使用r中的循环计算数组中的对
使用r中的循环计算数组中的对
提问于 2018-06-05 23:26:38
我在写一个简单的for循环时遇到了一些问题，条件是r。我得到了这个数组：
Temp <- c("A", "A", "B", "A", "C", "C", "A", "B")我想通过使用两个在循环期间递增的索引来计算这个数组中的对，必须遵循序列的顺序。
此数组的最终结果应为：
CountAA = 1
CountAB = 2
CountAC = 1
CountBA = 1
CountBB = 0
CountBC = 0
CountCA = 1
CountCB = 0
CountCC = 1我尝试过使用这段代码，但它给出了一个错误
"Error in if (Temp[i] == "A" & Temp[j] == "A") { :
argument is of length zero"代码
CountAA = 0
CountAB = 0
CountAC = 0
CountBA = 0
CountBB = 0
CountBC = 0
CountCA = 0
CountCB = 0
CountCC = 0
i = 1
j = 2
for (j in 1:length(Temp)-1){
if (Temp[i]=="A" & Temp[j]=="A"){
CountAA = CountAA + 1
i = i + 1
j = j + 1
}
if (Temp[i]=="A" & Temp[j]=="B"){
CountAB = CountAB + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="A" & Temp[j]=="C"){
CountAC = CountAC + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="B" & Temp[j]=="A"){
CountBA = CountBA + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="B" & Temp[j]=="B"){
CountBB = CountBB + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="B" & Temp[j]=="C"){
CountBC = CountBC + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="C" & Temp[j]=="A"){
CountCA = CountCA + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="C" & Temp[j]=="B"){
CountCB = CountCB + 1
i = i + 1
j = j + 1
}
if(Temp[i]=="C" & Temp[j]=="C"){
CountCC = CountCC + 1
i = i + 1
j = j + 1
}
}回答 4
Stack Overflow用户
回答已采纳
发布于 2018-06-05 23:32:49
在基本R中
# unique letter values
ut <- unique(Temp)
# expand to get a data.frame with all combinations
expnd <- data.frame(pair=do.call(paste0,expand.grid(ut,ut)),stringsAsFactors = FALSE)
# merge it with the table containing counts of all pair combinations
out <- merge(expnd, table(pair=paste0(head(Temp,-1),tail(Temp,-1))), all=TRUE)
# turn NAs into zeroes
out$Freq[is.na(out$Freq)] <- 0
# pair Freq
# 1 AA 1
# 2 AB 2
# 3 AC 1
# 4 BA 1
# 5 BB 0
# 6 BC 0
# 7 CA 1
# 8 CB 0
# 9 CC 1使用库tidyverse
library(tidyverse)
tibble(x=head(Temp,-1),y=tail(Temp,-1)) %>%
count(x,y) %>% # count combinations
complete(x,y) %>% # add missing combinations
replace_na(list(n=0)) %>% # make them zero
unite(pair,x,y,sep='') %>% # turn 2 columns into 1
arrange(pair) # sort
# # A tibble: 9 x 2
# pair n
# <chr> <dbl>
# 1 AA 1
# 2 AB 2
# 3 AC 1
# 4 BA 1
# 5 BB 0
# 6 BC 0
# 7 CA 1
# 8 CB 0
# 9 CC 1Stack Overflow用户
发布于 2018-06-06 00:04:53
下面是一个简单的基本解决方案：
table(sapply(1:(length(Temp) - 1), function(x) paste(Temp[x:(x+1)], collapse = "")))
AA AB AC BA CA CC
1 2 1 1 1 1如果您真的想要查看所有可能的排列，您可以使用任何将生成重复排列的包。下面我们使用gtools。
library(gtools)
## Same as above
vec <- table(sapply(1:(length(Temp) - 1), function(x) paste(Temp[x:(x+1)], collapse = "")))
## Generate all permutations
myNames <- apply(permutations(3, 2, unique(Temp), repeats.allowed = TRUE), 1, paste, collapse = "")
## Initialize return vector
res <- integer(length(myNames))
## Add names
names(res) <- myNames
## Subset on names
res[names(res) %in% names(vec)] <- vec
res
AA AB AC BA BB BC CA CB CC
1 2 1 1 0 0 1 0 1Stack Overflow用户
发布于 2018-06-05 23:45:10
你可以试试
library(tidyverse)
b <- table(sapply(seq_along(Temp), function(x) paste0(Temp[x], Temp[x+1]))[-length(Temp)])
expand.grid(unique(Temp), unique(Temp)) %>%
unite(Var1, Var1, Var2, sep = "") %>%
left_join(as.data.frame(b,stringsAsFactors = F)) %>%
mutate(Freq=ifelse(is.na(Freq), 0, Freq))
Var Freq
1 AA 1
2 BA 1
3 CA 1
4 AB 2
5 BB 0
6 CB 0
7 AC 1
8 BC 0
9 CC 1页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50703782
复制相关文章
相似问题
腾讯云开发者
