有效地找到数据表的第一个非零元素（对应列） [英] efficiently finding first nonzero element (corresponding column) of a data table

问题描述

关于以下类型的问题，有一些答案，但它们都效率低下，并且扩展性不佳。

There are some answers on stack to the below type of question, but they are all inefficient and do not scale well.

要重现它，假设我有如下数据：

To reproduce it, suppose I have data that looks like this:

    tempmat=matrix(c(1,1,0,4,1,0,0,4,0,1,0,4, 0,1,1,4, 0,1,0,5),5,4,byrow=T)
    tempmat=rbind(rep(0,4),tempmat)
    tempmat=data.table(tempmat)
    names(tempmat)=paste0('prod1vint',1:4)

这是数据的样子，尽管它要大得多，所以解决方案不能是应用或基于行的方法。

This is what the data look like, although it is MUCH bigger, so the solution cannot be an "apply" or row-wise based approach.

> tempmat
   prod1vint1 prod1vint2 prod1vint3 prod1vint4
1:          0          0          0          0
2:          1          1          0          4
3:          1          0          0          4
4:          0          1          0          4
5:          0          1          1          4
6:          0          1          0          5

我想要来标识第一个非零元素的列，因此输出将如下所示：

I want to identify the column of the first nonzero element, so the output would look like this:

> tempmat
   prod1vint1 prod1vint2 prod1vint3 prod1vint4 firstnonzero
1:          0          0          0          0           NA
2:          1          1          0          4            1
3:          1          0          0          4            1
4:          0          1          0          4            2
5:          0          1          1          4            2
6:          0          1          0          5            2

推荐答案

一种选择是使用 rowSums 和 max.col 指定 ties.method = first

One option is to use rowSums with max.col specifying ties.method = "first"

temp <- tempmat != 0
(NA^(rowSums(temp) == 0)) * max.col(temp, ties.method = "first")
#[1] NA  1  1  2  2  2

max.col 将给出列索引为每行的第一个最大值。但是，如果所有值均为0（如第1行），则返回1，因为0是该行的最大值。为避免这种情况，我们使用 rowSums 检查行中是否至少有一个非零值，并将其乘以 max.col 输出。

max.col would give column index of first maximum value in every row. However, this would return 1 in case all the values are 0 (like in 1st row) since 0 is the maximum value in the row. To avoid that we check if there is at least one non-zero value in the row using rowSums and multiply it to max.col output.

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