手动创建

Question

我正在尝试手动创建一个具有特定数据的：

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)

当我试图显示dataframe时，这会产生一个错误，所以我不知道如何做到这一点。

然而，火花文档似乎有点令我费解，当我试图遵循这些指令时，我也遇到了类似的错误。

有人知道怎么做吗？

Answer 1

简单的数据创建：

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|
+---+-----+

根据官方医生

• 当架构是列名列表时，将从数据推断每个列的类型。(↑上面的例子)
• 当模式是pyspark.sql.types.DataType或数据类型字符串时，它必须与实际数据匹配。(↓下面的例子)

# 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，您还可以从Pandas dataframe创建数据create，模式将从Pandas dataframe的类型中推断：

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|
+----+----+

Answer 2

以@Steven的回答为基础：

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格式。

我们现在可以简单地向它添加一行：

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)

Answer 3

这个答案演示了如何使用PySpark、create_df和toDF创建一个createDataFrame DataFrame。

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

df.show()

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

您还可以传递给createDataFrame一个RDD和模式，以更精确地构造DataFrames：

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|
+-----+---+

来自我的create_df项目的奎因允许两者兼而有之--它简洁且全面描述：

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|
+----+----+

与其他方法相比，toDF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|
+-----+---+