ValueError:无法将列转换为bool:请使用'&'代表“和"，"|"构建DataFrame布尔表达式时为'or'，为'〜'为'not' [英] ValueError: Cannot convert column into bool: please use '&' for 'and', '|' for 'or', '~' for 'not' when building DataFrame boolean expressions

问题描述

我在使用此代码删除带有pyspark的嵌套列时遇到此错误.为什么这不起作用?我正在尝试使用波浪号而不是not！=作为错误提示，但它也不起作用.那么在这种情况下您会怎么做?

I got this error while using this code to drop a nested column with pyspark. Why is this not working? I was trying to use a tilde instead of a not != as the error suggests but it doesnt work either. So what do you do in that case?

def drop_col(df, struct_nm, delete_struct_child_col_nm):
    fields_to_keep = filter(lambda x:  x != delete_struct_child_col_nm, df.select(" 
{}.*".format(struct_nm)).columns)
    fields_to_keep = list(map(lambda x:  "{}.{}".format(struct_nm, x), fields_to_keep))
    return df.withColumn(struct_nm, struct(fields_to_keep))

推荐答案

我构建了一个简单的示例，其中包含一个struct列和一些虚拟列:

I built a simple example with a struct column and a few dummy columns:

from pyspark import SQLContext
from pyspark.sql import SparkSession
from pyspark.sql.functions import monotonically_increasing_id, lit, col, struct
from pyspark.sql.types import StructType, StructField, StringType, IntegerType

spark = SparkSession.builder.getOrCreate()
sql_context = SQLContext(spark.sparkContext)
schema = StructType(
    [
        StructField('addresses',
                    StructType(
                        [StructField("state", StringType(), True),
                         StructField("street", StringType(), True),
                        StructField("country", StringType(), True),
                         StructField("code", IntegerType(), True)]
                    )
                    )
    ]
)

rdd = [({'state': 'pa', 'street': 'market', 'country': 'USA', 'code': 100},),
       ({'state': 'ca', 'street': 'baker',  'country': 'USA', 'code': 101},)]

df = sql_context.createDataFrame(rdd, schema)
df = df.withColumn('id', monotonically_increasing_id())
df = df.withColumn('name', lit('test'))

print(df.show())
print(df.printSchema())

输出:

+--------------------+-----------+----+
|           addresses|         id|name|
+--------------------+-----------+----+
|[pa, market, USA,...| 8589934592|test|
|[ca, baker, USA, ...|25769803776|test|
+--------------------+-----------+----+

root
 |-- addresses: struct (nullable = true)
 |    |-- state: string (nullable = true)
 |    |-- street: string (nullable = true)
 |    |-- country: string (nullable = true)
 |    |-- code: integer (nullable = true)
 |-- id: long (nullable = false)
 |-- name: string (nullable = false)

要删除整个struct列，只需使用drop函数:

To drop the whole struct column, you can simply use the drop function:

df2 = df.drop('addresses')
print(df2.show())

输出:

+-----------+----+
|         id|name|
+-----------+----+
| 8589934592|test|
|25769803776|test|
+-----------+----+

要删除特定字段，请在struct列中进行一些复杂操作-这里还有一些其他类似的问题:

To drop specific fields, in a struct column, it's a bit more complicated - there are some other similar questions here:

• 从Spark DataFrame中删除嵌套列
• 使用PySpark删除数据框的嵌套列

• Dropping a nested column from Spark DataFrame
• Dropping nested column of Dataframe with PySpark

无论如何，我发现它们有点复杂-我的方法是将原始列与要保留的struct字段的子集重新分配:

In any case, I found them to be a bit complicated - my approach would just be to reassign the original column with the subset of struct fields you want to keep:

columns_to_keep = ['country', 'code']

df = df.withColumn('addresses', struct(*[f"addresses.{column}" for column in columns_to_keep]))

输出:

+----------+-----------+----+
| addresses|         id|name|
+----------+-----------+----+
|[USA, 100]| 8589934592|test|
|[USA, 101]|25769803776|test|
+----------+-----------+----+

或者，如果您只想指定要删除的列而不是要保留的列:

Alternatively, if you just wanted to specify the columns you want to remove rather than the columns you want to keep:

columns_to_remove = ['country', 'code']
all_columns = df.select("addresses.*").columns
columns_to_keep = list(set(all_columns) - set(columns_to_remove))
df = df.withColumn('addresses', struct(*[f"addresses.{column}" for column in columns_to_keep]))

输出:

+------------+-----------+----+
|   addresses|         id|name|
+------------+-----------+----+
|[pa, market]| 8589934592|test|
| [ca, baker]|25769803776|test|
+------------+-----------+----+

希望这会有所帮助！

这篇关于ValueError:无法将列转换为bool:请使用'&'代表“和"，"|"构建DataFrame布尔表达式时为'or'，为'〜'为'not'的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！