我来自熊猫的背景，我习惯了从CSV文件读取数据到一个dataframe，然后简单地改变列名使用简单的命令有用的东西:
df.columns = new_column_name_list
然而，这在使用sqlContext创建的PySpark数据框架中是行不通的。
我能想到的唯一解决办法是:
df = sqlContext.read.format("com.databricks.spark.csv").options(header='false', inferschema='true', delimiter='\t').load("data.txt")
oldSchema = df.schema
for i,k in enumerate(oldSchema.fields):
k.name = new_column_name_list[i]
df = sqlContext.read.format("com.databricks.spark.csv").options(header='false', delimiter='\t').load("data.txt", schema=oldSchema)
这基本上是定义变量两次，首先推断模式，然后重命名列名，然后用更新的模式再次加载数据框架。
有没有更好更有效的方法来做到这一点，就像我们对熊猫做的那样?
我的Spark版本是1.5.0
我们可以使用各种方法重命名列名。
首先，让我们创建一个简单的数据框架。
df = spark.createDataFrame([("x", 1), ("y", 2)],
["col_1", "col_2"])
现在我们试着把col_1重命名为col_3。PFB的几个方法也一样。
# Approach - 1 : using withColumnRenamed function.
df.withColumnRenamed("col_1", "col_3").show()
# Approach - 2 : using alias function.
df.select(df["col_1"].alias("col3"), "col_2").show()
# Approach - 3 : using selectExpr function.
df.selectExpr("col_1 as col_3", "col_2").show()
# Rename all columns
# Approach - 4 : using toDF function. Here you need to pass the list of all columns present in DataFrame.
df.toDF("col_3", "col_2").show()
这是输出。
+-----+-----+
|col_3|col_2|
+-----+-----+
| x| 1|
| y| 2|
+-----+-----+
我希望这能有所帮助。
试试下面的方法。下面的方法允许您重命名多个文件的列
参考:https://www.linkedin.com/pulse/pyspark-methods-rename-columns-kyle-gibson/
df_initial = spark.read.load('com.databricks.spark.csv')
rename_dict = {
'Alberto':'Name',
'Dakota':'askdaosdka'
}
df_renamed = df_initial \
.select([col(c).alias(rename_dict.get(c, c)) for c in df_initial.columns])
rename_dict = {
'FName':'FirstName',
'LName':'LastName',
'DOB':'BirthDate'
}
return df.select([col(c).alias(rename_dict.get(c, c)) for c in df.columns])
df_renamed = spark.read.load('/mnt/datalake/bronze/testData') \
.transform(renameColumns)
有很多方法可以做到这一点:
Option 1. Using selectExpr.
data = sqlContext.createDataFrame([("Alberto", 2), ("Dakota", 2)],
["Name", "askdaosdka"])
data.show()
data.printSchema()
# Output
#+-------+----------+
#| Name|askdaosdka|
#+-------+----------+
#|Alberto| 2|
#| Dakota| 2|
#+-------+----------+
#root
# |-- Name: string (nullable = true)
# |-- askdaosdka: long (nullable = true)
df = data.selectExpr("Name as name", "askdaosdka as age")
df.show()
df.printSchema()
# Output
#+-------+---+
#| name|age|
#+-------+---+
#|Alberto| 2|
#| Dakota| 2|
#+-------+---+
#root
# |-- name: string (nullable = true)
# |-- age: long (nullable = true)
Option 2. Using withColumnRenamed, notice that this method allows you to "overwrite" the same column. For Python3, replace xrange with range.
from functools import reduce
oldColumns = data.schema.names
newColumns = ["name", "age"]
df = reduce(lambda data, idx: data.withColumnRenamed(oldColumns[idx], newColumns[idx]), xrange(len(oldColumns)), data)
df.printSchema()
df.show()
Option 3. using
alias, in Scala you can also use as.
from pyspark.sql.functions import col
data = data.select(col("Name").alias("name"), col("askdaosdka").alias("age"))
data.show()
# Output
#+-------+---+
#| name|age|
#+-------+---+
#|Alberto| 2|
#| Dakota| 2|
#+-------+---+
Option 4. Using sqlContext.sql, which lets you use SQL queries on DataFrames registered as tables.
sqlContext.registerDataFrameAsTable(data, "myTable")
df2 = sqlContext.sql("SELECT Name AS name, askdaosdka as age from myTable")
df2.show()
# Output
#+-------+---+
#| name|age|
#+-------+---+
#|Alberto| 2|
#| Dakota| 2|
#+-------+---+
