如何求R中非均匀类中所有值的两两差
我找到了简单向量的解决方案,但是有没有一种方法可以对一个类别中的所有元素使用dplyr或base R来产生所有成对的差异如何求R中非均匀类中所有值的两两差,r,dplyr,R,Dplyr,我找到了简单向量的解决方案,但是有没有一种方法可以对一个类别中的所有元素使用dplyr或base R来产生所有成对的差异 library(tidyverse) x = 1:10 y = rep(letters[1:5],each=2) z = rep(1:2,length.out =10) df = data.frame(x,y, z) df = rbind(df,c(11,"e",3)) df$verif = paste0(df$y,df$z) df$x = as.nume
library(tidyverse)
x = 1:10
y = rep(letters[1:5],each=2)
z = rep(1:2,length.out =10)
df = data.frame(x,y, z)
df = rbind(df,c(11,"e",3))
df$verif = paste0(df$y,df$z)
df$x = as.numeric(df$x)
df %>%
group_by(y) %>%
summarise(Diff = abs(x - lag(x)))
给出:
`summarise()` regrouping output by 'y' (override with `.groups` argument)
# A tibble: 11 x 2
# Groups: y [5]
y Diff
<chr> <dbl>
1 a NA
2 a 1
3 b NA
4 b 1
5 c NA
6 c 1
7 d NA
8 d 1
9 e NA
10 e 1
11 e 1
我现在需要得到,计算出的两两差异
又一次继续,弄得一团糟:
my.df=df %>%
group_by(y) %>%
summarise(Diff = combn(x,2,diff),
test = combn(verif,2,paste, simplify = FALSE)) %>%
mutate(test2 = paste0(test, collapse = "-"))
my.df
> my.df
# A tibble: 7 x 4
# Groups: y [5]
y Diff test test2
<chr> <dbl> <list> <chr>
1 a 1 <chr [2]> "c(\"a1\", \"a2\")"
2 b 1 <chr [2]> "c(\"b1\", \"b2\")"
3 c 1 <chr [2]> "c(\"c1\", \"c2\")"
4 d 1 <chr [2]> "c(\"d1\", \"d2\")"
5 e 1 <chr [2]> "c(\"e1\", \"e2\")-c(\"e1\", \"e3\")-c(\"e2\", \"e3\")"
6 e 2 <chr [2]> "c(\"e1\", \"e2\")-c(\"e1\", \"e3\")-c(\"e2\", \"e3\")"
7 e 1 <chr [2]> "c(\"e1\", \"e2\")-c(\"e1\", \"e3\")-c(\"e2\", \"e3\")"
my.df=df%>%
组别(y)%>%
总结(Diff=combn(x,2,Diff),
测试=combn(验证,2,粘贴,简化=FALSE))%>%
突变(test2=paste0(test,collapse=“-”))
我的.df
>我的.df
#一个tibble:7x4
#分组:y[5]
y差异测试2
1 a 1“c(\'a1\',\'a2\')”
2 b 1“c(\'b1\',\'b2\'”)
3 c 1“c(\'c1\',\'c2\')”
4 d 1“c(\“d1\”,“d2\”)
5 e 1“c(\“e1\”,“e2\”)-c(\“e1\”,“e3\”)-c(\“e2\”,“e3\”)
6 e 2“c”(“e1\”,“e2\”)-c(“e1\”,“e3\”)-c(“e2\”,“e3\”)
7 e 1“c(\“e1\”,“e2\”)-c(\“e1\”,“e3\”)-c(\“e2\”,“e3\”)
明白了:
library(tidyverse)
x = 1:10
y = rep(letters[1:5],each=2)
z = rep(1:2,length.out =10)
df = data.frame(x,y, z)
df = rbind(df,c(11,"e",3))
df$verif = paste0(df$y,df$z)
df$x = as.numeric(df$x)
my.df=df %>%
group_by(y) %>%
summarise(Diff = combn(x,2,diff),
test = combn(verif,2,paste, simplify = FALSE)) %>%
mutate(test2 = unlist(lapply(test, function(x)paste(x,collapse="-")))) %>%
select(-test)
这是输出
my.df
# A tibble: 7 x 3
# Groups: y [5]
y Diff test2
<chr> <dbl> <chr>
1 a 1 a1-a2
2 b 1 b1-b2
3 c 1 c1-c2
4 d 1 d1-d2
5 e 1 e1-e2
6 e 2 e1-e3
7 e 1 e2-e3
my.df
#一个tibble:7x3
#分组:y[5]
y差异测试2
a1-a2
2 b 1 b1-b2
3 c 1 c1-c2
4 d 1 d1-d2
5 e 1 e1-e2
6 e 2 e1-e3
7 e 1 e2-e3
您可以执行以下操作:
library(tidyverse)
df %>%
group_by(y) %>%
summarise(result = combn(seq_along(x), 2, function(i)
list(test1 = diff(x[i]), #The difference
test2 = paste0(verif[i], collapse = '-')), # The pairs
simplify = FALSE),
.groups = 'drop') %>%
unnest_wider(result)
# A tibble: 7 x 3
y test1 test2
<chr> <dbl> <chr>
1 a 1 a1-a2
2 b 1 b1-b2
3 c 1 c1-c2
4 d 1 d1-d2
5 e 1 e1-e2
6 e 2 e1-e3
7 e 1 e2-e3
库(tidyverse)
df%>%
组别(y)%>%
总结(结果=combn(沿(x)、2、功能(i)的顺序)
列表(test1=diff(x[i]),差异
test2=paste0(verif[i],collapse='-'),#这些对
简化=错误),
.groups='drop')%>%
unnest_加宽(结果)
#一个tibble:7x3
y测试1测试2
a1-a2
2 b 1 b1-b2
3 c 1 c1-c2
4 d 1 d1-d2
5 e 1 e1-e2
6 e 2 e1-e3
7 e 1 e2-e3
您可以在summairse
步骤本身中执行此操作-df%>%groupby(y)%%>%summary(Diff=combn(x,2,Diff),test=combn(verif,2,paste0,collapse='-')
library(tidyverse)
df %>%
group_by(y) %>%
summarise(result = combn(seq_along(x), 2, function(i)
list(test1 = diff(x[i]), #The difference
test2 = paste0(verif[i], collapse = '-')), # The pairs
simplify = FALSE),
.groups = 'drop') %>%
unnest_wider(result)
# A tibble: 7 x 3
y test1 test2
<chr> <dbl> <chr>
1 a 1 a1-a2
2 b 1 b1-b2
3 c 1 c1-c2
4 d 1 d1-d2
5 e 1 e1-e2
6 e 2 e1-e3
7 e 1 e2-e3