Question

请考虑以下内容：

自定义函数CustomFun带有几个数字参数。参数名称存储在resp中，并与函数参数名称相对应。参数值存储在列val中。

data.frame拥有有关多个患者（id）的信息，因此数据需要按id分组。

问题：

我们如何将自定义函数应用于分组的data.frame或data.table，它们从相同数据结构中的列获取参数？

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(data.table)
#> 
#> Attaching package: 'data.table'
#> The following objects are masked from 'package:dplyr':
#> 
#>     between, first, last

# The data
df.x <- data.frame(id = rep(c(1:2), each = 5),
                resp = c("val.a", "val.b", "val.c", "val.d", "val.e"),
                val = c(10, 15, NA, NA, NA,
                        1, 5, NA, NA, NA))
df.x
#>    id  resp val
#> 1   1 val.a  10
#> 2   1 val.b  15
#> 3   1 val.c  NA
#> 4   1 val.d  NA
#> 5   1 val.e  NA
#> 6   2 val.a   1
#> 7   2 val.b   5
#> 8   2 val.c  NA
#> 9   2 val.d  NA
#> 10  2 val.e  NA

# A simple function (minimal replicable example)
CustomFun <- function(a,b){
        a+b
}

所需的输出：

# Desired output
df.x %>% mutate(res = c(25, 25, NA, NA, NA, 6, 6, NA, NA, NA))
#>    id  resp val res
#> 1   1 val.a  10  25
#> 2   1 val.b  15  25
#> 3   1 val.c  NA  NA
#> 4   1 val.d  NA  NA
#> 5   1 val.e  NA  NA
#> 6   2 val.a   1   6
#> 7   2 val.b   5   6
#> 8   2 val.c  NA  NA
#> 9   2 val.d  NA  NA
#> 10  2 val.e  NA  NA

自己的方法：

在没有组（id）的情况下，此方法有效。对于所有非NA或val，在val.a中没有val.b不会造成问题，因为可以在第二步中将它们过滤掉。

# Approach without the need of grouping: one id only, problem: NA also assigned to val in df.z[3:5, ]

# dplyr
df.z <- df.x %>% slice(1:5)
df.z
#>   id  resp val
#> 1  1 val.a  10
#> 2  1 val.b  15
#> 3  1 val.c  NA
#> 4  1 val.d  NA
#> 5  1 val.e  NA

df.z %>% mutate(test = CustomFun(a = df.z %>% filter(resp == "val.a") %>% pull(val),
     b = df.z %>% filter(resp == "val.b") %>% pull(val))
)
#>   id  resp val test
#> 1  1 val.a  10   25
#> 2  1 val.b  15   25
#> 3  1 val.c  NA   25
#> 4  1 val.d  NA   25
#> 5  1 val.e  NA   25

# data.table
setDT(df.z)[, .(test= CustomFun(a = setDT(df.z)[resp == "val.a", val],
                        b = setDT(df.z)[resp == "val.b", val])),
         by = .(id, val, resp)]
#>    id val  resp test
#> 1:  1  10 val.a   25
#> 2:  1  15 val.b   25
#> 3:  1  NA val.c   25
#> 4:  1  NA val.d   25
#> 5:  1  NA val.e   25

# NOT working for groups =====================================

# data.frame
df.x %>%
        group_by(id) %>% 
        mutate(test = CustomFun(a = df.x %>% filter(resp == "val.a") %>% pull(val),
                                 b = df.x %>% filter(resp == "val.b") %>% pull(val))
)
#> Error in mutate_impl(.data, dots): Column `test` must be length 5 (the group size) or one, not 2

# data.table
setDT(df.x)[, .(test= CustomFun(a = setDT(df.x)[resp == "val.a", val],
                                b = setDT(df.x)[resp == "val.b", val])),
            by = .(id, val, resp)]
#>     id val  resp test
#>  1:  1  10 val.a   25
#>  2:  1  10 val.a    6
#>  3:  1  15 val.b   25
#>  4:  1  15 val.b    6
#>  5:  1  NA val.c   25
#>  6:  1  NA val.c    6
#>  7:  1  NA val.d   25
#>  8:  1  NA val.d    6
#>  9:  1  NA val.e   25
#> 10:  1  NA val.e    6
#> 11:  2   1 val.a   25
#> 12:  2   1 val.a    6
#> 13:  2   5 val.b   25
#> 14:  2   5 val.b    6
#> 15:  2  NA val.c   25
#> 16:  2  NA val.c    6
#> 17:  2  NA val.d   25
#> 18:  2  NA val.d    6
#> 19:  2  NA val.e   25
#> 20:  2  NA val.e    6

^{由reprex package（v0.2.1）于2018-11-13创建}

非常感谢！

Answer 1

有2个不同的问题：您在data.table中添加了不需要的分组变量，并且两个版本中的数据子集都不正确。

data.table的调整：

setDT(df.x)[!is.na(val), test := CustomFun(a = val[resp == "val.a"],
                                           b = val[resp == "val.b"]), by = id]

无需按resp和val进行分组，只需按id进行分组。

对于dplyr，您可以这样做：

df.x %>%
  group_by(id) %>% 
  mutate(test = if_else(!is.na(val), CustomFun(a = val[resp == "val.a"],
                                               b = val[resp == "val.b"]), NA_real_)
  )

两种情况下的输出：

    id  resp val test
 1:  1 val.a  10   25
 2:  1 val.b  15   25
 3:  1 val.c  NA   NA
 4:  1 val.d  NA   NA
 5:  1 val.e  NA   NA
 6:  2 val.a   1    6
 7:  2 val.b   5    6
 8:  2 val.c  NA   NA
 9:  2 val.d  NA   NA
10:  2 val.e  NA   NA

Answer 2

我们可以按组对值进行子集设置（假设每个“ id”只有一个“ val.a”，“ val.b”并添加

library(dplyr)
df.x %>%
    group_by(id) %>%
    mutate(res = (val[resp == 'val.a'] + val[resp == 'val.b']) * NA^(is.na(val)))
# A tibble: 10 x 4
# Groups:   id [2]
#      id resp    val   res
#   <int> <fct> <dbl> <dbl>
# 1     1 val.a    10    25
# 2     1 val.b    15    25
# 3     1 val.c    NA    NA
# 4     1 val.d    NA    NA
# 5     1 val.e    NA    NA
# 6     2 val.a     1     6
# 7     2 val.b     5     6
# 8     2 val.c    NA    NA
# 9     2 val.d    NA    NA
#10     2 val.e    NA    NA

或者另一种选择是filter，按组进行summarize，然后与原始数据集合并

df.x %>% 
   filter(resp %in% c('val.a', 'val.b')) %>% 
   group_by(id) %>% 
   summarise(res = sum(val)) %>%
   right_join(df.x) %>%
   mutate(res = replace(res, is.na(val), NA))

使用R中的函数和条件函数参数进行分组和变异

2 个答案: