我的主要目标是将任何df的所有列都转换为字符串,这样比较就容易了。
我已经尝试过以下多种建议的方法。但无法成功:
target_df = target_df.select([col(c).cast("string") for c in target_df.columns])
这给出了错误:
pyspark.sql.utils.AnalysisException: "Can't extract value from SDV#155: need struct type but got string;"
我尝试过的下一个是:
target_df = target_df.select([col(c).cast(StringType()).alias(c) for c in columns_list])
错误:
pyspark.sql.utils.AnalysisException: "Can't extract value from SDV#27: need struct type but got string;"
下一个方法是:
for column in target_df.columns:
target_df = target_df.withColumn(column, target_df[column].cast('string'))
错误:
pyspark.sql.utils.AnalysisException: "Can't extract value from SDV#27: need struct type but got string;"
在强制转换之前存在的几行代码:
columns_list = source_df.columns.copy()
target_df = target_df.toDF(*columns_list)
我正在尝试的示例df模式:
root
|-- A: string (nullable = true)
|-- S: string (nullable = true)
|-- D: string (nullable = true)
|-- F: string (nullable = true)
|-- G: double (nullable = true)
|-- H: double (nullable = true)
|-- J: string (nullable = true)
|-- K: string (nullable = true)
|-- L: string (nullable = true)
|-- M: string (nullable = true)
|-- N: string (nullable = true)
|-- B: string (nullable = true)
|-- V: string (nullable = true)
|-- C: string (nullable = true)
|-- X: string (nullable = true)
|-- Y: string (nullable = true)
|-- U: double (nullable = true)
|-- I: string (nullable = true)
|-- R: string (nullable = true)
|-- T: string (nullable = true)
|-- Q: string (nullable = true)
|-- E: double (nullable = true)
|-- W: string (nullable = true)
|-- AS: string (nullable = true)
|-- DSC: string (nullable = true)
|-- DCV: string (nullable = true)
|-- WV: string (nullable = true)
|-- SDV: string (nullable = true)
|-- SDV.1: string (nullable = true)
|-- WDV: string (nullable = true)
|-- FWFV: string (nullable = true)
|-- ERBVSER: string (nullable = true)
答案 0 :(得分:1)
如建议的那样,错误出自名为.
的列中的点SDV.1
,在选择该列时必须用反引号将其括起来:
for column in target_df.columns:
target_df = target_df.withColumn(column, target_df['`{}`'.format(column)].cast('string'))
或
target_df = target_df.select([col('`{}`'.format(c)).cast(StringType()).alias(c) for c in columns_list])
答案 1 :(得分:0)
我认为您的方法没什么问题
>>> df = spark.createDataFrame([(1,25),(1,20),(1,20),(2,26)],['id','age'])
>>> df.show()
+---+---+
| id|age|
+---+---+
| 1| 25|
| 1| 20|
| 1| 20|
| 2| 26|
+---+---+
>>> df.printSchema()
root
|-- id: long (nullable = true)
|-- age: long (nullable = true)
>>> df.select([col(i).cast('string') for i in df.columns]).printSchema()
root
|-- id: string (nullable = true)
|-- age: string (nullable = true)
>>> df.select([col(i).cast('string') for i in df.columns]).show()
+---+---+
| id|age|
+---+---+
| 1| 25|
| 1| 20|
| 1| 20|
| 2| 26|
+---+---+