Question

在dplyr中有一种优雅的方法可以将NA处理为0（na.rm = TRUE）吗？

data <- data.frame(a=c(1,2,3,4), b=c(4,NA,5,6), c=c(7,8,9,NA))

data %>% mutate(sum = a + b + c)

a  b  c sum
1  4  7  12
2 NA  8  NA
3  5  9  17
4  6 NA  NA

but I like to get

a  b  c sum
1  4  7  12
2 NA  8  10
3  5  9  17
4  6 NA  10

即使我知道在许多其他情况下这不是理想的结果

Answer 1

你可以用这个：

library(dplyr)
data %>% 
  #rowwise will make sure the sum operation will occur on each row
  rowwise() %>% 
  #then a simple sum(..., na.rm=TRUE) is enough to result in what you need
  mutate(sum = sum(a,b,c, na.rm=TRUE))

输出：

Source: local data frame [4 x 4]
Groups: <by row>

      a     b     c   sum
  (dbl) (dbl) (dbl) (dbl)
1     1     4     7    12
2     2    NA     8    10
3     3     5     9    17
4     4     6    NA    10

Answer 2

另一种选择：

data %>%
  mutate(sum = rowSums(., na.rm = TRUE))

<强>基准

library(microbenchmark)
mbm <- microbenchmark(
steven = data %>% mutate(sum = rowSums(., na.rm = TRUE)), 
lyz    = data %>% rowwise() %>% mutate(sum = sum(a, b, c, na.rm=TRUE)),
nar    = apply(data, 1, sum, na.rm = TRUE),
akrun  = data %>% mutate_each(funs(replace(., which(is.na(.)), 0))) %>% mutate(sum=a+b+c),
frank  = data %>% mutate(sum = Reduce(function(x,y) x + replace(y, is.na(y), 0), ., 
                                     init=rep(0, n()))),
times = 10)

#Unit: milliseconds
#   expr         min          lq       mean     median         uq        max neval cld
# steven    9.493812    9.558736   18.31476   10.10280   22.55230   65.15325    10 a  
#    lyz 6791.690570 6836.243782 6978.29684 6915.16098 7138.67733 7321.61117    10   c
#    nar  702.537055  723.256808  799.79996  805.71028  849.43815  909.36413    10  b 
#  akrun   11.372550   11.388473   28.49560   11.44698   20.21214  155.23165    10 a  
#  frank   20.206747   20.695986   32.69899   21.12998   25.11939  118.14779    10 a

Answer 3

或者我们可以replace NA使用0，然后使用OP的代码

data %>% 
   mutate_each(funs(replace(., which(is.na(.)), 0))) %>%
   mutate(Sum= a+b+c)
   #or as @Frank mentioned
   #mutate(Sum = Reduce(`+`, .))

基于使用@StevenBeaupré数据的基准测试，它似乎也很有效。

Answer 4

使用新的dplyr 1.0.0，您可以将c_across与rowwise一起使用。

library(dplyr)

data %>%
  rowwise() %>%
  mutate(sum = sum(c_across(a:c), na.rm = TRUE))

#      a     b     c   sum
#  <dbl> <dbl> <dbl> <dbl>
#1     1     4     7    12
#2     2    NA     8    10
#3     3     5     9    17
#4     4     6    NA    10

Answer 5

试试这个

data$sum <- apply(data, 1, sum, na.rm = T)

结果data是

a  b  c sum
1 1  4  7  12
2 2 NA  8  10
3 3  5  9  17
4 4  6 NA  10

Answer 6

这里与Steven的方法类似，但是包括dplyr::select()来明确声明要包括/忽略的列（例如ID变量）。

data %>% 
  mutate(sum = rowSums(dplyr::select(., a, b, c), na.rm = TRUE))

它具有与实际大小的数据集相当的性能。我不确定为什么会这样，因为在这个瘦小的示例中实际上没有排除任何列。

具有1M行的更大数据集：

pick <- function() { sample(c(1:5, NA), 1000000, replace=T) }
data <- data.frame(a=pick(), b=pick(), c=pick())

结果：

Unit: milliseconds
     expr         min          lq        mean      median          uq         max neval cld
   steven    22.05847    22.96164    56.84822    28.85411    54.99691   174.58447    10 a  
wibeasley    25.10274    26.98303    30.66911    29.30630    30.63343    49.46048    10 a  
      lyz 10408.89904 10548.33756 10887.51930 10720.92372 11017.56256 12250.41370    10   c
      nar  1975.35941  2011.36445  2123.81705  2090.43174  2172.80501  2362.13658    10  b 
    akrun    31.27247    35.41943    81.33320    57.93900    63.59119   302.21059    10 a  
    frank    37.48265    38.72270    65.02965    41.62735    44.45775   261.79898    10 a

在dplyr行总和中忽略NA

6 个答案: