Question

我想通过对同一组的非NA值进行采样来填补每个组的NA值。

这是最接近我想要使用!is.na() Ignoring values or NAs in the sample function实现的目标。


> dput(data)
structure(list(len = c(NA, 45447.4157838775, 161037.71538108, 
78147.8550470324, 7193.48815617057, 1571.95459212405, 18191.381972185, 
20366.2132412031, 10014.987524596, 1403.72511829297, 5651.17842991513, 
6848.03271105711, 8043.32937011393, 8926.65133418451, 5808.44456603825, 
2208.14264175252, 1797.4936747033, 5325.76651327694, 2660.66730207955, 
5844.07912541444, 3956.40473896271, 959.873314407621, 3294.01472360025, 
5221.94864001864, 3781.51913857335, 7811.83819953768, 3387.20323328623, 
5514.92099458441, 5792.54371531706, 5643.98385143961, 15478.916809379, 
8401.66533205217, 7046.25074819247, 2734.73639821402, NA, 62332.3343404513, 
NA, 46563.1214718113, 25590.4020105238, 13015.3682275862, 4984.80432801441, 
NA), point = c(NA, 0, 8, 5, 2, 0, 9, 0, 0, 0, 3, 1, 0, 6, 1, 
1, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, NA, 
10, NA, 19, 6, 5, 0, NA), country = structure(c(1L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 
3L, 2L, 2L, 2L, 2L, 1L), .Label = c("WCY_____ES", "WCY_____FR", 
"WCY_____IT"), class = "factor"), group = c(1L, 2L, 2L, 2L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
4L, 4L, 4L, 4L, 4L)), row.names = c(NA, -42L), class = "data.frame")

library(dplyr)

data1 <- data %>% 
  group_by(group) %>%
  mutate(nulen = if_else(country == 'WCY_____FR', len, sample(len[!is.na(len)], 1, TRUE)),
         nupoint = if_else(country == 'WCY_____FR', point, sample(point[!is.na(point)], 1, TRUE)))

但是我得到Error in sample.int(length(x), size, replace, prob) : invalid first argument

已知分布和缺口填充之间应该没有显着差异。如果没有可从同一组中采样的值（其他值为NA或“ group”中只有一行），则应从整个数据集中获取采样。任何包装都可以。

Answer 1

这是个主意，

dd %>%
    mutate(len1 = replace(len, is.na(len), sample(len[!is.na(len)], 1, TRUE)),
           point1 = replace(point, is.na(point), sample(point[!is.na(point)], 1, TRUE))) %>%
    group_by(group) %>% 
    mutate(nulen = ifelse(all(is.na(len)) & country == 'WCY_____FR', len1, 
                          ifelse(is.na(len) & country == 'WCY_____FR', sample(len[!is.na(len)], 1, TRUE), len)))

给出，

len point country    group    len1 point1   nulen
     <dbl> <dbl> <fct>      <int>   <dbl>  <dbl>   <dbl>
 1     NA     NA WCY_____ES     1   1572.      0     NA 
 2  45447.     0 WCY_____FR     2  45447.      0  45447.
 3 161038.     8 WCY_____FR     2 161038.      8 161038.
 4  78148.     5 WCY_____FR     2  78148.      5  78148.
 5   7193.     2 WCY_____FR     3   7193.      2   7193.
 6   1572.     0 WCY_____FR     3   1572.      0   1572.
 7  18191.     9 WCY_____FR     3  18191.      9  18191.
 8  20366.     0 WCY_____FR     3  20366.      0  20366.
 9  10015.     0 WCY_____FR     3  10015.      0  10015.
10   1404.     0 WCY_____FR     3   1404.      0   1404.
# ... with 32 more rows

point也可以这样做。

dplyr group_by和sample，同时忽略NA

1 个答案: