Python PySpark：具有不同列的数据帧的动态联合

考虑如下所示的数组。我有3套阵列：

阵列1：

C1  C2  C3
1   2   3
9   5   6

阵列2：

C2 C3 C4
11 12 13
10 15 16

阵列3：

C1   C4
111  112
110  115

我需要如下输出，输入我可以得到C1，…，C4的任何一个值，但是当连接时，我需要得到正确的值，如果该值不存在，那么它应该是零

预期产出：

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

我已经编写了pyspark代码，但是我已经硬编码了新列及其原始列的值，我需要将下面的代码转换为方法重载，以便我可以将此脚本用作自动脚本。我只需要使用python/pyspark，不需要使用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()

方法重载示例：

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))

---

也许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
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()

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|
+---+---+---+---+

---

也许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
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()

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|
+---+---+---+---+

我会试试看

df = df1.join(df2, ['each', 'shared', 'col], how='full')

我会试试看

df = df1.join(df2, ['each', 'shared', 'col], how='full')

---

你说可能更好，但我还没有遇到过这样的代码，谢谢你的新代码！你在Scala做过类似的事情吗？或者在Scala中看到过这样做的代码吗？@JoshuaJames-这是Scala中的版本-你说可能更好，但我没有遇到过这样的代码，谢谢你的这段新代码！你在Scala做过类似的事情吗？或者看到Scala中的代码了吗？@JoshuaJames-这是Scala中的版本-