返回子集中所有成对比较的匹配项和唯一项的计数 [英] Return counts of matches and unique items for all pairwise comparisons within subsets
问题描述
我有一个植物plantsp和草食动物lepsp物种及其相互作用int1和int2的数据框，其中的采样嵌套在site，season和group中.我希望创建一个循环，在每个site和season子集中收集的每个group级别之间进行成对比较.对于每个成对比较，我将计算int1和int2之间的总匹配和唯一交互.我设计了以下步骤来解决此问题:
I have a data frame of plant plantsp and herbivore lepsp species and their interactions int1 and int2 with sampling nested in site, season and group . I wish to create a loop that makes pairwise comparisons among each level of group collected within each site and season subset. Fore each pairwise comparison I will calculate total MATCHING and UNIQUE interactions among int1 and int2. I have devised the following steps to break down this problem:
请考虑以下示例数据框df:
Consider the following example data frame df:
sub<-data.frame(site= rep(1, 8),
season=rep("wet", 8),
group= c(1,1,1,2,2,3,3,3),
plantsp= c("P1", "P1", "P2", "P1", "P2", "P1", "P2","P2"),
lepsp= c("L3", "L1", "L2", "L1", "L2", "L1", "L1","L2"),
psitsp=c(NA, "psit1", NA, NA,NA, NA,NA, NA))
sub2<-data.frame(site= rep(1, 8),
season=rep("dry", 8),
group= c(1,1,1,2,2,3,3,3),
plantsp= c("P1", "P1", "P2", "P1", "P2", "P1", "P2","P2"),
lepsp= c("L3", "L1", "L2", "L1", "L2", "L1", "L1","L2"),
psitsp=c(NA, "psit1", NA, NA,NA, NA,NA, NA))
sub3<-data.frame(site= rep(2, 8),
season=rep("wet", 8),
group= c(1,1,1,2,2,3,3,3),
plantsp= c("P1", "P1", "P2", "P1", "P2", "P1", "P2","P2"),
lepsp= c("L3", "L1", "L2", "L1", "L2", "L1", "L1","L2"),
psitsp=c(NA, "psit1", NA, NA,NA, NA,NA, NA))
sub4<-data.frame(site= rep(2, 8),
season=rep("dry", 8),
group= c(1,1,1,2,2,3,3,3),
plantsp= c("P1", "P1", "P2", "P1", "P2", "P1", "P2","P2"),
lepsp= c("L3", "L1", "L2", "L1", "L2", "L1", "L1","L2"),
psitsp=c(NA, "psit1", NA, NA,NA, NA,NA, NA))
df<- rbind(sub, sub2, sub3, sub4)
df$int1<- paste( df$plantsp, df$lepsp, sep="_")
df$int2<-paste( df$lepsp, df$psitsp, sep="_")
df
步骤1:通过site和season子集df.示例:
sub1<- split(df,list(df$site, df$season))
sub1
步骤2:通过group子集df.示例:
sub2 <- split(sub1[[1]], sub1[[1]][[3]])
sub2
步骤3:我们将把sub2中的每个列表元素称为一个组.示例:
Step 3: We will call each list element in sub2 a group. Example:
#group1
group1<-sub2[1]
group1
#group2
group2<-sub2[2]
group2
步骤4:我想在每个group之间进行成对比较.对于每个成对的比较，我想创建一个向量来总结int1和int2之间的UNIQUE和MATCHING元素的计数.对于所有子集，所有组之间的所有可能的成对比较将通过df进行迭代. group1和group2的示例:
Step 4: I want to make pairwise comparisons among each group. For each pairwise comparison I want to create vectors that summarize counts of UNIQUE and MATCHING elements among int1 and int2. This will be iterated though df for all possible pairwise comparisons among all groups for all subsets. Example for group1 and group2:
#CALCULATE MATCHING ELEMENTS
#Count matches in `int1` among both levels of `group`
match1<- length(intersect(sub2[[1]][[7]], sub2[[2]][[7]])) # P1_L1 & P2_L2
match1
#Count matches in `int2` among both levels of `group`. Exclude `int1` or `int2` with NAs
temp<-lapply(sub2, na.omit)
temp
match2<- length(intersect(temp[[1]][[8]],temp[[2]][[8]]))
match2
#SUM `match1` and `match2` and put result into vector called `vecA`.
#`vecA`: represents vector of sums of the counts of MATCHING items in
# both groups within `int1` AND `int2` columns.
vecA<-sum(match1, match2)
vecA
#CALCULATE UNIQUE ELEMENTS TO GROUP1
#Count unique items in `int1` within the first level of `group`
unique_int1<- df[1,] # P1_L3
unique_int1<- length(unique_int1$int1)
#Count unique items in `int2` within the first level of `group`
unique_int2<- df[2,] #L1_psit1
unique_int2<- length(unique_int2$int2)
#SUM `unique_int1` and `unique_int2` and put result into vector called
#`vecB`.`vecB`: represents vector of sums of `int1` AND `int2` that
#are UNIQUE to `group1` in the pairwise comparison
vecB<-sum(unique_int1, unique_int2)
vecB
#CALCULATE UNIQUE ELEMENTS TO GROUP2
#Count unique items in `int1` to `group2`
unique_int1<- 0
#Count unique items in `int2` within the first level of `group`
unique_int2<- 0
#SUM `unique1_int1` and `unique1_int2` and put result into vector
#called `vecC`.`vecC`: represents vector of sums of `int1` AND `int2`
#that are UNIQUE to `group2` in the pairwise comparison
vecC<-sum(unique_int1, unique_int2)
vecC
给定df的所有子集的所有成对比较的预期结果为:
The expected result for all pairwise comparisons for all subsets given df and the steps above is:
result1<-data.frame(site= c(rep(1, 6),rep(2, 6)),
season=c(rep("wet", 3), rep("dry", 3), rep("wet", 3), rep("dry", 3)),
group_pairs= c("1_2", "2_3", "1_3", "1_2", "2_3", "1_3","1_2", "2_3", "1_3", "1_2", "2_3", "1_3"),
vecA= c(2,2,2,2,2,2,2,2,2,2,2,2),
vecB= c(2,0,2,2,0,2,2,0,2,2,0,2),
vecC=c(0,1,0,0,1,0,0,1,0,0,1,0))
步骤5:请执行上述步骤，但仅适用于两个水平都低于group的物种.
Step 5: Conduct steps above but ONLY for species present in BOTH levels of group.
#CALCULATE MATCHING ELEMENTS
#If `plantsp` OR `lepsp` match among both levels of `group`,count matches in `int1`.
match1<- length(intersect(sub2[[1]][[7]], sub2[[2]][[7]]))
match1
# If `lepsp` OR `psitsp` match among both levels of `group`, count matches in `int2`. Remove NAs
temp<-lapply(sub2, na.omit)
temp
match2<- length(intersect(temp[[1]][[8]], temp[[2]][[8]]))
match2
#SUM `match1` and `match2` above and put result into vector called `vecD`. `vecD`: vector of sums of MATCHING items in `int1` and `int2` after subsetting for those species both levels of group share.
vecD<- sum(match1, match2)
#CALCULATE UNIQUE ELEMENTS TO GROUP1
# If `plantsp` OR `lepsp` match among both levels of `group`, count unique items in `int1`. This is represented by the P1_L3 interaction in `int1`
unique_int1<-1
# If `lepsp` and `psitsp` match among both levels of `group`, count unique items in `int2`. This is represented by the L1_psit1 interaction in `int2`
unique_int2<-1
# SUM `unique_int1` and `unique_int2` above and put result into vector called `vecE`. `vecE`: vector of sums of UNIQUE items to the FIRST level of `group` included in the pairwise comparison after after subsetting for those species both levels of group share.
