PySpark:具有不同列的数据帧的动态联合

2024-05-16 13:08:04 发布

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考虑如下所示的数组。我有三套阵法:

阵列1:

C1  C2  C3
1   2   3
9   5   6

阵列2:

^{pr2}$

阵列3:

C1   C4
111  112
110  115

我需要如下的输出,输入可以得到C1,…,C4的任何一个值,但是在连接时,我需要得到正确的值,如果这个值不存在,那么它应该是零。在

预期产量:

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的值及其原始值,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作为自动脚本使用。我只需要使用python/pyspark而不是pandas。在

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))

Tags: importnoneaddpysparkc2c1c3lit
1条回答
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1楼 · 发布于 2024-05-16 13:08:04

也许有很多更好的方法可以做到这一点,但也许下面的方法对将来的任何人都是有用的。在

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:

^{pr2}$

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