如果R中有0,如何填写前面的数字？

Question

我有一串数字:

n1 = c(1, 1, 0, 6, 0, 0, 10, 10, 11, 12, 0, 0, 19, 23, 0, 0)

我需要用它前面的相应数字替换0来得到:

n2 = c(1, 1, 1, 6, 6, 6, 10, 10, 11, 12, 12, 12, 19, 23, 23, 23)

我如何从n1到n2？

提前致谢!

Answer 1

n2 <- n1[cummax(seq_along(n1) * (n1 != 0))]

Answer 2

试试na.locf()包zoo:

library(zoo)
n1 <- c(1, 1, 0, 6, 0, 0, 10, 10, 11, 12, 0, 0, 19, 23, 0, 0)
n1[n1 == 0] <- NA
na.locf(n1)
## [1]  1  1  1  6  6  6 10 10 11 12 12 12 19 23 23 23

此函数将每个函数替换NA为最新的非函数NA.这就是为什么我在应用它之前替换了所有0s NA.

这里有一个讨论的一个类似(但不相同)的问题.

编辑:如果n1最终由NAs 组成,请尝试例如

n1 <- c(1, 1, 0, 6, 0, 0, 10, NA, 11, 12, 0, 0, 19, NA, 0, 0)
wh_na <- which(is.na(n1))
n1[n1 == 0] <- NA
n2 <- na.locf(n1)
n2[wh_na] <- NA
n2
##  [1]  1  1  1  6  6  6 10 NA 11 12 12 12 19 NA 19 19

EDIT2:这种c(1,NA,0)退货方式c(1,NA,1).另外两个玩家给出了c(1,NA,NA).换句话说,这里我们用最后一个非缺失的非零值替换0.选择您喜欢的选项.

编辑3:受@ Thell的Rcpp解决方案的启发,我想添加另一个 - 这次使用"纯"R/C API.

library('inline')
sexp0 <- cfunction(signature(x="numeric"), "
   x = Rf_coerceVector(x, INTSXP); // will not work for factors
   R_len_t n = LENGTH(x);
   SEXP ret;
   PROTECT(ret = Rf_allocVector(INTSXP, n));
   int lval = NA_INTEGER;
   int* xin = INTEGER(x);
   int* rin = INTEGER(ret);
   for (R_len_t i=0; i<n; ++i, ++xin, ++rin) {
      if (*xin == 0)
         *rin = lval;
      else {
         lval = *xin;
         *rin = lval;
      }
   }
   UNPROTECT(1);
   return ret;
", language="C++")

在这种情况下,我们会得到c(1,NA,NA)的c(1,NA,0).一些基准:

library(microbenchmark)
set.seed(1L)
n1 <- sample(c(0:10), 1e6, TRUE)
microbenchmark(sexp0(n1), rollValue(n1), n1[cummax(seq_along(n1) * (n1 != 0))])
## Unit: milliseconds
##                                   expr       min        lq    median        uq       max neval
##                              sexp0(n1)  2.468588  2.494233  3.198711  4.216908  63.21236   100
##                          rollValue(n1)  8.151000  9.359731 10.603078 12.760594  75.88901   100
##  n1[cummax(seq_along(n1) * (n1 != 0))] 32.899420 36.956711 39.673726 45.419449 106.48180   100

Answer 3

这是一个解决方案data.table:

require(data.table) ## >= 1.9.2
idx = which(!n1 %in% 0L)
DT <- data.table(val=n1[idx], idx=idx)
setattr(DT, 'sorted', "idx")
n1 = DT[J(seq_along(n1)), roll=Inf]$val
#  [1]  1  1  1  6  6  6 10 10 11 12 12 12 19 23 23 23

更大数据的基准:

require(zoo)
require(data.table)

set.seed(1L)
n1 = sample(c(0:10), 1e6, TRUE)

## data.table
dt_fun <- function(n1) {
    idx = which(!n1 %in% 0L)
    DT <- data.table(val=n1[idx], idx=idx)
    setattr(DT, 'sorted', "idx")
    DT[J(seq_along(n1)), roll=Inf]$val
}

# na.locf from zoo - gagolews
zoo_fun <- function(n1) {
    wh_na <- which(is.na(n1))
    n1[n1 == 0] <- NA
    n2 <- na.locf(n1)
    n2[wh_na] <- NA
    n2
}

## rle - thelatemail
rle_fun <- function(n1) {
    r <- rle(n1)
    r$values[which(r$values==0)] <- r$values[which(r$values==0)-1]
    inverse.rle(r)
}

flodel_fun <- function(n1) n1[cummax(seq_along(n1) * (n1 != 0))]

require(microbenchmark)
microbenchmark(a1 <- dt_fun(n1), 
               a2 <- zoo_fun(n1), 
               a3 <- rle_fun(n1), 
               a4 <- flodel_fun(n1), times=10L)

这是基准测试结果:

# Unit: milliseconds
#                  expr       min        lq    median        uq       max neval
#      a1 <- dt_fun(n1) 155.49495 164.04133 199.39133 243.22995 289.80908    10
#     a2 <- zoo_fun(n1) 596.33039 632.07841 671.51439 682.85950 697.33500    10
#     a3 <- rle_fun(n1) 356.95103 377.61284 383.63109 406.79794 495.09942    10
#  a4 <- flodel_fun(n1)  51.52259  55.54499  56.20325  56.39517  60.15248    10

Answer 4

因为rle是一切的答案:

#make an example including an NA value
n1 <- c(1, 1, 0, 6, NA, 0, 10, 10, 11, 12, 0, 0, 19, 23, 0, 0)
r <- rle(n1)
r$values[which(r$values==0)] <- r$values[which(r$values==0)-1]
inverse.rle(r)
# [1]  1  1  1  6 NA NA 10 10 11 12 12 12 19 23 23 23

跳过NAs 的版本是:

n1 <- c(1, 1, 0, 6, NA, 0, 10, 10, 11, 12, 0, 0, 19, 23, 0, 0)
r <- rle(n1[!is.na(n1)])
r$values[which(r$values==0)] <- r$values[which(r$values==0)-1]
n1[!is.na(n1)] <- inverse.rle(r)
n1
# [1]  1  1  1  6 NA  6 10 10 11 12 12 12 19 23 23 23

Answer 5

不要忘记Rcpp的简单性和性能提升......

使用Arun的样本量我得到......

Unit: milliseconds
                                  expr       min        lq    median        uq      max neval
                         rollValue(n1)  3.998953  4.105954  5.803294  8.774286 36.52492   100
 n1[cummax(seq_along(n1) * (n1 != 0))] 17.634569 18.295344 20.698524 23.104847 74.72795   100

.cpp来源的文件很简单......

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::plugins("cpp11")]]

// [[Rcpp::export]]
NumericVector rollValue(const NumericVector v) {
  auto out = clone(v);
  auto tmp = v[0];
  for( auto & e : out) {
    if( e == 0 ) {
      e = tmp;
      continue;
    }
    tmp = e;
  }
  return out;
}