两个答案似乎都可以，但有一个例外 - 如果您有数字列，它将被转换为字符串列.为避免这种情况，请尝试:

Both answers seems to be ok with one exception - if you have numeric column, it will be converted to string column. To avoid this, try:

import org.apache.spark.sql.types._
import org.apache.spark.sql.functions._
val fields = df.schema.fields
val stringFields = df.schema.fields.filter(f => f.dataType == StringType)
val nonStringFields = df.schema.fields.filter(f => f.dataType != StringType).map(f => f.name).map(f => col(f))

val stringFieldsTransformed = stringFields .map (f => f.name).map(f => upper(col(f)).as(f))
val df = sourceDF.select(stringFieldsTransformed ++ nonStringFields: _*)

当您有非字符串字段(即数字字段)时，现在类型也是正确的).如果您知道每一列都是 String 类型，请使用其他答案之一 - 在这种情况下它们是正确的 :)

Now types are correct also when you have non-string fields, i.e. numeric fields). If you know that each column is of String type, use one of the other answers - they are correct in that cases :)

PySpark 中的 Python 代码:

Python code in PySpark:

from pyspark.sql.functions import *
from pyspark.sql.types import *
sourceDF = spark.createDataFrame([(1, "a")], ['n', 'n1'])
 fields = sourceDF.schema.fields
stringFields = filter(lambda f: isinstance(f.dataType, StringType), fields)
nonStringFields = map(lambda f: col(f.name), filter(lambda f: not isinstance(f.dataType, StringType), fields))
stringFieldsTransformed = map(lambda f: upper(col(f.name)), stringFields) 
allFields = [*stringFieldsTransformed, *nonStringFields]
df = sourceDF.select(allFields)