我有一个PySpark dataframe,如下所示。我需要将dataframe行折叠成包含column:value对的Python dictionary行。最后,将字典转换为Python list of tuples,如下所示。我使用的是Spark 2.4。DataFrame:>>> myDF.show() +---+---+---+---+ |fname |age|location | dob | +---+---+---+---+ | John|...
pyspark dataframe Column alias 重命名列(name) df = spark.createDataFrame( [(2, "Alice"), (5, "Bob")], ["age", "name"])df.select(df.age.alias("age2")).show()+---+|age2|+---+| 2|| 5|+---+ astype alias cast 修改列类型 data.schemaStructType([StructField('name', String...
@udf(returnType=StringType()) def convertCase(str): resStr="" arr = str.split(" ") for x in arr: resStr= resStr + x[0:1].upper() + x[1:len(x)] + " " return resStr df.withColumn("Cureated Name", convertCase(col("Name"))).show(truncate=False) 1. 2. 3. 4. 5. ...
我在dataframe中有一列作为对象列表(结构数组),如 column: [{key1:value1}, {key2:value2}, {key3:value3}] 我想将此列拆分为单独的列,在同一行中键名作为列名,值作为列值。最终结果如 key1:value1, key2:value2, key3:value3 如何在pyspark中实现这一点? E.g. 要创建dataframe的示例数据: my_new...
功能:选择DataFrame中的指定列(通过传入参数进行指定) 语法: 可传递: ·可变参数的cols对象,cols对象可以是Column对象来指定列或者字符串列名来指定列 ·List[Column]对象或者List[str]对象, 用来选择多个列 网页链接 功能:过滤DataFrame内的数据,返回一个过滤后的DataFrame ...
Another way to traverse a PySpark DataFrame is to iterate over its columns. We can access the columns of a DataFrame using thecolumnsattribute, which returns a list of column names. We can then iterate over this list to access individual columns: ...
示例二 from pyspark.sql import Row from pyspark.sql.functions import explode eDF = spark.createDataFrame([Row( a=1, intlist=[1, 2, 3], mapfield={"a": "b"})]) eDF.select(explode(eDF.intlist).alias("anInt")).show() +---+ |anInt| +---+ | 1| | 2| | 3| +---+ isin...
功能:选择DataFrame中的指定列(通过传入参数进行指定) 语法: 可传递: · 可变参数的cols对象,cols对象可以是Column对象来指定列或者字符串列名来指定列 · List[Column]对象或者List[str]对象, 用来选择多个列 DSL - filter和where 功能:过滤DataFrame内的数据,返回一个过滤后的DataFrame 语法: df.filter() df.whe...
解决toDF()跑出First 100 rows类型无法确定的异常,可以采用将Row内每个元素都统一转格式,或者判断格式处理的方法,解决包含None类型时转换成DataFrame出错的问题: @staticmethod def map_convert_none_to_str(row): dict_row = row.asDict() for key in dict_row: ...
"""Converts JSON columns to complex types Args: df: Spark dataframe col_dtypes (dict): dictionary of columns names and their datatype Returns: Spark dataframe """ selects = list() for column in df.columns: if column in col_dtypes.keys(): ...