vecE<- sum(unique_int1, unique_int2)
#CALCULATE UNIQUE ELEMENTS TO GROUP2
# If `plantsp` OR `lepsp` match among both levels of `group`, count unique items in `int1`.
unique_int1<-0
# If `lepsp` and `psitsp` match among both levels of `group`, count unique items in `int2`.
unique_int2<-0
# SUM `unique_int1` and `unique_int2` above and put result into vector called `vecF`. `vecF`: vector of sums of UNIQUE items to the SECOND level of `group` included in the pairwise comparison after after subsetting for those species both levels of group share.
vecE<- sum(unique_int1, unique_int2)
给定df的所有子集的所有成对比较的预期结果为:
The expected result for all pairwise comparisons for all subsets given df and the steps above is:
result2<-data.frame(site= c(rep(1, 6),rep(2, 6)),
season=c(rep("wet", 3), rep("dry", 3), rep("wet", 3), rep("dry", 3)),
group_pairs= c("1_2", "2_3", "1_3", "1_2", "2_3", "1_3","1_2", "2_3", "1_3", "1_2", "2_3", "1_3"),
vecD= c(2,2,2,2,2,2,2,2,2,2,2,2),
vecE= c(0,0,0,0,0,0,0,0,0,0,0,0),
vecF=c(0,1,1,0,1,1,0,1,1,0,1,1))
A similar question is posted here, however this approach is unique for all pairwise comparisons among groups.
推荐答案
这里使用 data.table .
library(data.table)
dt <- as.data.table(df)
dt[,
{
groups <- combn(unique(group), 2)
group_pairs = apply(groups, 2, paste, collapse = '_')
vecA = apply(groups, 2, FUN = function(x) length(intersect(group, x[1])) + length(intersect(group, x[2])))
#apply(groups, 2, function(x) .SD[group %in% x, print(.SD)])
list(group_pairs = group_pairs, vecA = vecA)
}
,
by = .(site, season)]
site season group_pairs vecA
1: 1 wet 1_2 2
2: 1 wet 1_3 2
3: 1 wet 2_3 2
4: 1 dry 1_2 2
5: 1 dry 1_3 2
6: 1 dry 2_3 2
7: 2 wet 1_2 2
8: 2 wet 1_3 2
9: 2 wet 2_3 2
10: 2 dry 1_2 2
11: 2 dry 1_3 2
12: 2 dry 2_3 2
vecA的注意事项我窃取了您的代码.不幸的是，您的代码对vecB没有类似的解释，依此类推.它只是简单地声明unique_int1 <- 1; unique_int2 <- 1; vecB<-sum(unique_int1, unique_int2)而没有方程式.
Note for vecA I steal your code. Unfortunately, your code doesn't have similar explanations for vecB and so on. It's just simply states unique_int1 <- 1; unique_int2 <- 1; vecB<-sum(unique_int1, unique_int2) with no equation.
这是group1的数据本身:
> group1
$`1`
site season group plantsp lepsp psitsp int1 int2
1 1 wet 1 P1 L3 <NA> P1_L3 L3_NA
2 1 wet 1 P1 L1 psit1 P1_L1 L1_psit1
3 1 wet 1 P2 L2 <NA> P2_L2 L2_NA
如果在我的代码中取消注释apply行，则会得到以下打印输出(为简洁起见被截断):
If you uncomment out the apply line in my code, you get the following printout (truncated for brevity):
#site 1, season == wet
group plantsp lepsp psitsp int1 int2
1: 1 P1 L3 <NA> P1_L3 L3_NA
2: 1 P1 L1 psit1 P1_L1 L1_psit1
3: 1 P2 L2 <NA> P2_L2 L2_NA
4: 2 P1 L1 <NA> P1_L1 L1_NA
5: 2 P2 L2 <NA> P2_L2 L2_NA
也许您可以使用该apply()语句并运行它.
Maybe you can take that apply() statement and run with it.
这篇关于返回子集中所有成对比较的匹配项和唯一项的计数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
