1. The problem is that isin was added to Spark in version 1.5.0 and therefore not yet avaiable in your version of Spark as seen in the documentation of isin here. There is a similar function in in the Scala API that was introduced in 1.3.0 which has a similar functionality (there are some differences in the input since in only accepts columns).Dec 31, 2018 · PySpark: TypeError: 'str' object is not callable in dataframe operations. 1 *PySpark* TypeError: int() argument must be a string or a number, not 'Column' 3. TypeError: 'NoneType' object is not iterable Is a python exception (as opposed to a spark error), which means your code is failing inside your udf . Your issue is that you have some null values in your DataFrame.The Jars for geoSpark are not correctly registered with your Spark Session. There's a few ways around this ranging from a tad inconvenient to pretty seamless. For example, if when you call spark-submit you specify: --jars jar1.jar,jar2.jar,jar3.jar. then the problem will go away, you can also provide a similar command to pyspark if that's your ...I imported a df into Databricks as a pyspark.sql.dataframe.DataFrame. Within this df I have 3 columns (which I have verified to be strings) that I wish to concatenate. I have tried to use a simple "+" function first, eg.from pyspark.sql.functions import max as spark_max linesWithSparkGDF = linesWithSparkDF.groupBy(col("id")).agg(spark_max(col("cycle"))) Solution 3: use the PySpark create_map function Instead of using the map function, we can use the create_map function. The map function is a Python built-in function, not a PySpark function.Sep 5, 2022 · I am performing outlier detection in my pyspark dataframe. For that I am using an custom outlier function from here def find_outliers(df): # Identifying the numerical columns in a spark datafr... 1. Change DataType using PySpark withColumn () By using PySpark withColumn () on a DataFrame, we can cast or change the data type of a column. In order to change data type, you would also need to use cast () function along with withColumn (). The below statement changes the datatype from String to Integer for the salary column.from pyspark.sql import SparkSession spark = SparkSession.builder.getOrCreate () # ... here you get your DF # Assuming the first column of your DF is the JSON to parse my_df = spark.read.json (my_df.rdd.map (lambda x: x [0])) Note that it won't keep any other column present in your dataset.class PySparkValueError (PySparkException, ValueError): """ Wrapper class for ValueError to support error classes. """ class PySparkTypeError (PySparkException, TypeError): """ Wrapper class for TypeError to support error classes. """ class PySparkAttributeError (PySparkException, AttributeError): """ Wrapper class for AttributeError to support ... Dec 2, 2022 · I imported a df into Databricks as a pyspark.sql.dataframe.DataFrame. Within this df I have 3 columns (which I have verified to be strings) that I wish to concatenate. I have tried to use a simple "+" function first, eg. If you are using the RDD[Row].toDF() monkey-patched method you can increase the sample ratio to check more than 100 records when inferring types: # Set sampleRatio smaller as the data size increases my_df = my_rdd.toDF(sampleRatio=0.01) my_df.show()Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about TeamsAug 8, 2016 · So you could manually convert the numpy.float64 to float like. df = sqlContext.createDataFrame ( [ (float (tup [0]), float (tup [1]) for tup in preds_labels], ["prediction", "label"] ) Note pyspark will then take them as pyspark.sql.types.DoubleType. This is true for string as well. So if you created your list strings using numpy , try to ... Aug 14, 2022 · Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams *PySpark* TypeError: int() argument must be a string or a number, not 'Column' Hot Network Questions Can a group generated by its involutions, the product of every two of which has order a power of 2, have an element of odd order?Dec 9, 2022 · I am trying to install Pyspark in Google Colab and I got the following error: TypeError: an integer is required (got type bytes) I tried using latest spark 3.3.1 and it did not resolve the problem. 1 Answer. You have to perform an aggregation on the GroupedData and collect the results before you can iterate over them e.g. count items per group: res = df.groupby (field).count ().collect () Thank you Bernhard for your comment. But actually I'm creating some index & returning it.Apr 7, 2022 · By using the dir function on the list, we can see its method and attributes.One of which is the __getitem__ method. Similarly, if you will check for tuple, strings, and dictionary, __getitem__ will be present. I've installed OpenJDK 13.0.1 and python 3.8 and spark 2.4.4. Instructions to test the install is to run .\\bin\\pyspark from the root of the spark installation. I'm not sure if I missed a step in ... May 20, 2019 · This is where I am running into TypeError: TimestampType can not accept object '2019-05-20 12:03:00' in type <class 'str'> or TypeError: TimestampType can not accept object 1558353780000000000 in type <class 'int'>. I have tried converting the column to different date formats in python, before defining the schema but can seem to get the import ... May 22, 2020 · 1 Answer. Sorted by: 2. You can use sql expr using F.expr. from pyspark.sql import functions as F condition = "type_txt = 'clinic'" input_df1 = input_df.withColumn ( "prm_data_category", F.when (F.expr (condition), F.lit ("clinic")) .when (F.col ("type_txt") == 'office', F.lit ("office")) .otherwise (F.lit ("other")), ) Share. Follow. Sep 5, 2022 · I am performing outlier detection in my pyspark dataframe. For that I am using an custom outlier function from here def find_outliers(df): # Identifying the numerical columns in a spark datafr... Dec 31, 2018 · PySpark: TypeError: 'str' object is not callable in dataframe operations. 1 *PySpark* TypeError: int() argument must be a string or a number, not 'Column' 3. Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about TeamsTypeError: element in array field Category: Can not merge type <class 'pyspark.sql.types.StringType'> and <class 'pyspark.sql.types.DoubleType'> 0 TypeError: a float is required pysparkTeams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about TeamsPyspark - TypeError: 'float' object is not subscriptable when calculating mean using reduceByKey. Ask Question Asked 5 years, 6 months ago. Modified 5 years, 6 months ...SparkSession.createDataFrame, which is used under the hood, requires an RDD / list of Row / tuple / list / dict * or pandas.DataFrame, unless schema with DataType is provided. Try to convert float to tuple like this: myFloatRdd.map (lambda x: (x, )).toDF () or even better: from pyspark.sql import Row row = Row ("val") # Or some other column ...1 Answer. In the document of createDataFrame you can see the data field must be: data: Union [pyspark.rdd.RDD [Any], Iterable [Any], ForwardRef ('PandasDataFrameLike')] Ah, I get it, to make this answer clearer. (1,) is a tuple, (1) is an integer. Hence it fulfills the iterable requirement.Sep 5, 2022 · I am performing outlier detection in my pyspark dataframe. For that I am using an custom outlier function from here def find_outliers(df): # Identifying the numerical columns in a spark datafr... I imported a df into Databricks as a pyspark.sql.dataframe.DataFrame. Within this df I have 3 columns (which I have verified to be strings) that I wish to concatenate. I have tried to use a simple "+" function first, eg.pyspark / python 3.6 (TypeError: 'int' object is not subscriptable) list / tuples. 2. TypeError: tuple indices must be integers, not str using pyspark and RDD. 0.Jan 31, 2023 · The issue here is with F.lead() call. Third parameter (default value) is not of Column type, but this is just some constant value. If you want to use Column for default value use coalesce(): This question already has answers here : How to fix 'TypeError: an integer is required (got type bytes)' error when trying to run pyspark after installing spark 2.4.4 (8 answers) Closed 2 years ago. Created a conda environment: conda create -y -n py38 python=3.8 conda activate py38. Installed Spark from Pip: Next thing I need to do is derive the year from "REPORT_TIMESTAMP". I have tried various approaches, for instance: jsonDf.withColumn ("YEAR", datetime.fromtimestamp (to_timestamp (jsonDF.reportData.timestamp).cast ("integer")) that ended with "TypeError: an integer is required (got type Column) I also tried:PySpark: TypeError: 'str' object is not callable in dataframe operations. 1 *PySpark* TypeError: int() argument must be a string or a number, not 'Column' 3.May 20, 2019 · This is where I am running into TypeError: TimestampType can not accept object '2019-05-20 12:03:00' in type <class 'str'> or TypeError: TimestampType can not accept object 1558353780000000000 in type <class 'int'>. I have tried converting the column to different date formats in python, before defining the schema but can seem to get the import ... Next thing I need to do is derive the year from "REPORT_TIMESTAMP". I have tried various approaches, for instance: jsonDf.withColumn ("YEAR", datetime.fromtimestamp (to_timestamp (jsonDF.reportData.timestamp).cast ("integer")) that ended with "TypeError: an integer is required (got type Column) I also tried:However once I test the function. TypeError: Invalid argument, not a string or column: DataFrame [Name: string] of type <class 'pyspark.sql.dataframe.DataFrame'>. For column literals, use 'lit', 'array', 'struct' or 'create_map' function. I´ve been trying to fix this problem through different approaches but I cant make it work and I know very ...总结. 在本文中，我们介绍了PySpark中的TypeError: ‘JavaPackage’对象不可调用错误，并提供了解决方案和示例代码进行说明。. 当我们遇到这个错误时，只需要正确地调用相应的函数，并遵循正确的语法即可解决问题。. 学习正确使用PySpark的函数调用方法，将会帮助 ...4 Answers. Sorted by: 43. It's because, you've overwritten the max definition provided by apache-spark, it was easy to spot because max was expecting an iterable. To fix this, you can use a different syntax, and it should work: linesWithSparkGDF = linesWithSparkDF.groupBy (col ("id")).agg ( {"cycle": "max"}) Or, alternatively:总结. 在本文中，我们介绍了PySpark中的TypeError: ‘JavaPackage’对象不可调用错误，并提供了解决方案和示例代码进行说明。. 当我们遇到这个错误时，只需要正确地调用相应的函数，并遵循正确的语法即可解决问题。. 学习正确使用PySpark的函数调用方法，将会帮助 ...Oct 13, 2020 · PySpark error: TypeError: Invalid argument, not a string or column. 0. Py(Spark) udf gives PythonException: 'TypeError: 'float' object is not subscriptable. 3. This is where I am running into TypeError: TimestampType can not accept object '2019-05-20 12:03:00' in type <class 'str'> or TypeError: TimestampType can not accept object 1558353780000000000 in type <class 'int'>. I have tried converting the column to different date formats in python, before defining the schema but can seem to get the import ...This question already has answers here : How to fix 'TypeError: an integer is required (got type bytes)' error when trying to run pyspark after installing spark 2.4.4 (8 answers) Closed 2 years ago. Created a conda environment: conda create -y -n py38 python=3.8 conda activate py38. Installed Spark from Pip: Nov 23, 2021 · Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Aug 29, 2019 · from pyspark.sql.functions import col, trim, lower Alternatively, double-check whether the code really stops in the line you said, or check whether col, trim, lower are what you expect them to be by calling them like this: col should return. function pyspark.sql.functions._create_function.._(col) Reading between the lines. You are. reading data from a CSV file. and get . TypeError: StructType can not accept object in type <type 'unicode'> This happens because you pass a string not an object compatible with struct. Aug 29, 2016 · TypeError: 'JavaPackage' object is not callable on PySpark, AWS Glue 0 sc._jvm.org.apache.spark.streaming.kafka.KafkaUtilsPythonHelper() TypeError: 'JavaPackage' object is not callable when using Apr 22, 2021 · pyspark: TypeError: IntegerType can not accept object in type <type 'unicode'> 3 Getting int() argument must be a string or a number, not 'Column'- Apache Spark Sep 20, 2018 · If parents is indeed an array, and you can access the element at index 0, you have to modify your comparison to something like: df_categories.parents[0] == 0 or array_contains(df_categories.parents, 0) depending on the position of the element you want to check or if you just want to know whether the value is in the array Solution for TypeError: Column is not iterable. PySpark add_months () function takes the first argument as a column and the second argument is a literal value. if you try to use Column type for the second argument you get “TypeError: Column is not iterable”. In order to fix this use expr () function as shown below. Mar 4, 2022 · PySpark error: TypeError: Invalid argument, not a string or column. Hot Network Questions Is a garlic bulb which is coloured brown on the outside safe to eat? ... Jun 19, 2022 · When running PySpark 2.4.8 script in Python 3.8 environment with Anaconda, the following issue occurs: TypeError: an integer is required (got type bytes). The environment is created using the following code: Oct 13, 2020 · PySpark error: TypeError: Invalid argument, not a string or column. 0. Py(Spark) udf gives PythonException: 'TypeError: 'float' object is not subscriptable. 