R 使用来自更新的先前值的信息更新分组 df 中的值
R update values within a grouped df with information from updated previous value
我想根据此函数在不同时间点 (timepoint
) 有条件地改变组 (id
) 内的变量 (var1, var2
),使用以前的 updated/muated 值:
change_function <- function(value,pastvalue,timepoint){
if(timepoint==1){valuenew=value} else
if(value==0){valuenew=pastvalue-1}
if(value==1){valuenew=pastvalue}
if(value==2){valuenew=pastvalue+1}
return(valuenew)
}
pastvalue
是 MUTATED/UPDATED 值 timepoint
-1 for timepoint 2:4
这是一个示例和输出文件:
``` r
#example data
df <- data.frame(id=c(1,1,1,1,2,2,2,2),timepoint=c(1,2,3,4,1,2,3,4),var1=c(1,0,1,2,2,2,1,0),var2=c(2,0,1,2,3,2,1,0))
df
#> id timepoint var1 var2
#> 1 1 1 1 2
#> 2 1 2 0 0
#> 3 1 3 1 1
#> 4 1 4 2 2
#> 5 2 1 2 3
#> 6 2 2 2 2
#> 7 2 3 1 1
#> 8 2 4 0 0
#desired output
output <- data.frame(id=c(1,1,1,1,2,2,2,2),timepoint=c(1,2,3,4,1,2,3,4),var1=c(1,0,0,1,2,3,3,2),var2=c(2,1,1,2,3,4,4,3))
output
#> id timepoint var1 var2
#> 1 1 1 1 2
#> 2 1 2 0 1
#> 3 1 3 0 1
#> 4 1 4 1 2
#> 5 2 1 2 3
#> 6 2 2 3 4
#> 7 2 3 3 4
#> 8 2 4 2 3
```
<sup>Created on 2020-11-23 by the [reprex package](https://reprex.tidyverse.org) (v0.3.0)</sup>
我的方法:使用我的函数 dplyr::mutate_at
library(dplyr)
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(.=change_function(.,dplyr::lag(.),timepoint)))
但是,这不起作用,因为 if/else 未矢量化
更新 1:
使用嵌套的 ifelse 函数无法提供所需的输出,因为它没有使用更新的 pastvalue
:
change_function <- function(value,pastvalue,timepoint){
ifelse((timepoint==1),value,
ifelse((value==0),pastvalue-1,
ifelse((value==1),pastvalue,
ifelse((value==2),pastvalue+1,NA))))
}
library(dplyr)
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(.=change_function(.,dplyr::lag(.),timepoint)))
id TimePoint var1 var2 var1_. var2_.
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 2 1 2
2 1 2 0 0 0 1
3 1 3 1 1 0 0
4 1 4 2 2 2 2
5 2 1 2 3 2 3
6 2 2 2 2 3 4
7 2 3 1 1 2 2
8 2 4 0 0 0 0
更新二:
根据评论,可以使用purrr:accumulate
感谢 akrun,我得到了正确的函数:
# write a vectorized function
change_function <- function(prev, new) {
change=if_else(new==0,-1,
if_else(new==1,0,1))
if_else(is.na(new), new, prev + change)
}
# use purrr:accumulate
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(accumulate(.,change_function)))
# A tibble: 8 x 4
# Groups: id [2]
id timepoint var1 var2
<dbl> <dbl> <dbl> <dbl>
1 1 1 1 2
2 1 2 0 1
3 1 3 0 1
4 1 4 1 2
5 2 1 2 3
6 2 2 3 4
7 2 3 3 4
8 2 4 2 3
我想根据此函数在不同时间点 (timepoint
) 有条件地改变组 (id
) 内的变量 (var1, var2
),使用以前的 updated/muated 值:
change_function <- function(value,pastvalue,timepoint){
if(timepoint==1){valuenew=value} else
if(value==0){valuenew=pastvalue-1}
if(value==1){valuenew=pastvalue}
if(value==2){valuenew=pastvalue+1}
return(valuenew)
}
pastvalue
是 MUTATED/UPDATED 值 timepoint
-1 for timepoint 2:4
这是一个示例和输出文件:
``` r
#example data
df <- data.frame(id=c(1,1,1,1,2,2,2,2),timepoint=c(1,2,3,4,1,2,3,4),var1=c(1,0,1,2,2,2,1,0),var2=c(2,0,1,2,3,2,1,0))
df
#> id timepoint var1 var2
#> 1 1 1 1 2
#> 2 1 2 0 0
#> 3 1 3 1 1
#> 4 1 4 2 2
#> 5 2 1 2 3
#> 6 2 2 2 2
#> 7 2 3 1 1
#> 8 2 4 0 0
#desired output
output <- data.frame(id=c(1,1,1,1,2,2,2,2),timepoint=c(1,2,3,4,1,2,3,4),var1=c(1,0,0,1,2,3,3,2),var2=c(2,1,1,2,3,4,4,3))
output
#> id timepoint var1 var2
#> 1 1 1 1 2
#> 2 1 2 0 1
#> 3 1 3 0 1
#> 4 1 4 1 2
#> 5 2 1 2 3
#> 6 2 2 3 4
#> 7 2 3 3 4
#> 8 2 4 2 3
```
<sup>Created on 2020-11-23 by the [reprex package](https://reprex.tidyverse.org) (v0.3.0)</sup>
我的方法:使用我的函数 dplyr::mutate_at
library(dplyr)
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(.=change_function(.,dplyr::lag(.),timepoint)))
但是,这不起作用,因为 if/else 未矢量化
更新 1:
使用嵌套的 ifelse 函数无法提供所需的输出,因为它没有使用更新的 pastvalue
:
change_function <- function(value,pastvalue,timepoint){
ifelse((timepoint==1),value,
ifelse((value==0),pastvalue-1,
ifelse((value==1),pastvalue,
ifelse((value==2),pastvalue+1,NA))))
}
library(dplyr)
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(.=change_function(.,dplyr::lag(.),timepoint)))
id TimePoint var1 var2 var1_. var2_.
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 2 1 2
2 1 2 0 0 0 1
3 1 3 1 1 0 0
4 1 4 2 2 2 2
5 2 1 2 3 2 3
6 2 2 2 2 3 4
7 2 3 1 1 2 2
8 2 4 0 0 0 0
更新二:
根据评论,可以使用purrr:accumulate
感谢 akrun,我得到了正确的函数:
# write a vectorized function
change_function <- function(prev, new) {
change=if_else(new==0,-1,
if_else(new==1,0,1))
if_else(is.na(new), new, prev + change)
}
# use purrr:accumulate
df %>%
group_by(id) %>%
mutate_at(.vars=vars(var1,var2),
.funs=funs(accumulate(.,change_function)))
# A tibble: 8 x 4
# Groups: id [2]
id timepoint var1 var2
<dbl> <dbl> <dbl> <dbl>
1 1 1 1 2
2 1 2 0 1
3 1 3 0 1
4 1 4 1 2
5 2 1 2 3
6 2 2 3 4
7 2 3 3 4
8 2 4 2 3