仅对非 NA 元素求和，但如果所有 NA 则返回 NA [英] sum non NA elements only, but if all NA then return NA

问题描述

我想我已经在评论中得到了很好的答案，但我会改写问题以供将来参考.

I think I already got really good answers on the comments, but I will rephrase the question for future reference.

我正在尝试使用 data.table 按组求和.问题是某些组只有有 NA.对于这些组，我希望总和返回 NA.但是，如果有一组的值与 NA 不同，我想得到非 NA 值的总和.

I am trying to sum by groups using data.table. The problem is that some groups only have NA. For these groups I would like the sum to return NA. However, if there is one group that has one value different that NA, I would like to get the sum of the non-NA values.

A <- data.table(col1= c('A','A','B','B','C','C'),  
                col2= c(NA,NA,2,3,NA,4))

这没有添加参数 na.rm = T，C 组在应该返回 4 时返回 NA.

This without adding the argument na.rm = T, group C returns NA when it should return 4.

A[, sum(col2), by = .(col1)]
   col1 V1
1:    A NA
2:    B  5
3:    C NA

但是，添加 na.rm = T 在应该返回 NA 的情况下在 A 组中返回 0.

However, adding na.rm = T returns 0 in group A when it should return NA.

A[, sum(col2, na.rm = T), by = .(col1)]
   col1 V1
1:    A  0
2:    B  5
3:    C  4

我最喜欢的方法是sandipan在评论中建议的方法，类似于我在下面写的函数:

The approach that i like the best is the one that sandipan suggested in the comments, which is akin to the function I wrote below:

ifelse(all(is.na(col2)), NA, sum(col2, na.rm = T)

我创建了一个函数来解决它，但我不确定是否有一个已经内置的方法来解决这个问题:

I created a function to get around it, but I am not sure whether there is an already built-in way to get around this:

sum.na <- function(df){

  if (all(is.na(df))){

    suma <- NA
  }  
  else {    
    suma <- sum(df, na.rm = T)
  }

  return(suma)
}

推荐答案

根据其他用户的建议，我将发布我的问题的答案.该解决方案由@sandipan 在上面的评论中提供:

Following the suggestions from other users, I will post the answer to my question. The solution was provided by @sandipan in the comments above:

如问题中所述，如果您需要对包含 NA 的一列的值求和，有两种好方法:

As noted in the question, if you need to sum the values of one column which contains NAs,there are two good approaches:

1) 使用 ifelse:

1) using ifelse:

A[, (ifelse(all(is.na(col2)), col2[NA_integer_], sum(col2, na.rm = T))), 
  by = .(col1)]

2) 按照@Frank 的建议定义一个函数:

2) define a function as suggested by @Frank:

suma = function(x) if (all(is.na(x))) x[NA_integer_] else sum(x, na.rm = TRUE)

A[, suma(col2), by = .(col1)]

请注意，正如@Frank 指出的那样，我添加了 NA_integer_，因为我不断收到有关类型的错误.

Note that I added NA_integer_ as @Frank pointed out because I kept getting errors about the types.

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