3. TypeError: unsupported operand type (s) for +: 'int' and 'str' Now, this does not make sense to me, since I see the types are fine for aggregation in printSchema () as you can see above. So, I tried converting it to integer just incase: mydf_converted = mydf.withColumn ("converted",mydf ["bytes_out"].cast (IntegerType ()).alias ("bytes_converted"))TypeError: StructType can not accept object 'string indices must be integers' in type <class 'str'> I tried many posts on Stackoverflow, like Dealing with non-uniform JSON columns in spark dataframe Non of it worked.May 22, 2020 · 1 Answer. Sorted by: 2. You can use sql expr using F.expr. from pyspark.sql import functions as F condition = "type_txt = 'clinic'" input_df1 = input_df.withColumn ( "prm_data_category", F.when (F.expr (condition), F.lit ("clinic")) .when (F.col ("type_txt") == 'office', F.lit ("office")) .otherwise (F.lit ("other")), ) Share. Follow. Dec 2, 2022 · I imported a df into Databricks as a pyspark.sql.dataframe.DataFrame. Within this df I have 3 columns (which I have verified to be strings) that I wish to concatenate. I have tried to use a simple "+" function first, eg. This is where I am running into TypeError: TimestampType can not accept object '2019-05-20 12:03:00' in type <class 'str'> or TypeError: TimestampType can not accept object 1558353780000000000 in type <class 'int'>. I have tried converting the column to different date formats in python, before defining the schema but can seem to get the import ...*PySpark* TypeError: int() argument must be a string or a number, not 'Column' Hot Network Questions Can a group generated by its involutions, the product of every two of which has order a power of 2, have an element of odd order?Jun 19, 2022 · When running PySpark 2.4.8 script in Python 3.8 environment with Anaconda, the following issue occurs: TypeError: an integer is required (got type bytes). The environment is created using the following code: Jan 31, 2023 · The issue here is with F.lead() call. Third parameter (default value) is not of Column type, but this is just some constant value. If you want to use Column for default value use coalesce(): 1. The Possible Issues faced when running Spark on Windows is, of not giving proper Path or by using Python 3.x to run Spark. So, Do check Path Given for spark i.e /usr/local/spark Proper or Not. Do set Python Path to Python 2.x (remove Python 3.x). Share. Improve this answer. Follow. edited Aug 3, 2017 at 9:25.1 Answer. Sorted by: 5. Row is a subclass of tuple and tuples in Python are immutable hence don't support item assignment. If you want to replace an item stored in a tuple you have rebuild it from scratch: ## replace "" with placeholder of your choice tuple (x if x is not None else "" for x in row) If you want to simply concatenate flat schema ...Mar 9, 2018 · You cannot use flatMap on an Int object. flatMap can be used in collection objects such as Arrays or list.. You can use map function on the rdd type that you have RDD[Integer] ... However once I test the function. TypeError: Invalid argument, not a string or column: DataFrame [Name: string] of type <class 'pyspark.sql.dataframe.DataFrame'>. For column literals, use 'lit', 'array', 'struct' or 'create_map' function. I´ve been trying to fix this problem through different approaches but I cant make it work and I know very ...Pyspark - How do you split a column with Struct Values of type Datetime? 1 Converting a date/time column from binary data type to the date/time data type using PySparkNov 30, 2022 · 1 Answer. In the document of createDataFrame you can see the data field must be: data: Union [pyspark.rdd.RDD [Any], Iterable [Any], ForwardRef ('PandasDataFrameLike')] Ah, I get it, to make this answer clearer. (1,) is a tuple, (1) is an integer. Hence it fulfills the iterable requirement. 1 Answer. Sorted by: 5. Row is a subclass of tuple and tuples in Python are immutable hence don't support item assignment. If you want to replace an item stored in a tuple you have rebuild it from scratch: ## replace "" with placeholder of your choice tuple (x if x is not None else "" for x in row) If you want to simply concatenate flat schema ...Pyspark, TypeError: 'Column' object is not callable 1 pyspark.sql.utils.AnalysisException: THEN and ELSE expressions should all be same type or coercible to a common typeAug 29, 2016 · TypeError: 'JavaPackage' object is not callable on PySpark, AWS Glue 0 sc._jvm.org.apache.spark.streaming.kafka.KafkaUtilsPythonHelper() TypeError: 'JavaPackage' object is not callable when using Aug 8, 2016 · So you could manually convert the numpy.float64 to float like. df = sqlContext.createDataFrame ( [ (float (tup [0]), float (tup [1]) for tup in preds_labels], ["prediction", "label"] ) Note pyspark will then take them as