如何获得R中两个向量之间元素的第n个匹配？

Question

match返回其第一个和第二个参数之间的第一个匹配位置：

match(c("a","c"), c("a", "a", "b", "c", "c", "c")) # 1 4

指定除第一个之外的匹配的最佳方法是什么？ 例如，我们希望第二场比赛为"a" ，第三场比赛为"c" （所以我们得到： 2 6 ）。

更新：效率低下的解决方案会进行n次查找：

value_index_query <- data.frame(value = c("a", "c"), index = c(2, 3))
id <-  c("a", "a", "b", "c", "c", "c")
apply(value_index_query, 1, function(value_index) {
  value <- value_index[1]
  index <- as.integer(value_index[2])
  which(id == value)[index]
})

Answer 1

这也使用mapply通过哪个（。）[n]操作串联运行两列。

with(value_index_query,  
     mapply( function(target, nth) which(id==target)[nth], 
               target=value, nth=index) )
[1] 2 6

Answer 2

这是一个data.table解决方案，我们将id向量与映射表连接起来。 然后我们可以使用.EACHI进行分组，从每个组的.I获取index 。

library(data.table)
## 'dti' would be your 'value_index_query' with the 'value' column renamed
dti <- data.table(id = c("a", "c"), index = c(2, 3))
## join it with 'id' and take 'index' by group
data.table(id)[dti, .I[index], by = .EACHI, on = "id"]$V1
# [1] 2 6

我们可以把它放到一个函数中：

viq <- function(id, value, index) {
    dti <- data.table(id = value, index = index)
    data.table(id)[dti, .I[index], by = .EACHI, on = "id"]$V1
}

id <- c("a", "a", "b", "c", "c", "c")

viq(id, c("a", "c"), 2:3)
# [1] 2 6
viq(id, c("a", "c"), c(2, 4))
# [1]  2 NA
viq(id, c("a", "b", "c"), c(2, 1, 4))
# [1]  2  3 NA
viq(id, c("a", "b", "c"), c(2, 1, 3))
# [1] 2 3 6

Answer 3

用grep一次一个。

vec <- c("a", "a", "b", "c", "c", "c")
aa <-grep("a", vec)[2] #2nd
cc <-grep("c", vec)[3] #3rd
c(aa,cc)
#[1] 2 6

Answer 4

这是一种dplyr方式

library(dplyr)

test = data_frame(value = c("a","c"), order = c(2, 3))
original = data_frame(value =  c("a", "a", "b", "c", "c", "c"))

original %>%
  mutate(ID = 1:n()) %>%
  right_join(test) %>%
  group_by(value) %>%
  slice(order %>% first)

Answer 5

那这个呢？：

mapply(function(x,y) x[[y]], x = sapply(v1, function(x) which(x == v2)), y = c(2,3))
a c 
2 6 

Answer 6

为了比较，一个（可能不是理想的，我还在学习）Rcpp解决方案与其他三个主要方法的一些时间安排。

library(Rcpp)
library(microbenchmark)
library(data.table)
library(dplyr)

foo_mapply <- function(value,id,index){
    mapply( function(target, nth, id) which(id==target)[nth], 
                            target=value, nth=index,MoreArgs = list(id = id))
}

foo_dt <- function(dti,id){
    data.table(id)[dti, .I[index], by = .EACHI, on = "id"]$V1
}

foo_dplyr <- function(test,original){
    original %>%
        mutate(ID = 1:n()) %>%
        right_join(test,by = "value") %>%
        group_by(value) %>%
        slice(order %>% first)
}

cppFunction('IntegerVector nmatch(CharacterVector value,CharacterVector id,IntegerVector index){
                        int nvalue = value.size();
            int nid = id.size();
            int completed = 0;
            IntegerVector match_count(nvalue,0);
            IntegerVector out(nvalue,IntegerVector::get_na());

            for (int i = 0; i < nid; ++i){
              for (int j = 0; j < nvalue; ++j){
                if (value[j] == id[i]){
                  match_count[j] = match_count[j] + 1;
                  if (match_count[j] == index[j]){
                    out[j] = i + 1;
                    completed++;
                  }
                }
              }
              if (completed == nvalue){
                break;
              }
            }
            return out;
                        }')

