我有以下示例输出：
country country-year year a b
1 France France2000 2000 NA NA
2 France France2001 2001 1000 1000
3 France France2002 2002 NA NA
4 France France2003 2003 1600 2200
5 France France2004 2004 NA NA
6 UK UK2000 2000 1000 1000
7 UK UK2001 2001 NA NA
8 UK UK2002 2002 1000 1000
9 UK UK2003 2003 NA NA
10 UK UK2004 2004 NA NA
11 Germany UK2000 2000 NA NA
12 Germany UK2001 2001 NA NA
13 Germany UK2002 2002 NA NA
14 Germany UK2003 2003 NA NA
15 Germany UK2004 2004 NA NA
我想对数据I进行插值（但不作推断），并删除列a和b均为NA的列。换句话说，我想删除所有无法插值的列；在示例中：
1 France France2000 NA NA
5 France France2004 NA NA
9 UK UK2003 NA NA
10 UK UK2004 NA NA
11 Germany UK2000 NA NA
12 Germany UK2001 NA NA
13 Germany UK2002 NA NA
14 Germany UK2003 NA NA
15 Germany UK2004 NA NA
有两种选择几乎可以满足我的要求：
library(tidyverse)
library(zoo)
df %>%
group_by(country) %>%
mutate_at(vars(a:b),~na.fill(.x,c(NA, "extend", NA))) %>%
filter(!is.na(a) | !is.na(b))
AND
df%>%
group_by(Country)%>%
mutate_if(is.numeric,~if(all(is.na(.x))) NA else na.fill(.x,"extend"))
是否可以将这些代码组合起来，就像这样：
df <- df%>%
group_by(country)%>%
mutate_at(vars(a:b),~if(all(is.na(.x))) NA else(.x,c(NA, "extend", NA)))
filter(!is.na(df$a | df$a))
所需的输出：
country country-year a b
2 France France2001 1000 1000
3 France France2002 1300 1600
4 France France2003 1600 2200
6 UK UK2000 1000 1000
7 UK UK2001 0 0
8 UK UK2002 1000 1000
答案 0 :(得分：1)
我知道这并不能直接回答如何组合mutate_if和mutate_at的问题，但这可以解决您的一般问题：
我首先摆脱所有缺少a和b的国家/地区，然后为每个国家/地区确定不缺少的最小和最大Year。过滤掉这些之后，我使用na.fill。
library(dplyr)
library(readr)
library(zoo)
country_data %>%
mutate(Year = parse_number(`country-year`)) %>%
group_by(country) %>%
mutate(not_all_na = any(!(is.na(a) & is.na(b)))) %>%
filter(not_all_na) %>%
mutate(Year_min_not_na = min(Year[!(is.na(a) & is.na(b))]),
Year_max_not_na = max(Year[!(is.na(a) & is.na(b))])) %>%
filter(Year >= Year_min_not_na, Year <= Year_max_not_na) %>%
mutate_at(vars(a:b), ~na.fill(.x, "extend"))
# A tibble: 6 x 8
# Groups: country [2]
# country `country-year` a b Year not_all_na Year_min_not_na Year_max_not_na
# <fct> <fct> <dbl> <dbl> <dbl> <lgl> <dbl> <dbl>
# 1 France France2001 1000 1000 2001 TRUE 2001 2003
# 2 France France2002 1300 1600 2002 TRUE 2001 2003
# 3 France France2003 1600 2200 2003 TRUE 2001 2003
# 4 UK UK2000 1000 1000 2000 TRUE 2000 2002
# 5 UK UK2001 1000 1000 2001 TRUE 2000 2002
# 6 UK UK2002 1000 1000 2002 TRUE 2000 2002
数据
country_data <-
structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L),
.Label = c("France", "Germany", "UK"), class = "factor"),
country.year = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 6L, 7L, 8L, 9L, 10L),
.Label = c("France2000", "France2001", "France2002", "France2003",
"France2004", "UK2000", "UK2001", "UK2002", "UK2003", "UK2004"),
class = "factor"),
a = c(NA, 1000L, NA, 1600L, NA, 1000L, NA, 1000L, NA, NA, NA, NA, NA, NA, NA),
b = c(NA, 1000L, NA, 2200L, NA, 1000L, NA, 1000L, NA, NA, NA, NA, NA, NA, NA)),
class = "data.frame", row.names = c(NA, -15L))
答案 1 :(得分：1)
不幸的是，@ kath onyl的解决方案在给定的示例中有效，但如果仅一列包含数据，则失败，例如：
country country-year year a b
France France2000 2000 NA NA
France France2001 2001 1000 1000
France France2002 2002 NA NA
France France2003 2003 1600 2200
France France2004 2004 NA NA
UK UK2000 2000 1000 1000
UK UK2001 2001 NA NA
UK UK2002 2002 1000 1000
UK UK2003 2003 NA NA
UK UK2004 2004 NA NA
Germany UK2000 2000 NA NA
Germany UK2001 2001 NA 500
Germany UK2002 2002 NA NA
Germany UK2003 2003 NA 1100
Germany UK2004 2004 NA NA
不幸的是，OP问题的答案是否定的，您不能混合使用mutate_at和mutate_if（没有函数可以指定.predicate和.vars）
但是，您可以在mutate_at中使用的函数中使用预测函数。所以这是我使用包含预测函数的mutate_at解决方案：
df %>%
group_by(country) %>%
# Interpolate if at least two non-null values are present
mutate_at(vars(a,b), funs(if(sum(!is.na(.))<2) {NA_real_} else{approx(year, ., year)$y})) %>%
# keep only rows with original or interpolated values in either column a or b
filter_at(vars(a,b), any_vars(!is.na(.)))
