = dfs.foldLeft(collection.mutable.ArrayBuffer.empty[String])((a, b) => a ++ b.columns).distinct

        val appendMissingColumns = (df: DataFrame) => {
          val columns = df.columns.toSet
          df.select(allColumns.map(c => if (columns.contains(c)) col(c) else lit(null).as(c)): _*)
        }

        dfs.tail.foldLeft(appendMissingColumns(dfs.head))((a, b) => a.union(appendMissingColumns(b)))
    }

解决方案 20:

这是我的 pyspark 版本:

from functools import reduce
from pyspark.sql.functions import lit

def concat(dfs):
    # when the dataframes to combine do not have the same order of columns
    # https://datascience.stackexchange.com/a/27231/15325
    return reduce(lambda df1, df2: df1.union(df2.select(df1.columns)), dfs) 

def union_all(dfs):
    columns = reduce(lambda x, y : set(x).union(set(y)), [ i.columns for i in dfs ]  )

    for i in range(len(dfs)):
        d = dfs[i]
        for c in columns:
            if c not in d.columns:
                d = d.withColumn(c, lit(None))
        dfs[i] = d

    return concat(dfs)

解决方案 21:

或者您可以使用完全连接。

list_of_files = ['test1.parquet', 'test2.parquet']

def merged_frames():
  if list_of_files:
    frames = [spark.read.parquet(df.path) for df in list_of_files]
    if frames:
      df = frames[0]
      if frames[1]:
        var = 1
        for element in range(len(frames)-1):
          result_df = df.join(frames[var], 'primary_key', how='full')
          var += 1
    display(result_df)

解决方案 22:

如果您正在从文件加载,我想您只需使用带有文件列表的读取功能即可。

    # file_paths is list of files with different schema
    df = spark.read.option("mergeSchema", "true").json(file_paths)

生成的数据框将具有合并的列。

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