我需要用组中以前的非NaN值替换NaN值。
这里是一个例子:
+-------+------------+-------+
| ts_id | date | value |
+-------+------------+-------+
| 2 | 01/10/2014 | 18 |
| 2 | 01/11/2014 | 15 |
| 2 | 01/12/2014 | NaN |
| 2 | 01/01/2015 | NaN |
| 2 | 01/02/2015 | NaN |
| 3 | 01/03/2015 | 19 |
| 3 | 01/04/2015 | 20 |
| 3 | 01/10/2015 | 12 |
| 3 | 01/11/2015 | 17 |
| 3 | 01/12/2015 | NaN |
| 3 | 01/01/2016 | NaN |
| 3 | 01/08/2016 | 7 |
| 3 | 01/09/2016 | NaN |
| 3 | 01/10/2016 | NaN |
| 3 | 01/11/2016 | NaN |
| 3 | 01/12/2016 | NaN |
| 3 | 01/01/2017 | NaN |
+-------+------------+-------+
数据:
data <- structure(list(ts_id = c(2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3), date = structure(c(16344, 16375, 16405, 16436,
16467, 16495, 16526, 16709, 16740, 16770, 16801, 17014, 17045,
17075, 17106, 17136, 17167), class = "Date"), value = c(18, 15,
NaN, NaN, NaN, 19, 20, 12, 17, NaN, NaN, 7, NaN, NaN, NaN, NaN,
NaN)), row.names = c(NA, -17L), vars = "ts_id", drop = TRUE, indices = list(
0:16), group_sizes = 17L, biggest_group_size = 17L, labels = structure(list(
ts_id = 3L), row.names = c(NA, -1L), class = "data.frame", vars = "ts_id", drop = TRUE), class = "data.frame")
在每个组中(由ts_id标识),我可以在任何给定日期获得NaN值。我需要用最新的非NaN值替换每个NaN。
结果应如下所示:
+-------+------------+-------+
| ts_id | date | value |
+-------+------------+-------+
| 2 | 01/10/2014 | 18 |
| 2 | 01/11/2014 | 15 |
| 2 | 01/12/2014 | 15 |
| 2 | 01/01/2015 | 15 |
| 2 | 01/02/2015 | 15 |
| 3 | 01/03/2015 | 19 |
| 3 | 01/04/2015 | 20 |
| 3 | 01/10/2015 | 12 |
| 3 | 01/11/2015 | 17 |
| 3 | 01/12/2015 | 17 |
| 3 | 01/01/2016 | 17 |
| 3 | 01/08/2016 | 7 |
| 3 | 01/09/2016 | 7 |
| 3 | 01/10/2016 | 7 |
| 3 | 01/11/2016 | 7 |
| 3 | 01/12/2016 | 7 |
| 3 | 01/01/2017 | 7 |
+-------+------------+-------+
谢谢。
答案 0 :(得分:4)
您可以使用此:
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答案 1 :(得分:2)
遵循与na.locf
相同的逻辑,但是将其保留在'verse中,我们可以做到,
library(tidyverse)
data %>%
group_by(ts_id) %>%
mutate(value = replace(value, is.nan(value), NA)) %>%
fill(value)