r语言 - 如何比较二乘二的多个测量值的增减?

Question

我有一个数据，我想通过取差值来比较二乘二的测量结果，如果差值大于0.2，则添加另一列为"是"，否则为"否"。就像图像中的示例数据一样。最后，对于每个人如果至少有一个是，那么最后是是。我已经手动填写了第一行，但我想为我所有的个人(1000个人)做同样的事情。输入图片描述

library(readxl)
dd <- read_excel("dd.xlsx")
dd <- dput(dd)
structure(list(ID = c(1, 2, 3, 4, 5, 6), m1 = c("2.1", "1.4","NA", "4.0", "2.5", "NA"), 
m2 = c("2.8", "1.5", "NA", "4.0", "3.8", "1.1"),
m3 = c("3.5", "1.5", "NA", "4.0", "NA", "1.3"), 
m4 = c("NA", "1.8", "1.8", "4.0", "NA", "1.3"), 
m5 = c("NA","1.5", "2.9", "NA", "NA", "1.5"), 
m6 = c("NA", "NA", "3.5", "NA", "NA", "1.2")), 
class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L))

Answer 1

这种方法使用更长的枢轴来估计列，然后使用rename更宽的枢轴将得到您想要的结果(参见第二个解决方案)。但是，如果您只需要Final列的ID，只需这样做:

1. 通过ID获取Final的简单方法

dd %>% 
pivot_longer(cols = !ID) %>%
mutate(value=as.numeric(value)) %>% 
group_by(ID) %>% 
summarize(Final = any(value-lag(value)>0.2))

2. 扩展方法，如果你真的想要所有的中间变量

dd %>% 
pivot_longer(cols = !ID) %>%
mutate(value=as.numeric(value)) %>% 
group_by(ID) %>% 
mutate(diff=value-lag(value),
gt=if_else(diff>0.2, "Yes", "No")) %>% 
pivot_wider(id_cols = ID,names_from = name,values_from = value:gt) %>% 
rowwise() %>% 
mutate(Final = any(c_across(starts_with("gt"))=="Yes")) %>% 
select(!c(diff_m1,gt_m1)) %>% 
rename_with(~c("ID", "m1","m2","m3","m4","m5","m6",
"m2-m1", "m3-m2", "m4-m3", "m5-m4", "m6-m5",
"r(m2-m1)", "r(m3-m2)", "r(m4-m3)", "r(m5-m4)", "r(m6-m5)",
"Final"), .cols = everything()
)

输出:

ID    m1    m2    m3    m4    m5    m6 `m2-m1` `m3-m2` `m4-m3` `m5-m4` `m6-m5` `r(m2-m1)` `r(m3-m2)` `r(m4-m3)`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl> <chr>      <chr>      <chr>     
1     1   2.1   2.8   3.5  NA    NA    NA     0.7       0.7    NA      NA      NA   Yes        Yes        NA        
2     2   1.4   1.5   1.5   1.8   1.5  NA     0.100     0       0.3    -0.3    NA   No         No         Yes       
3     3  NA    NA    NA     1.8   2.9   3.5  NA        NA      NA       1.1     0.6 NA         NA         NA        
4     4   4     4     4     4    NA    NA     0         0       0      NA      NA   No         No         No        
5     5   2.5   3.8  NA    NA    NA    NA     1.3      NA      NA      NA      NA   Yes        NA         NA        
6     6  NA     1.1   1.3   1.3   1.5   1.2  NA         0.2     0       0.2    -0.3 NA         No         No        
# … with 3 more variables: `r(m5-m4)` <chr>, `r(m6-m5)` <chr>, Final <lgl>

Answer 2

这是另一种tidyverse方法:

library(dplyr)
library(tibble)
dd %>%
type.convert(as.is =TRUE) %>% 
select(-1) %>% 
mutate(across(-1, .names = "{.col}-{names(.)[match(.col, 
names(.))-1]}") - across(-last_col())) %>% 
mutate(across(contains("-"), ~ifelse(is.na(.), NA, "YES"), .names = "r({.col})")) %>% 
mutate(Final = ifelse(rowSums(dd == "YES") > 0, "No", "Yes")) %>% 
add_column(ID = dd$ID, .before = "m1")

ID    m1    m2    m3    m4    m5    m6 `m2-m1` `m3-m2` `m4-m3` `m5-m4` `m6-m5` `r(m2-m1)` `r(m3-m2)` `r(m4-m3)` `r(m5-m4)` `r(m6-m5)` Final
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl> <chr>      <chr>      <chr>      <chr>      <chr>      <chr>
1     1   2.1   2.8   3.5  NA    NA    NA     0.7       0.7    NA      NA      NA   YES        YES        NA         NA         NA         Yes  
2     2   1.4   1.5   1.5   1.8   1.5  NA     0.100     0       0.3    -0.3    NA   YES        YES        YES        YES        NA         Yes  
3     3  NA    NA    NA     1.8   2.9   3.5  NA        NA      NA       1.1     0.6 NA         NA         NA         YES        YES        Yes  
4     4   4     4     4     4    NA    NA     0         0       0      NA      NA   YES        YES        YES        NA         NA         Yes  
5     5   2.5   3.8  NA    NA    NA    NA     1.3      NA      NA      NA      NA   YES        NA         NA         NA         NA         Yes  
6     6  NA     1.1   1.3   1.3   1.5   1.2  NA         0.2     0       0.2    -0.3 NA         YES        YES        YES        YES        Yes