[英]Scala: How to do union of data frames in the loop
I would like to do union of data frame in the recursive method.我想在递归方法中进行数据框的联合。
I am doing some calculations in the recursive method and filtering the data and storing in one variable.我正在用递归方法进行一些计算并过滤数据并存储在一个变量中。 In 2nd iteration i will do some calculation and again i will store the data in same variable.when i am calling the method second time my first result is getting vanished.Ideally i have to store the result in one temp variable and i need to do union of all the result till the recursive method gets completed its execution.
在第二次迭代中,我将进行一些计算,并再次将数据存储在同一个变量中。当我第二次调用该方法时,我的第一个结果消失了。理想情况下,我必须将结果存储在一个临时变量中,我需要这样做所有结果的并集,直到递归方法完成其执行。
Iteration1 output in df: df 中的迭代 1 输出:
Col1
14
35
Iteration2 output in df: df中的迭代2输出:
Col1
18
20
Now i need the final output as,现在我需要最终输出,
Col1
14
35
18
20
Code:代码:
def myRecursiveMethod(first: List[List[String]],
Inputcolumnsdummy: List[List[String]],
secondInputcolumns: List[List[String]] = {
val ongoingResult = doSomeCalculation(first,Inputcolumnsdummy, secondInputcolumns)
}
I want my code should be something like below,我希望我的代码应该如下所示,
def myRecursiveMethod(first: List[List[String]],
Inputcolumnsdummy: List[List[String]],
secondInputcolumns: List[List[String]]) = {
val ongoingResult = doSomeCalculation(first, Inputcolumnsdummy, secondInputcolumns)
Val temp = temp.union(ongoingResult)
}
You should try: use union
like this: df1.union(df2)
or df1.union(computation(df2,...))
.你应该尝试:像这样使用
union
: df1.union(df2)
或df1.union(computation(df2,...))
。
Exemple below:示例如下:
def doCompute(df: DataFrame): DataFrame = {
val tmp: DataFrame = ... // TODO: call to your computation method
tmp.show()
df.union(tmp)
}
val df1: DataFrame = ...
val df2: DataFrame = ...
val df3: DataFrame = ...
var union_df: DataFrame = df1.union(doCompute(df2)).union(doCompute(df3))
One thing I did not understand in your question is how is your function myRecursiveMethod
recursive?我在你的问题中不明白的一件事是你的函数
myRecursiveMethod
是如何递归的? A recursive function calls itself, by definition.根据定义,递归函数调用自身。 Not sure your question is really clear.
不确定你的问题是否真的清楚。
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