基准

Question

假设我有矢量z

z<-c(6,7,10,11,12,13,17,20,31,32,33,40,56,57,58,59)

我想要一些输出，只要有连续系列就会存储值作为系列计数，否则如果没有系列则简单计算该数字

 [1]  2 4 1 1 3 1 4

这里

2表示6,7

4表示10,11,12,13

1 for 17

1 for 20

3表示31,32,33

1 for 40

4表示56,57,58,59

希望你理解这个问题。我已经在stackoverflow上看到了一个例子，其中一个可以计算总数没有出现类似数字，就像没有。在问题中1和2的，但没有发现任何类似的东西。请提供解决方案

Akrun和bgoltst为我的帖子提供了准确的答案。现在我想在没有使用循环的预构建函数的情况下解决它。任何想法的人？

Answer 1

尝试

 unname(tapply(z, cumsum(c(TRUE, diff(z)!=1)), FUN= length))
 #[1] 2 4 1 1 3 1 4

或者紧凑版本

lengths(split(z, cumsum(c(0,diff(z)!=1))))

或者我们可以做到

tabulate(cumsum(c(TRUE,z[-1]-z[-length(z)] !=1)))

Answer 2

rle(c(0,cumsum(diff(z)!=1)))$lengths;
## [1] 2 4 1 1 3 1 4

基准

library(microbenchmark);

bgoldst <- function(z) rle(c(0,cumsum(diff(z)!=1)))$lengths;
akrun1 <- function(z) unname(tapply(z,cumsum(c(TRUE,diff(z)!=1)),length));
akrun2 <- function(z) unname(lengths(split(z, cumsum(c(0,diff(z)!=1)))));
akrun3 <- function(z) tabulate(cumsum(c(TRUE,z[-1L]-z[-length(z)]!=1)));
loop <- function(z) { res <- integer(); if (length(z)==0L) return(res); ri <- 1L; res[ri] <- 1L; for (zi in seq(2L,len=length(z)-1L)) if (z[zi]==z[zi-1L]+1L) res[ri] <- res[ri]+1L else { ri <- ri+1L; res[ri] <- 1L; }; res; };

expected <- c(2L,4L,1L,1L,3L,1L,4L);
identical(expected,bgoldst(z));
## [1] TRUE
identical(expected,structure(akrun1(z),dim=NULL));
## [1] TRUE
identical(expected,akrun2(z));
## [1] TRUE
identical(expected,akrun3(z));
## [1] TRUE
identical(expected,loop(z));
## [1] TRUE

microbenchmark(bgoldst(z),akrun1(z),akrun2(z),akrun3(z),loop(z));
## Unit: microseconds
##        expr     min       lq      mean   median       uq     max neval
##  bgoldst(z)  29.081  35.9240  41.52996  40.2000  43.6215 112.901   100
##   akrun1(z) 139.416 152.2450 163.97971 161.6535 169.7790 301.068   100
##   akrun2(z)  94.940 103.0640 110.23655 107.7685 116.3220 168.924   100
##   akrun3(z)   4.277   6.4150   7.37772   7.6980   8.1260  18.817   100
##     loop(z)  42.338  50.4635  58.54198  54.7400  64.3625 136.422   100

比例测试：

set.seed(1L);
N <- 1e5L; z <- sort(sample(seq(1L,N*3L),N));
expected <- bgoldst(z);
identical(expected,structure(akrun1(z),dim=NULL));
## [1] TRUE
identical(expected,akrun2(z));
## [1] TRUE
identical(expected,akrun3(z));
## [1] TRUE
identical(expected,loop(z));
## [1] TRUE

microbenchmark(bgoldst(z),akrun1(z),akrun2(z),akrun3(z),loop(z),times=10L);
## Unit: milliseconds
##        expr         min          lq        mean      median          uq        max neval
##  bgoldst(z)    7.260254    8.391395   11.045106    9.945911   11.704845   17.99818    10
##   akrun1(z)  193.259087  198.464899  215.520530  203.924951  220.394912  300.59812    10
##   akrun2(z)  217.264925  228.902627  263.216340  250.776189  285.028577  400.09614    10
##   akrun3(z)    2.322153    3.299338    6.076256    3.928843    7.569875   18.12305    10
##     loop(z) 2392.752491 2463.121401 2545.236552 2494.690056 2527.570319 2882.29511    10

统计数字

2 个答案:

基准