我正在查看padr的文档:
https://cran.r-project.org/web/packages/padr/vignettes/padr.html。
稍作修改,对数据进行线性插值(zoo::na.approx()),会产生一个错误:
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()在以下方面的成果:
# 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现在,,我要插值42到88线性。我认为实现这一目标的最好方法是在padr::fill_by_function()中使用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)但我得到了以下错误:
Error in inds[i] <- which(colnames_x == as.character(cols[[i]])) :
replacement has length zero有什么办法开始解决这个问题吗?
发布于 2017-04-10 22:20:36
您只需要mutate来执行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 NAhttps://stackoverflow.com/questions/43332417
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