pyspark：ValueError：推断后无法确定某些类型

使用Pyspark处理大规模数据集时，有时会遇到"ValueError: Cannot infer schema for some types"的错误。这个错误通常是由于数据集中存在无法确定类型的值而导致的。在这篇文章中，我们将探讨这个错误的原因，并介绍如何解决这个问题。

python
from pyspark.sql.types import StructType, StructField, StringType, IntegerType
schema = StructType([
    StructField("name", StringType(), nullable=True),
    StructField("age", IntegerType(), nullable=True),
    StructField("city", StringType(), nullable=True)
])
data = [("John", 25, "New York"), ("Alice", 30, "Los Angeles"), ("Bob", 35, "San Francisco")]
df = spark.createDataFrame(data, schema)

python
df = df.fillna({"age": 0})  # 将缺失值填充为0
df = df.filter(df["age"] > 0)  # 过滤掉年龄小于等于0的异常值

python
df = df.withColumn("age", df["age"].cast(IntegerType()))  # 将年龄列的数据类型转换为整数类型

python
from pyspark.sql import SparkSession
from pyspark.sql.types import StructType, StructField, StringType, IntegerType
spark = SparkSession.builder.appName("ValueErrorExample").getOrCreate()
data = [("John", 25, "New York"), ("Alice", 30, "Los Angeles"), ("Bob", 35, "San Francisco"), ("Mike", None, "Chicago")]
schema = StructType([
    StructField("name", StringType(), nullable=True),
    StructField("age", IntegerType(), nullable=True),
    StructField("city", StringType(), nullable=True)
])
df = spark.createDataFrame(data, schema)
df.show()

python
df = df.fillna({"age": 0})
df.show()