29

我正在尝试在给定某些数据的情况下手动创建 pyspark 数据框：

row_in=[(1566429545575348),(40.353977),(-111.701859)]
rdd=sc.parallelize(row_in)
schema = StructType([StructField("time_epocs", DecimalType(),    True),StructField("lat", DecimalType(),True),StructField("long", DecimalType(),True)])
df_in_test=spark.createDataFrame(rdd,schema)

当我尝试显示数据框时，这会出错，所以我不知道该怎么做。

但是，这里的 Spark 文档对我来说似乎有点令人费解，当我尝试遵循这些说明时遇到了类似的错误。

有谁知道如何做到这一点？

4

6 回答 6

70

简单的数据框创建：

df = spark.createDataFrame(
    [
        (1, "foo"),  # create your data here, be consistent in the types.
        (2, "bar"),
    ],
    ["id", "label"]  # add your column names here
)

df.printSchema()
root
 |-- id: long (nullable = true)
 |-- label: string (nullable = true)

df.show()
+---+-----+                                                                     
| id|label|
+---+-----+
|  1|  foo|
|  2|  bar|
+---+-----+

根据官方文档：

  • 当 schema 是列名列表时，将从数据中推断出每列的类型。（上例↑）
  • 当模式是pyspark.sql.types.DataType或数据类型字符串时，它必须与真实数据匹配。（以下示例↓）
# Example with a datatype string
df = spark.createDataFrame(
    [
        (1, "foo"),  # Add your data here
        (2, "bar"),
    ],  
    "id int, label string",  # add column names and types here
)

# Example with pyspark.sql.types
from pyspark.sql import types as T
df = spark.createDataFrame(
    [
        (1, "foo"),  # Add your data here
        (2, "bar"),
    ],
    T.StructType(  # Define the whole schema within a StructType
        [
            T.StructField("id", T.IntegerType(), True),
            T.StructField("label", T.StringType(), True),
        ]
    ),
)


df.printSchema()
root
 |-- id: integer (nullable = true)  # type is forced to Int
 |-- label: string (nullable = true)

此外，您可以从 Pandas 数据框创建数据框，架构将从 Pandas 数据框的类型推断：

import pandas as pd
import numpy as np


pdf = pd.DataFrame(
    {
        "col1": [np.random.randint(10) for x in range(10)],
        "col2": [np.random.randint(100) for x in range(10)],
    }
)


df= spark.createDataFrame(pdf)

df.show()
+----+----+
|col1|col2|
+----+----+
|   6|   4|
|   1|  39|
|   7|   4|
|   7|  95|
|   6|   3|
|   7|  28|
|   2|  26|
|   0|   4|
|   4|  32|
+----+----+
于 2019-09-16T15:40:31.820 回答
6

详细说明/建立@Steven的答案：

field = [
    StructField("MULTIPLIER", FloatType(), True),
    StructField("DESCRIPTION", StringType(), True),
]
schema = StructType(field)
multiplier_df = sqlContext.createDataFrame(sc.emptyRDD(), schema)

将创建一个空白数据框。

我们现在可以简单地添加一行：

l = [(2.3, "this is a sample description")]
rdd = sc.parallelize(l)
multiplier_df_temp = spark.createDataFrame(rdd, schema)
multiplier_df = wtp_multiplier_df.union(wtp_multiplier_df_temp)
于 2019-09-20T18:00:03.233 回答
4

createDataFrame此答案演示了如何使用、create_df和来创建 PySpark DataFrame toDF。

df = spark.createDataFrame([("joe", 34), ("luisa", 22)], ["first_name", "age"])

df.show()

+----------+---+
|first_name|age|
+----------+---+
|       joe| 34|
|     luisa| 22|
+----------+---+

您还可以传递createDataFrameRDD 和模式以更精确地构造 DataFrame：

from pyspark.sql import Row
from pyspark.sql.types import *

rdd = spark.sparkContext.parallelize([
    Row(name='Allie', age=2),
    Row(name='Sara', age=33),
    Row(name='Grace', age=31)])

schema = schema = StructType([
   StructField("name", StringType(), True),
   StructField("age", IntegerType(), False)])

df = spark.createDataFrame(rdd, schema)

df.show()

+-----+---+
| name|age|
+-----+---+
|Allie|  2|
| Sara| 33|
|Grace| 31|
+-----+---+

create_df从我的Quinn项目中可以实现两全其美 - 它简洁且完全描述性：

from pyspark.sql.types import *
from quinn.extensions import *

df = spark.create_df(
    [("jose", "a"), ("li", "b"), ("sam", "c")],
    [("name", StringType(), True), ("blah", StringType(), True)]
)

df.show()

+----+----+
|name|blah|
+----+----+
|jose|   a|
|  li|   b|
| sam|   c|
+----+----+

toDF与其他方法相比没有任何优势：

from pyspark.sql import Row

rdd = spark.sparkContext.parallelize([
    Row(name='Allie', age=2),
    Row(name='Sara', age=33),
    Row(name='Grace', age=31)])
df = rdd.toDF()
df.show()

+-----+---+
| name|age|
+-----+---+
|Allie|  2|
| Sara| 33|
|Grace| 31|
+-----+---+
于 2021-02-24T03:40:18.337 回答
1

带格式

from pyspark.sql import SparkSession
from pyspark.sql.types import StructField, StructType, IntegerType, StringType

spark = SparkSession.builder.getOrCreate()
df = spark.createDataFrame(
    [
        (1, "foo"),
        (2, "bar"),
    ],
    StructType(
        [
            StructField("id", IntegerType(), False),
            StructField("txt", StringType(), False),
        ]
    ),
)
print(df.dtypes)
df.show()
于 2021-03-09T11:47:56.613 回答
1

扩展@Steven 的答案：

data = [(i, 'foo') for i in range(1000)] # random data

columns = ['id', 'txt']    # add your columns label here

df = spark.createDataFrame(data, columns)

注意：当schema是列名列表时，将从数据中推断出每列的类型。

如果您想专门定义架构，请执行以下操作：

from pyspark.sql.types import StructType, StructField, IntegerType, StringType
schema = StructType([StructField("id", IntegerType(), True), StructField("txt", StringType(), True)])
df1 = spark.createDataFrame(data, schema)

输出：

>>> df1
DataFrame[id: int, txt: string]
>>> df
DataFrame[id: bigint, txt: string]
于 2020-07-26T06:15:49.580 回答
-1

对于初学者，从文件导入数据的完整示例：

from pyspark.sql import SparkSession
from pyspark.sql.types import (
    ShortType,
    StringType,
    StructType,
    StructField,
    TimestampType,
)

import os

here = os.path.abspath(os.path.dirname(__file__))


spark = SparkSession.builder.getOrCreate()
schema = StructType(
    [
        StructField("id", ShortType(), nullable=False),
        StructField("string", StringType(), nullable=False),
        StructField("datetime", TimestampType(), nullable=False),
    ]
)

# read file or construct rows manually
df = spark.read.csv(os.path.join(here, "data.csv"), schema=schema, header=True)
于 2020-08-19T02:50:12.620 回答