结合 Spark 数据帧实现代码的更有效方法

如何解决结合 Spark 数据帧实现代码的更有效方法

有没有更有效的方法来组合 Spark 数据帧而不使用 for 循环？在 this 帖子中，答案使用 for 循环，如果您有多个数据框，这似乎需要很长时间。代码如下：

from pyspark.sql import SparkSession
from pyspark.sql.functions import lit

spark = SparkSession.builder\
    .appName("DynamicFrame")\
    .getorCreate()

df01 = spark.createDataFrame([(1,2,3),(9,5,6)],("C1","C2","C3"))
df02 = spark.createDataFrame([(11,12,13),(10,15,16)],("C2","C3","C4"))
df03 = spark.createDataFrame([(111,112),(110,115)],"C4"))

dataframes = [df01,df02,df03]

# Create a list of all the column names and sort them
cols = set()
for df in dataframes:
    for x in df.columns:
        cols.add(x)
cols = sorted(cols)

# Create a dictionary with all the dataframes
dfs = {}
for i,d in enumerate(dataframes):
    new_name = 'df' + str(i)  # New name for the key,the dataframe is the value
    dfs[new_name] = d
    # Loop through all column names. Add the missing columns to the dataframe (with value 0)
    for x in cols:
        if x not in d.columns:
            dfs[new_name] = dfs[new_name].withColumn(x,lit(0))
    dfs[new_name] = dfs[new_name].select(cols)  # Use 'select' to get the columns sorted

# Now put it al together with a loop (union)
result = dfs['df0']      # Take the first dataframe,add the others to it
dfs_to_add = dfs.keys()  # List of all the dataframes in the dictionary
dfs_to_add.remove('df0') # Remove the first one,because it is already in the result
for x in dfs_to_add:
    result = result.union(dfs[x])
result.show()

可以使用某种递归技术吗？