pyspark.sql.types.DoubleType. This is true for string as well. So if you created your list strings using numpy , try to ... Apr 7, 2022 · By using the dir function on the list, we can see its method and attributes.One of which is the __getitem__ method. Similarly, if you will check for tuple, strings, and dictionary, __getitem__ will be present. May 26, 2021 · OUTPUT:-Python TypeError: int object is not subscriptableThis code returns “Python,” the name at the index position 0. We cannot use square brackets to call a function or a method because functions and methods are not subscriptable objects. *PySpark* TypeError: int() argument must be a string or a number, not 'Column' Hot Network QuestionsIt returns "TypeError: StructType can not accept object 60651 in type <class 'int'>". Here you can see better: # Create a schema for the dataframe schema = StructType ( [StructField ('zipcd', IntegerType (), True)] ) # Convert list to RDD rdd = sc.parallelize (zip_cd) #solution: close within []. Another problem for the solution, if I do that ...I imported a df into Databricks as a pyspark.sql.dataframe.DataFrame. Within this df I have 3 columns (which I have verified to be strings) that I wish to concatenate. I have tried to use a simple "+" function first, eg.PySpark: TypeError: 'str' object is not callable in dataframe operations. 1 *PySpark* TypeError: int() argument must be a string or a number, not 'Column' 3.TypeError: StructType can not accept object '_id' in type <class 'str'> and this is how I resolved it. I am working with heavily nested json file for scheduling , json file is composed of list of dictionary of list etc.Jan 31, 2023 · The issue here is with F.lead() call. Third parameter (default value) is not of Column type, but this is just some constant value. If you want to use Column for default value use coalesce(): You could also try: import pyspark from pyspark.sql import SparkSession sc = pyspark.SparkContext ('local [*]') spark = SparkSession.builder.getOrCreate () . . . spDF.createOrReplaceTempView ("space") spark.sql ("SELECT name FROM space").show () The top two lines are optional to someone to try this snippet in local machine. Share.class DecimalType (FractionalType): """Decimal (decimal.Decimal) data type. The DecimalType must have fixed precision (the maximum total number of digits) and scale (the number of digits on the right of dot).The Jars for geoSpark are not correctly registered with your Spark Session. There's a few ways around this ranging from a tad inconvenient to pretty seamless. For example, if when you call spark-submit you specify: --jars jar1.jar,jar2.jar,jar3.jar. then the problem will go away, you can also provide a similar command to pyspark if that's your ...Jan 31, 2023 · The issue here is with F.lead() call. Third parameter (default value) is not of Column type, but this is just some constant value. If you want to use Column for default value use coalesce(): Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about TeamsSolution for TypeError: Column is not iterable. PySpark add_months () function takes the first argument as a column and the second argument is a literal value. if you try to use Column type for the second argument you get “TypeError: Column is not iterable”. In order to fix this use expr () function as shown below.1. Change DataType using PySpark withColumn () By using PySpark withColumn () on a DataFrame, we can cast or change the data type of a column. In order to change data type, you would also need to use cast () function along with withColumn (). The below statement changes the datatype from String to Integer for the salary column.TypeError: StructType can not accept object 'string indices must be integers' in type <class 'str'> I tried many posts on Stackoverflow, like Dealing with non-uniform JSON columns in spark dataframe Non of it worked.
This is where I am running into TypeError: TimestampType can not accept object '2019-05-20 12:03:00' in type <class 'str'> or TypeError: TimestampType can not accept object 1558353780000000000 in type <class 'int'>. I have tried converting the column to different date formats in python, before defining the schema but can seem to get the import .... Programs
