如何在dplyr中使用带有na.Abrox（）[线性插值]的fill_by_函数（）

我正在浏览padr的文档：

https://cran.r-project.org/web/packages/padr/vignettes/padr.html

稍微更改vignette示例以对数据使用线性插值（

zoo:：na.approx（）

）会产生错误：

library(tidyverse)
library(padr)
library(zoo)

set.seed(123)

emergency %>% 
  filter(title == 'EMS: DEHYDRATION') %>% 
  thicken(interval = 'day') %>% 
  group_by(time_stamp_day) %>% 
  summarise(nr = n() + as.integer(runif(1, 1, 999)) ) %>% 
  pad()

结果：

# A tibble: 307 × 2
   time_stamp_day    nr
           <date> <int>
1      2015-12-12    79
2      2015-12-13    42
3      2015-12-14    NA
4      2015-12-15    NA
5      2015-12-16    NA
6      2015-12-17    NA
7      2015-12-18    88
8      2015-12-19    NA
9      2015-12-20    NA
10     2015-12-21    NA
# ... with 297 more rows

但我得到了以下错误：

Error in inds[i] <- which(colnames_x == as.character(cols[[i]])) : 
  replacement has length zero

---

IND中的错误[i]您只需
变异
即可执行
近似值
：

library(tibble);library(zoo)
emergency <- as_tibble(read.table(text="time_stamp_day    nr
1      2015-12-12    79
2      2015-12-13    42
3      2015-12-14    NA
4      2015-12-15    NA
5      2015-12-16    NA
6      2015-12-17    NA
7      2015-12-18    88
8      2015-12-19    NA
9      2015-12-20    NA
10     2015-12-21    NA",header=TRUE,stringsAsFactors=FALSE))

emergency %>% mutate(nr=na.approx(nr,na.rm =FALSE))

# A tibble: 10 × 2
   time_stamp_day    nr
            <chr> <dbl>
1      2015-12-12  79.0
2      2015-12-13  42.0
3      2015-12-14  51.2
4      2015-12-15  60.4
5      2015-12-16  69.6
6      2015-12-17  78.8
7      2015-12-18  88.0
8      2015-12-19    NA
9      2015-12-20    NA
10     2015-12-21    NA

库（tibble）；图书馆（动物园）
紧急变异百分比（nr=na.近似值（nr，na.rm=FALSE））
#一个tibble:10×2
时间戳日编号
1      2015-12-12  79.0
2      2015-12-13  42.0
3      2015-12-14  51.2
4      2015-12-15  60.4
5      2015-12-16  69.6
6      2015-12-17  78.8
7      2015-12-18  88.0
8 2015-12-19北美
9 2015-12-20北美
10 2015-12-21北美
如果有人能添加#padr标签，我将不胜感激。谢谢@P——这当然有道理。我正在创建一个新变量，我应该使用mutate。
library(tibble);library(zoo)
emergency <- as_tibble(read.table(text="time_stamp_day    nr
1      2015-12-12    79
2      2015-12-13    42
3      2015-12-14    NA
4      2015-12-15    NA
5      2015-12-16    NA
6      2015-12-17    NA
7      2015-12-18    88
8      2015-12-19    NA
9      2015-12-20    NA
10     2015-12-21    NA",header=TRUE,stringsAsFactors=FALSE))

emergency %>% mutate(nr=na.approx(nr,na.rm =FALSE))

# A tibble: 10 × 2
   time_stamp_day    nr
            <chr> <dbl>
1      2015-12-12  79.0
2      2015-12-13  42.0
3      2015-12-14  51.2
4      2015-12-15  60.4
5      2015-12-16  69.6
6      2015-12-17  78.8
7      2015-12-18  88.0
8      2015-12-19    NA
9      2015-12-20    NA
10     2015-12-21    NA