时间结果如下：

> #One with all matches relatively early
> set.seed(123)
> value <- c("a","b", "c")
> index <- c(150,50,500)
> id <-  sample(letters[1:5],10000,replace = TRUE)
> dti <- data.table(id = value,index = index)
> test = data_frame(value = value, order = index)
> original = data_frame(value =  id)
> 
> microbenchmark(nmatch(value = value, id = id,index = index),
+                            foo_mapply(value = value,id = id,index = index),
+                            foo_dt(dti = dti,id = id),
+                            foo_dplyr(test = test,original = original))
Unit: microseconds
                                              expr      min        lq      mean    median        uq      max neval  cld
     nmatch(value = value, id = id, index = index)  118.326  121.9060  124.2930  122.8535  124.5040  167.713   100 a   
 foo_mapply(value = value, id = id, index = index)  863.281  873.1505  949.8326  878.8535  896.7795 2119.411   100  b  
                        foo_dt(dti = dti, id = id) 1860.678 1927.0990 2038.5965 1985.2720 2082.7900 3761.116   100   c 
       foo_dplyr(test = test, original = original) 2862.143 2943.7280 3175.9202 2986.2385 3121.7685 4502.976   100    d

> #One with a match that forces us nearer the end of the list
> set.seed(123)
> value <- c("a","b", "c")
> index <- c(150,50,2000)
> id <-  sample(letters[1:5],10000,replace = TRUE)
> dti <- data.table(id = value,index = index)
> test = data_frame(value = value, order = index)
> original = data_frame(value =  id)
> 
> microbenchmark(nmatch(value = value, id = id,index = index),
+                            foo_mapply(value = value,id = id,index = index),
+                            foo_dt(dti = dti,id = id),
+                            foo_dplyr(test = test,original = original))
Unit: microseconds
                                              expr      min        lq      mean    median        uq       max neval cld
     nmatch(value = value, id = id, index = index)  469.208  473.4735  481.0698  475.1040  487.7145   560.031   100 a  
 foo_mapply(value = value, id = id, index = index)  861.797  872.6845  949.6749  882.5335  903.1255  2091.864   100 a  
                        foo_dt(dti = dti, id = id) 1821.554 1924.5690 2022.2231 1977.5970 2082.6035  3300.399   100  b 
       foo_dplyr(test = test, original = original) 2875.626 2945.7560 3681.2624 2995.7900 3100.3235 53508.339   100   c

Answer 7

有了这个设置

set.seed(123)
id <-  sample(letters[1:5], 10000, replace = TRUE)
value <- c("a", "b", "c")
index <- c(150, 50, 500)

索引然后拆分id向量

index_by_id <- split(seq_along(id), id)

将值与id_by_value的条目id_by_value

value_idx <- match(value, names(index_by_id))

选择每个匹配的第i个元素

mapply(`[`, index_by_id[value_idx], index)

并作为一个功能：

f1 <- function(id, value, index) {
    index_by_id <- split(seq_along(id), id)
    value_idx <- match(value, names(index_by_id))
    mapply(`[`, index_by_id[value_idx], index)
}

当value很长但有几个等级时，这会很快，例如，

f0 <- function(id, value, index)
    mapply(function(target, nth) which(id==target)[nth], value, index)

viq <- function(id, value, index) {
    dti <- data.table(id = value, index = index)
    data.table(id)[dti, .I[index], by = .EACHI, on = "id"]$V1
}

和

> value <- rep(value, 100)
> identical(f0(id, value, index), f1(id, value, index))
[1] TRUE
> all.equal(f0(id, value, index), viq(id, value, index),
+           check.attributes=FALSE)
[1] TRUE
> microbenchmark(f0(id, value, index), f1(id, value, index),
+                viq(id, value, index))
Unit: milliseconds
                  expr       min        lq      mean    median        uq
  f0(id, value, index) 53.166878 54.909566 56.917717 55.336116 56.503741
  f1(id, value, index)  1.682265  1.716843  1.883576  1.755070  1.831189
 viq(id, value, index)  4.304148  4.381708  4.667590  4.656087  4.757184
       max neval
 99.621742   100
  3.291769   100
  6.590130   100

Answer 8

@ 42-答案的变体

mapply(
  function(value, index) which(value == id)[index], 
  value = value_index_query$value, 
  index = value_index_query$index
)