替换所有以独立于类型的变量属性为条件的负值
Replacing all negative values conditional on the attributes of the variable independent of type
我有一个非常大的混合数据集（字符变量、数值变量、因子），其中负值通常表示缺失值，请参阅 Scales，但并非总是如此，请参阅 Profit :
Country Ccode Year Profit Scale ID Happiness_d Power_d ID_d
<chr> <fcr> <dbl> <dbl> <labelled> <dbl> <dbl> <dbl> <dbl>
1 France FR 2000 1000 NA 1 40000. 160000. 1.67
2 France FR 2001 -1200 1 1 80000. 320000. 1.67
3 France FR 2000 1400 0 2 40000. 160000. 1.67
4 France FR 2001 1600 3 2 80000. 320000. 1.67
5 UK UK 2000 -1000 -9 3 40000. 160000. 1.67
6 UK UK 2001 1000 2 3 80000. 320000. 1.67
7 UK UK 2000 1000 4 4 40000. 160000. 1.67
8 UK UK 2001 1000 0 4 80000. 320000. 1.67
我想使用 NA 替换所有负值：
df[df< 0] <- NA
问题是，虽然它旨在删除代表 NA 的负值，例如 Scale，但它会在示例数据集中删除 Profit 中的负数，这显然不是 NA .
因此，我想让结果以变量的范围为条件。 Scale变量的结构如下：
Class 'labelled' atomic [1:135894] NA NA 2 NA NA NA NA NA NA NA ...
..- attr(*, "label")= chr "Do You Use Technology Licensed From A Foreign-Owned Company?"
..- attr(*, "format.stata")= chr "%24.0g"
..- attr(*, "labels")= Named num [1:3] -9 1 2
.. ..- attr(*, "names")= chr [1:3] "Don't Know (Spontaneous)" "Yes" "No"
> names(New_Comprehensive_June_25_2018$e6)
我发现使用 haven 库 link 您可以从中得出因子水平；
..- attr(*, "labels")= Named num [1:3] -9 1 2
与 get_values()。
get_values(df$Scale)
[1] -9 1 2
是否可以让解决方案只去除这些负面因素而不是其他负面价值？
..- attr(*, "labels")= Named num [1:3] -9 1 2
明确地说，所需的输出为：
Country Ccode Year Profit Scale ID Happiness_d Power_d ID_d
<chr> <fcr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 France FR 2000 1000 NA 1 40000. 160000. 1.67
2 France FR 2001 -1200 1 1 80000. 320000. 1.67
3 France FR 2000 1400 0 2 40000. 160000. 1.67
4 France FR 2001 1600 3 2 80000. 320000. 1.67
5 UK UK 2000 -1000 **NA** 3 40000. 160000. 1.67
6 UK UK 2001 1000 2 3 80000. 320000. 1.67
7 UK UK 2000 1000 4 4 40000. 160000. 1.67
8 UK UK 2001 1000 0 4 80000. 320000. 1.67
dput 示例（请注意变量 Scale 实际上并不存在：
h7a = structure(c(1, -9, 2, 3, 1, 3, -9, 2, 3, 1, 2, 1, 3,
3, 2, 2, 1, 2, 2, 1, 2, -9, 1, 4, 3, 3, 1, 1, 1, 1, 3, 4,
3, 1, 2, 2, 1, 2, 1, NA, 2, 1, 2, 4, 3, 1, 3, 4, 4, 3, 2,
4, 1, 1, 2, 3, 2, 2, 2, 2, 1, 2, 1, 3, 4, 3, 1, 3, 1, 2,
3, 3, 3, 1, 1, 4, -9, 4, 3, 1, 2, 3, 1, -9, 1, 4, 1, 3, 1,
-9, 1, 1, 1, 1, 2, 3, 1, 3, 1, 2, 1, 2, 3, 4, 3, 3, 2, 4,
3, 3, 1, -9, 1, -7, 3, 1, 1, 2, 1, 2, -7, 2, 3, 1, 3, -7,
3, 4, 3, 2, 3, NA, 3, 3, 3, 1, 1, 2, 2, -9, 3, 1, 1, 2, 1,
1, -9, -9, -9, 2, -9, 1, 2, 3, 2, 3, 3, 3, 3, 1, 2, -9, 4,
3, 3, 1, 2, 2, 4, 4, 4, 3, 2, 1, 2, 2, 2, 2, 2, 3, 2, 2,
-9, 4, 4, 4, 2, 1, -7, 2, 2, 1, 1, 2, 1, 2, 2, 4, 2, 3, -7,
3, 3, 3, 4, 2, 4, 2, NA, 1, 3, 1, 2, 3, 4, 3, -9, 3, 3, 4,
3, 2, 4, 1, 3, 1, 3, 4, 3, 1, 3, 3, 3, NA, 1, 3, 3, -7, 1,
1, 3, 2, 1, 4), label = "The Court System Is Fair, Impartial And Uncorrupted", format.stata = "%24.0g", class = "labelled", labels = structure(c(NA,
NA, 1, 2, 3, 4), .Names = c("Don't Know (Spontaneous)", "Does Not Apply",
"Strongly disagree", "Tend to disagree", "Tend to agree",
"Strongly agree"))),
这是一个您可以应用于您的数据集的简单示例。
# example data
df = data.frame(a = c("A","A","B"),
x = c(1,2,3),
y = c(NA,3,-7),
z = c(200,300,-400))
library(dplyr)
df %>% mutate_if(is.numeric, ~ifelse(between(min(., na.rm = T), -9, -1) & .<0, NA, .))