答案 2 :(得分：0)
这是我的看法：
library(data.table)
library(tidyverse)
library(zoo)
df <- fread("
n country country-year a b
1 France France2000 NA NA
2 France France2001 1000 1000
3 France France2002 NA NA
4 France France2003 1600 2200
5 France France2004 NA NA
6 UK UK2000 1000 1000
7 UK UK2001 NA NA
8 UK UK2002 1000 1000
9 UK UK2003 NA NA
10 UK UK2004 NA NA
11 Germany UK2000 NA NA
12 Germany UK2001 NA NA
13 Germany UK2002 NA NA
14 Germany UK2003 NA NA
15 Germany UK2004 NA NA
") %>% select(-n)
# Clean data
df <- df %>%
mutate(year = str_extract_all(`country-year`, "[0-9]{4}$", simplify = T)) %>%
select(country, year, a, b)
# Remove all rows NA in a and b if there is no earlier
# or later row with value for a and b
# I hope this was what you meant with extrapolate :)
df <- df %>%
group_by(country) %>%
filter(year >= min(year[!is.na(a) | !is.na(b)]),
year <= max(year[!is.na(a) | !is.na(b)])) %>%
ungroup()
# Intrapolate
df %>%
mutate_at(vars(a:b), ~na.fill(., "extend"))
结果：
# A tibble: 6 x 4
country year a b
<chr> <chr> <dbl> <dbl>
1 France 2001 1000. 1000.
2 France 2002 1300. 1600.
3 France 2003 1600. 2200.
4 UK 2000 1000. 1000.
5 UK 2001 1000. 1000.
6 UK 2002 1000. 1000.
答案 3 :(得分：0)
这里还有另外两种使用filter和slice的方法。第一种方法应该最接近OP寻找的东西：
library(dplyr)
library(zoo)
df %>%
group_by(country) %>%
mutate_if(is.numeric, na.approx, na.rm = FALSE) %>%
filter(!is.na(a|b))
或使用slice：
df %>%
group_by(country) %>%
filter(any(!is.na(a|b))) %>%
slice(min(which(!is.na(a|b))):max(which(!is.na(a|b)))) %>%
mutate_if(is.numeric, na.approx)
结果：
# A tibble: 6 x 4
# Groups: country [2]
country country.year a b
<fct> <fct> <dbl> <dbl>
1 France France2001 1000 1000
2 France France2002 1300 1600
3 France France2003 1600 2200
4 UK UK2000 1000 1000
5 UK UK2001 1000 1000
6 UK UK2002 1000 1000
数据：
df <- structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 3L,
3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L), .Label = c("France", "Germany",
"UK"), class = "factor"), country.year = structure(c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 6L, 7L, 8L, 9L, 10L), .Label = c("France2000",
"France2001", "France2002", "France2003", "France2004", "UK2000",
"UK2001", "UK2002", "UK2003", "UK2004"), class = "factor"), a = c(NA,
1000L, NA, 1600L, NA, 1000L, NA, 1000L, NA, NA, NA, NA, NA, NA,
NA), b = c(NA, 1000L, NA, 2200L, NA, 1000L, NA, 1000L, NA, NA,
NA, NA, NA, NA, NA)), .Names = c("country", "country.year", "a",
"b"), class = "data.frame", row.names = c("1", "2", "3", "4",
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"))
答案 4 :(得分：0)
dplyr 0.8.3的灵感来自于：
library(dplyr)
(iris [1:3,]
%>% mutate_at(c("Petal.Width"),
list(~ifelse(Sepal.Width == 3.5,
.+10,
.+100)
)
)
)
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1 5.1 3.5 1.4 10.2 setosa
#> 2 4.9 3.0 1.4 100.2 setosa
#> 3 4.7 3.2 1.3 100.2 setosa
带有一个新的列toto：
library(dplyr)
(iris [1:3,]
%>% mutate_at(c("Petal.Width"),
list(toto=~ifelse(Sepal.Width == 3.5,
.+10,
.+100)
)
)
)
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species toto
#> 1 5.1 3.5 1.4 0.2 setosa 10.2
#> 2 4.9 3.0 1.4 0.2 setosa 100.2
#> 3 4.7 3.2 1.3 0.2 setosa 100.2
由reprex package（v0.2.1）于2019-07-30创建