我可以合并 dplyr mutate_at & mutate_if 语句吗？

Question

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

我想插入数据 I（但不是外推），并删除列 a 和 b 均为 NA 的列。换句话说，我想删除所有我无法插入的列；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

有两个选项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))

和

df%>% 
  group_by(Country)%>% 
  mutate_if(is.numeric,~if(all(is.na(.x))) NA else na.fill(.x,"extend"))

是否可以结合这些代码，做这样的事情：

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))

期望的输出：

   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

Answer 1

我知道这并没有直接回答如何结合 mutate_if 和 mutate_at 的问题，但这解决了您的一般问题：

我首先去掉所有a和b都缺失的国家，然后为每个国家确定最小和最大年份，这是不缺失的。过滤完这些后，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))

Answer 2

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.

Answer 3

这是另外两种使用 filter 和 slice 的方法。第一种方法应该最接近 OP 的要求：

library(dplyr)
library(zoo)

df %>%
  group_by(country) %>%
  mutate_if(is.numeric, na.approx, na.rm = FALSE) %>%
  filter(!is.na(a|b))

或 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"))

Answer 4

不幸的是，@kath onyl 的解决方案在给定示例中有效，但如果只有一列包含数据，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

同样不幸的是，OPs 问题的答案是否定的，你不能混合使用 mutate_at 和 mutate_if（没有允许你指定 .predicate 和 .vars 的函数）

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 的解决方案：

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(.)))

Answer 5

dplyr 0.8.3 灵感来自：

community.rstudio.com: How to mutate_at/mutate_if multiple columns using condition on other column outside .vars ?

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

新列总计：

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

我可以合并 dplyr mutate_at & mutate_if 语句吗？

Can I combine a dplyr mutate_at & mutate_if statement?

interpolation

r

dplyr