# a x y z
# 1 A 1 NA 200
# 2 A 2 3 300
# 3 B 3 NA -400
您可以更新 (mutate) 仅当该列为数字且该列的最小值介于 -9 和 -1 之间时。更新是用 NA.
替换负值
这假设您只有整数值。如果没有，您可以使用 between(..., -9, 0).
Base-R 解决方案：
# Find negative value from 3rd column onwards, replace it with NA
# and bind with Country,Ccode and Profit columns.
cbind(df[,c(1,2,4)],do.call(cbind, lapply(df[,-c(1,2,4)], function(x) ifelse(x<0,NA,x))))
输出：
Country Ccode Profit Year Scale ID Happiness_d Power_d ID_d
1 France FR 1000 2000 NA 1 40000 160000 1.67
2 France FR -1200 2001 1 1 80000 320000 1.67
3 France FR 1400 2000 0 2 40000 160000 1.67
4 France FR 1600 2001 3 2 80000 320000 1.67
5 UK UK -1000 2000 NA 3 40000 160000 1.67
6 UK UK 1000 2001 2 3 80000 320000 1.67
7 UK UK 1000 2000 4 4 40000 160000 1.67
8 UK UK 1000 2001 0 4 80000 320000 1.67
我有一个非常大的混合数据集（字符变量、数值变量、因子），其中负值通常表示缺失值，请参阅 Scales，但并非总是如此，请参阅 Profit :
Country Ccode Year Profit Scale ID Happiness_d Power_d ID_d
<chr> <fcr> <dbl> <dbl> <labelled> <dbl> <dbl> <dbl> <dbl>
1 France FR 2000 1000 NA 1 40000. 160000. 1.67
2 France FR 2001 -1200 1 1 80000. 320000. 1.67
3 France FR 2000 1400 0 2 40000. 160000. 1.67
4 France FR 2001 1600 3 2 80000. 320000. 1.67
5 UK UK 2000 -1000 -9 3 40000. 160000. 1.67
6 UK UK 2001 1000 2 3 80000. 320000. 1.67
7 UK UK 2000 1000 4 4 40000. 160000. 1.67
8 UK UK 2001 1000 0 4 80000. 320000. 1.67
我想使用 NA 替换所有负值：
df[df< 0] <- NA
问题是，虽然它旨在删除代表 NA 的负值，例如 Scale，但它会在示例数据集中删除 Profit 中的负数，这显然不是 NA .
因此，我想让结果以变量的范围为条件。 Scale变量的结构如下：
Class 'labelled' atomic [1:135894] NA NA 2 NA NA NA NA NA NA NA ...
..- attr(*, "label")= chr "Do You Use Technology Licensed From A Foreign-Owned Company?"
