获取CSV到Spark数据帧

我在Spark上使用python,并希望将csv放入数据帧 .

Spark SQL的documentation奇怪地没有提供CSV作为源的解释 .

我找到了Spark-CSV,但是我对文档的两个部分有问题:

  • "This package can be added to Spark using the --jars command line option. For example, to include it when starting the spark shell: $ bin/spark-shell --packages com.databricks:spark-csv_2.10:1.0.3" 每次启动pyspark或spark-submit时,我真的需要添加这个参数吗?它似乎非常不优雅 . 有没有办法在python中导入它而不是每次重新加载它?

  • df = sqlContext.load(source="com.databricks.spark.csv", header="true", path = "cars.csv") 即使我这样做,这也行不通 . "source"参数在这行代码中代表什么?我如何简单地在linux上加载本地文件,比如"/Spark_Hadoop/spark-1.3.1-bin-cdh4/cars.csv"?

回答(8)

2 years ago

将csv文件读入RDD,然后从原始RDD生成RowRDD .

创建由与步骤1中创建的RDD中的行结构匹配的StructType表示的模式 .

通过SQLContext提供的createDataFrame方法将模式应用于行的RDD .

lines = sc.textFile("examples/src/main/resources/people.txt")
parts = lines.map(lambda l: l.split(","))
# Each line is converted to a tuple.
people = parts.map(lambda p: (p[0], p[1].strip()))

# The schema is encoded in a string.
schemaString = "name age"

fields = [StructField(field_name, StringType(), True) for field_name in schemaString.split()]
schema = StructType(fields)

# Apply the schema to the RDD.
schemaPeople = spark.createDataFrame(people, schema)

来源:SPARK PROGRAMMING GUIDE

2 years ago

from pyspark.sql.types import StringType
from pyspark import SQLContext
sqlContext = SQLContext(sc)

Employee_rdd = sc.textFile("\..\Employee.csv")
               .map(lambda line: line.split(","))

Employee_df = Employee_rdd.toDF(['Employee_ID','Employee_name'])

Employee_df.show()

2 years ago

随着更新版本的Spark(我相信,1.4),这变得更加容易 . 表达式 sqlContext.read 为您提供DataFrameReader实例,其中包含 .csv() 方法:

df = sqlContext.read.csv("/path/to/your.csv")

请注意,您还可以通过将关键字参数 header=True 添加到 .csv() 调用来指示csv文件具有标头 . 还有一些其他选项可供使用,并在上面的链接中进行了描述 .

2 years ago

如果您不介意额外的包依赖项,可以使用Pandas来解析CSV文件 . 它处理内部逗号就好了 .

依赖关系:

from pyspark import SparkContext
from pyspark.sql import SQLContext
import pandas as pd

立即将整个文件读入Spark DataFrame:

sc = SparkContext('local','example')  # if using locally
sql_sc = SQLContext(sc)

pandas_df = pd.read_csv('file.csv')  # assuming the file contains a header
# If no header:
# pandas_df = pd.read_csv('file.csv', names = ['column 1','column 2']) 
s_df = sql_sc.createDataFrame(pandas_df)

或者,更有数据意识的是,您可以将数据块化为Spark RDD然后DF:

chunk_100k = pd.read_csv('file.csv', chunksize=100000)

for chunky in chunk_100k:
    Spark_temp_rdd = sc.parallelize(chunky.values.tolist())
    try:
        Spark_full_rdd += Spark_temp_rdd
    except NameError:
        Spark_full_rdd = Spark_temp_rdd
    del Spark_temp_rdd

Spark_DF = Spark_full_rdd.toDF(['column 1','column 2'])

2 years ago

遵循Spark 2.0,建议使用Spark会话:

from pyspark.sql import SparkSession
from pyspark.sql import Row

# Create a SparkSession
spark = SparkSession \
    .builder \
    .appName("basic example") \
    .config("spark.some.config.option", "some-value") \
    .getOrCreate()

def mapper(line):
    fields = line.split(',')
    return Row(ID=int(fields[0]), field1=str(fields[1].encode("utf-8")), field2=int(fields[2]), field3=int(fields[3]))

lines = spark.sparkContext.textFile("file.csv")
df = lines.map(mapper)

# Infer the schema, and register the DataFrame as a table.
schemaDf = spark.createDataFrame(df).cache()
schemaDf.createOrReplaceTempView("tablename")

2 years ago

for Pyspark, assuming that the first row of the csv file contains a header

spark = SparkSession.builder.appName('chosenName').getOrCreate()
df=spark.read.csv('fileNameWithPath', mode="DROPMALFORMED",inferSchema=True, header = True)

2 years ago

我遇到了类似的问题 . 解决方案是添加一个名为“PYSPARK_SUBMIT_ARGS”的环境变量,并将其值设置为“--packages com.databricks:spark-csv_2.10:1.4.0 pyspark-shell” . 这适用于Spark的Python交互式shell .

确保将spark-csv的版本与安装的Scala版本相匹配 . 使用Scala 2.11,它是spark-csv_2.11,使用Scala 2.10或2.10.5,它是spark-csv_2.10 .

希望它有效 .

2 years ago

基于Aravind的答案,但更短,例如:

lines = sc.textFile("/path/to/file").map(lambda x: x.split(","))
df = lines.toDF(["year", "month", "day", "count"])