Met*_*ata 9 python apache-spark pyspark
我正在尝试将字典:转换
data_dict = {'t1': '1', 't2': '2', 't3': '3'}为数据框:
key | value|
----------------
t1 1
t2 2
t3 3
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为此,我尝试了:
schema = StructType([StructField("key", StringType(), True), StructField("value", StringType(), True)])
ddf = spark.createDataFrame(data_dict, schema)
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但我收到以下错误:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/Cellar/apache-spark/2.4.5/libexec/python/pyspark/sql/session.py", line 748, in createDataFrame
rdd, schema = self._createFromLocal(map(prepare, data), schema)
File "/usr/local/Cellar/apache-spark/2.4.5/libexec/python/pyspark/sql/session.py", line 413, in _createFromLocal
data = list(data)
File "/usr/local/Cellar/apache-spark/2.4.5/libexec/python/pyspark/sql/session.py", line 730, in prepare
verify_func(obj)
File "/usr/local/Cellar/apache-spark/2.4.5/libexec/python/pyspark/sql/types.py", line 1389, in verify
verify_value(obj)
File "/usr/local/Cellar/apache-spark/2.4.5/libexec/python/pyspark/sql/types.py", line 1377, in verify_struct
% (obj, type(obj))))
TypeError: StructType can not accept object 't1' in type <class 'str'>
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所以我尝试了这个,没有指定任何模式,而只是列数据类型:
ddf = spark.createDataFrame(data_dict, StringType()&ddf = spark.createDataFrame(data_dict, StringType(), StringType())
但两者都会产生一个具有一列的数据框,该列是字典的关键,如下所示:
+-----+
|value|
+-----+
|t1 |
|t2 |
|t3 |
+-----+
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谁能告诉我如何将字典转换为 PySpark 中的 Spark 数据框?
ern*_*t_k 10
您可以使用data_dict.items()列出键/值对:
spark.createDataFrame(data_dict.items()).show()
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哪个打印
+---+---+
| _1| _2|
+---+---+
| t1| 1|
| t2| 2|
| t3| 3|
+---+---+
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当然,您可以指定您的架构:
spark.createDataFrame(data_dict.items(),
schema=StructType(fields=[
StructField("key", StringType()),
StructField("value", StringType())])).show()
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导致
+---+-----+
|key|value|
+---+-----+
| t1| 1|
| t2| 2|
| t3| 3|
+---+-----+
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