2

我正在浏览 padr 的文档：

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

稍微更改小插图示例以zoo::na.approx()在数据上使用线性插值 ( ) 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()

结果是：

# 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

现在我想线性插值 42 到 88。我认为最好的方法是使用zoo::na.approx()inside padr::fill_by_function()：

emergency %>% 
 filter(title == 'EMS: DEHYDRATION') %>% 
 thicken(interval = 'day') %>% 
 group_by(time_stamp_day) %>% 
 summarise(nr = n() + as.integer(runif(1, 1, 99)) ) %>% 
 pad() %>% 
 fill_by_function(nr, na.approx)

但我收到以下错误：

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

关于如何开始解决这个问题的任何想法？

4

1 回答 1

3

你只需mutate要这样做na.approx：

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
于 2017-04-10T22:20:36.283 回答