Dec 9, 2022 · I am trying to install Pyspark in Google Colab and I got the following error: TypeError: an integer is required (got type bytes) I tried using latest spark 3.3.1 and it did not resolve the problem. Edit: RESOLVED I think the problem is with the multi-dimensional arrays generated from Elmo inference. I averaged all the vectors and then used the final average vector for all words in the sentenc...Dec 1, 2019 · TypeError: field date: DateType can not accept object '2019-12-01' in type <class 'str'> I tried to convert stringType to DateType using to_date plus some other ways but not able to do so. Please advise If a field only has None records, PySpark can not infer the type and will raise that error. Manually defining a schema will resolve the issue >>> from pyspark.sql.types import StructType, StructField, StringType >>> schema = StructType([StructField("foo", StringType(), True)]) >>> df = spark.createDataFrame([[None]], schema=schema) >>> df.show ... Jul 10, 2019 · I built a fasttext classification model in order to do sentiment analysis for facebook comments (using pyspark 2.4.1 on windows). When I use the prediction model function to predict the class of a sentence, the result is a tuple with the form below: File "/.../3.8/lib/python3.8/runpy.py", line 183, in _run_module_as_main mod_name, mod_spec, code = _get_module_details(mod_name, _Error) File "/.../3.8/lib/python3.8 ...However once I test the function. TypeError: Invalid argument, not a string or column: DataFrame [Name: string] of type <class 'pyspark.sql.dataframe.DataFrame'>. For column literals, use 'lit', 'array', 'struct' or 'create_map' function. I´ve been trying to fix this problem through different approaches but I cant make it work and I know very ...Aug 13, 2018 · You could also try: import pyspark from pyspark.sql import SparkSession sc = pyspark.SparkContext ('local [*]') spark = SparkSession.builder.getOrCreate () . . . spDF.createOrReplaceTempView ("space") spark.sql ("SELECT name FROM space").show () The top two lines are optional to someone to try this snippet in local machine. Share. PySpark: TypeError: 'str' object is not callable in dataframe operations. 3. cannot resolve column due to data type mismatch PySpark. 0. I'm encountering Pyspark ...from pyspark.sql.functions import * is bad . It goes without saying that the solution was to either restrict the import to the needed functions or to import pyspark.sql.functions and prefix the needed functions with it.Mar 4, 2022 · PySpark error: TypeError: Invalid argument, not a string or column. Hot Network Questions Is a garlic bulb which is coloured brown on the outside safe to eat? ... It returns "TypeError: StructType can not accept object 60651 in type <class 'int'>". Here you can see better: # Create a schema for the dataframe schema = StructType ( [StructField ('zipcd', IntegerType (), True)] ) # Convert list to RDD rdd = sc.parallelize (zip_cd) #solution: close within []. Another problem for the solution, if I do that ...1. The problem is that isin was added to Spark in version 1.5.0 and therefore not yet avaiable in your version of Spark as seen in the documentation of isin here. There is a similar function in in the Scala API that was introduced in 1.3.0 which has a similar functionality (there are some differences in the input since in only accepts columns).10. Its because you are trying to apply the function contains to the column. The function contains does not exist in pyspark. You should try like. Try this: import pyspark.sql.functions as F df = df.withColumn ("AddCol",F.when (F.col ("Pclass").like ("3"),"three").otherwise ("notthree")) Or if you just want it to be exactly the number 3 you ...You could also try: import pyspark from pyspark.sql import SparkSession sc = pyspark.SparkContext ('local [*]') spark = SparkSession.builder.getOrCreate () . . . spDF.createOrReplaceTempView ("space") spark.sql ("SELECT name FROM space").show () The top two lines are optional to someone to try this snippet in local machine. Share.from pyspark.sql.functions import col, trim, lower Alternatively, double-check whether the code really stops in the line you said, or check whether col, trim, lower are what you expect them to be by calling them like this: col should return. function pyspark.sql.functions._create_function.._(col)I've installed OpenJDK 13.0.1 and python 3.8 and spark 2.4.4. Instructions to test the install is to run .\\bin\\pyspark from the root of the spark installation. I'm not sure if I missed a step in ... The psdf.show() does not work although DataFrame looks to be created. I wonder what is the cause of this. The environment is Pyspark:3.2.1-hadoop3.2 Hadoop:3.2.1 JDK: 18.0.1.1 local The code is the总结. 在本文中，我们介绍了PySpark中的TypeError: ‘JavaPackage’对象不可调用错误，并提供了解决方案和示例代码进行说明。. 当我们遇到这个错误时，只需要正确地调用相应的函数，并遵循正确的语法即可解决问题。. 学习正确使用PySpark的函数调用方法，将会帮助 ....