..- attr(*, "format.stata")= chr "%24.0g"
..- attr(*, "labels")= Named num [1:3] -9 1 2
.. ..- attr(*, "names")= chr [1:3] "Don't Know (Spontaneous)" "Yes" "No"
> names(New_Comprehensive_June_25_2018$e6)
我发现使用 haven 库 link 您可以从中得出因子水平；
..- attr(*, "labels")= Named num [1:3] -9 1 2
与 get_values()。
get_values(df$Scale)
[1] -9 1 2
是否可以让解决方案只去除这些负面因素而不是其他负面价值？
..- attr(*, "labels")= Named num [1:3] -9 1 2
明确地说，所需的输出为：
Country Ccode Year Profit Scale ID Happiness_d Power_d ID_d
<chr> <fcr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 France FR 2000 1000 NA 1 40000. 160000. 1.67
2 France FR 2001 -1200 1 1 80000. 320000. 1.67
3 France FR 2000 1400 0 2 40000. 160000. 1.67
4 France FR 2001 1600 3 2 80000. 320000. 1.67
5 UK UK 2000 -1000 **NA** 3 40000. 160000. 1.67
6 UK UK 2001 1000 2 3 80000. 320000. 1.67
7 UK UK 2000 1000 4 4 40000. 160000. 1.67
8 UK UK 2001 1000 0 4 80000. 320000. 1.67
dput 示例（请注意变量 Scale 实际上并不存在：
h7a = structure(c(1, -9, 2, 3, 1, 3, -9, 2, 3, 1, 2, 1, 3,
3, 2, 2, 1, 2, 2, 1, 2, -9, 1, 4, 3, 3, 1, 1, 1, 1, 3, 4,
3, 1, 2, 2, 1, 2, 1, NA, 2, 1, 2, 4, 3, 1, 3, 4, 4, 3, 2,
4, 1, 1, 2, 3, 2, 2, 2, 2, 1, 2, 1, 3, 4, 3, 1, 3, 1, 2,
3, 3, 3, 1, 1, 4, -9, 4, 3, 1, 2, 3, 1, -9, 1, 4, 1, 3, 1,
-9, 1, 1, 1, 1, 2, 3, 1, 3, 1, 2, 1, 2, 3, 4, 3, 3, 2, 4,
3, 3, 1, -9, 1, -7, 3, 1, 1, 2, 1, 2, -7, 2, 3, 1, 3, -7,
3, 4, 3, 2, 3, NA, 3, 3, 3, 1, 1, 2, 2, -9, 3, 1, 1, 2, 1,
1, -9, -9, -9, 2, -9, 1, 2, 3, 2, 3, 3, 3, 3, 1, 2, -9, 4,
3, 3, 1, 2, 2, 4, 4, 4, 3, 2, 1, 2, 2, 2, 2, 2, 3, 2, 2,
-9, 4, 4, 4, 2, 1, -7, 2, 2, 1, 1, 2, 1, 2, 2, 4, 2, 3, -7,
3, 3, 3, 4, 2, 4, 2, NA, 1, 3, 1, 2, 3, 4, 3, -9, 3, 3, 4,
3, 2, 4, 1, 3, 1, 3, 4, 3, 1, 3, 3, 3, NA, 1, 3, 3, -7, 1,
1, 3, 2, 1, 4), label = "The Court System Is Fair, Impartial And Uncorrupted", format.stata = "%24.0g", class = "labelled", labels = structure(c(NA,
NA, 1, 2, 3, 4), .Names = c("Don't Know (Spontaneous)", "Does Not Apply",
"Strongly disagree", "Tend to disagree", "Tend to agree",
"Strongly agree"))),
这是一个您可以应用于您的数据集的简单示例。
# example data
df = data.frame(a = c("A","A","B"),
x = c(1,2,3),
y = c(NA,3,-7),
z = c(200,300,-400))
library(dplyr)
df %>% mutate_if(is.numeric, ~ifelse(between(min(., na.rm = T), -9, -1) & .<0, NA, .))
# a x y z
# 1 A 1 NA 200
# 2 A 2 3 300
# 3 B 3 NA -400
您可以更新 (mutate) 仅当该列为数字且该列的最小值介于 -9 和 -1 之间时。更新是用 NA.
这假设您只有整数值。如果没有，您可以使用 between(..., -9, 0).
Base-R 解决方案：
# Find negative value from 3rd column onwards, replace it with NA
# and bind with Country,Ccode and Profit columns.
cbind(df[,c(1,2,4)],do.call(cbind, lapply(df[,-c(1,2,4)], function(x) ifelse(x<0,NA,x))))
输出：
Country Ccode Profit Year Scale ID Happiness_d Power_d ID_d
1 France FR 1000 2000 NA 1 40000 160000 1.67
2 France FR -1200 2001 1 1 80000 320000 1.67
3 France FR 1400 2000 0 2 40000 160000 1.67
4 France FR 1600 2001 3 2 80000 320000 1.67
5 UK UK -1000 2000 NA 3 40000 160000 1.67
6 UK UK 1000 2001 2 3 80000 320000 1.67
7 UK UK 1000 2000 4 4 40000 160000 1.67
8 UK UK 1000 2001 0 4 80000 320000 1.67