英文:
How to pad variable-length vectors with NULLs to a fixed length?
问题
我有一组可变长度的整数向量，我想通过在末尾插入NULL值来使它们具有相同的固定长度。我如何实现这个目标？例如，我有一个包含可变长度向量的元组 "a"，包括 [1..3]、[1..5] 和 [1..6]。我如何将它们转换为长度为6的向量？
col1 col2 col3
1 1 1
2 2 2
3 3 3
null 4 4
null 5 5
null null 6
英文:
I have a set of variable-length integer vectors and I want to align them by inserting NULL values at the end to make them have the same fiexed length. How can I achieve this? For example, I have a tuple “a“ containing variable-length vectors [1..3], [1..5], and [1..6]. How can I transform them into vectors of length 6?
col1 col2 col3
1 1 1
2 2 2
3 3 3
null 4 4
null 5 5
null null 6
答案1
得分: 1
方法1： 使用 append!：
a = [1..3, 1..5, 1..6]
def f(mutable x, N):
x.copy().append!(array(DOUBLE, N - size(x), N - size(x), NULL))
each(f{, 6}, a)
输出：
col1 col2 col3
1 1 1
2 2 2
3 3 3
4 4
5 5
6
方法2： 向每个向量添加 6 个 NULL 值，以确保它们的长度大于或等于 6。然后提取每个向量的前六个元素：
a = [1..3, 1..5, 1..6]
def fillNull(x, n):
take(x < -take(00i, n), n)
each(fillNull{, 6}, a)
英文:
Method 1: Use append!:
a=[1..3,1..5,1..6]
def f(mutable x, N): x.copy().append!(array(DOUBLE,N - size(x),N - size(x),NULL))
each(f{,6},a)
Output:
col1 col2 col3
1 1 1
2 2 2
3 3 3
4 4
5 5
6
Method 2: Add 6 NULL values to each of these vectors to ensure their length is greater than or equal to 6. Then extract the first six elements of each vector:
a = [1..3, 1..5, 1..6]
def fillNull(x,n): take(x<-take(00i, n), n)
each(fillNull{, 6}, a)
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