可能有很多更好的方法可以做到这一点，但也许下面的内容对未来的任何人都有用.

from pyspark.sql import SparkSession从 pyspark.sql.functions 导入点亮spark = SparkSession.builder.appName("动态框架").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)], ("C1", "C4"))数据帧 = [df01, df02, df03]# 创建所有列名的列表并对其进行排序cols = 设置()对于数据帧中的 df:对于 df.columns 中的 x:cols.add(x)cols = 排序(列)# 创建一个包含所有数据框的字典dfs = {}对于 enumerate(dataframes) 中的 i, d:new_name = 'df' + str(i) # 键的新名称，数据帧是值dfs[new_name] = d# 遍历所有列名.将缺失的列添加到数据框中(值为 0)对于列中的 x:如果 x 不在 d.columns 中:dfs[new_name] = dfs[new_name].withColumn(x, lit(0))dfs[new_name] = dfs[new_name].select(cols) # 使用'select'对列进行排序# 现在用一个循环(联合)把它放在一起result = dfs['df0'] # 取第一个数据帧，添加其他数据帧dfs_to_add = dfs.keys() # 字典中所有数据框的列表dfs_to_add.remove('df0') # 删除第一个，因为它已经在结果中对于 dfs_to_add 中的 x:结果 = result.union(dfs[x])结果.show()

输出:

+---+---+---+---+|C1|C2|C3|C4|+---+---+---+---+|1|2|3|0||9|5|6|0||0|11|12|13||0|10|15|16||111|0|0|112||110|0|0|115|+---+---+---+---+

Consider the arrays as shown here. I have 3 sets of array:

Array 1:

C1  C2  C3
1   2   3
9   5   6

Array 2:

C2 C3 C4
11 12 13
10 15 16

Array 3:

C1   C4
111  112
110  115

I need the output as following, the input I can get any one value for C1, ..., C4 but while joining I need to get correct values and if the value is not there then it should be zero.

Expected output:

C1 C2 C3 C4
1  2  3  0
9  5  6  0
0  11 12 13
0 10 15 16
111 0 0 112
110 0 0 115

I have written pyspark code but I have hardcoded the value for the new column and its RAW, I need to convert the below code to method overloading, so that I can use this script as automatic one. I need to use only python/pyspark not pandas.

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

sqlContext = pyspark.SQLContext(pyspark.SparkContext())

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

df01_add = df01.withColumn("C4", lit(0)).select("c1","c2","c3","c4")
df02_add = df02.withColumn("C1", lit(0)).select("c1","c2","c3","c4")
df03_add = df03.withColumn("C2", lit(0)).withColumn("C3", lit(0)).select("c1","c2","c3","c4")

df_uni = df01_add.union(df02_add).union(df03_add)
df_uni.show()

Method Overloading Example:

class Student:
     def ___Init__ (self,m1,m2):
         self.m1 = m1
         self.m2 = m2

     def sum(self,c1=None,c2=None,c3=None,c4=None):
         s = 0
         if c1!= None and c2 != None and c3 != None:
            s = c1+c2+c3
         elif c1 != None and c2 != None:
             s = c1+c2
         else:
            s = c1
         return s

print(s1.sum(55,65,23))

There are probably plenty of better ways to do it, but maybe the below is useful to anyone in the future.

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)], ("C1", "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()

Output:

+---+---+---+---+
| C1| C2| C3| C4|
+---+---+---+---+
|  1|  2|  3|  0|
|  9|  5|  6|  0|
|  0| 11| 12| 13|
|  0| 10| 15| 16|
|111|  0|  0|112|
|110|  0|  0|115|
+---+---+---+---+