基于其他列(在R中)的特定字符的新数据集列

Question

在我的数据集中，我想创建一个新列，该列将以其他两个列中的字符为条件。如果longdescriptions .desc. en_us中有一个单词Plage，同时externalCode以数字1开头，则在新列中添加值a。如果longdescriptions .desc. en_us没有单词Plage，同时externalCode以数字1开头，则在新列中添加值B。否则为空或NA。

df <- structure(list(X.OPERATOR. = c(" Clear and Delete", NA, NA, NA, 
NA, "<p>Je voornaamste taken:</p>"), externalCode = c("Job Profile.GUID", 
"1008141", "1008168", "1008170", "1008170", NA), longDesciptions.sectionId = c("sectionId", 
"199624017", "200226564", "200226592", "200226594", NA), longDesciptions.sectionType = c("sectionType", 
"LONGDESCRIPTION", "LONGDESCRIPTION", "LONGDESCRIPTION", "LONGDESCRIPTION", 
NA), longDesciptions.desc.en_US = c("US English", "Class: 06, Plage: C, Function code:", 
"Class: 03", "Class: 03", "<p>Als Legal Counsel maak je deel uit van het departement Secretariaat-Generaal. Je ondersteunt zowel de secretaris-generaal en de directie alsook de verschillende entiteiten van Elia groep, zowel op nationaal als internationaal niveau.</p>", 
NA), longDesciptions.desc.defaultValue = c("Default Value", "Class: 06, Plage: C, Function code:", 
"Class: 03", "Class: 03", NA, NA), longDesciptions.desc.en_GB = c("English (United Kingdom)", 
"Class: 06, Plage: C, Function code:", "Class: 03", "Class: 03", 
NA, NA), longDesciptions.desc.de_DE = c("German (Germany)", NA, 
NA, NA, NA, NA), longDesciptions.desc.fr_FR = c("French (France)", 
"Classe: 06, Plage: C, Code de la fonction:", "Classe: 03", "Classe: 03", 
NA, NA), longDesciptions.desc.nl_NL = c("Dutch (Netherlands)", 
"Klasse: 06, Plage: C, Functiecode:", "Klasse: 03", "Klasse: 03", 
NA, NA), longDesciptions.status = c("status(Valid Values : A/I   A for Active  I for Inactive  )", 
"A", "A", "A", NA, NA), longDesciptions.externalCode = c("externalCode", 
"1035137", "1035330", "1035330", NA, NA), longDesciptions.subModule = c("subModule", 
NA, NA, NA, NA, NA), NA. = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..1 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), NA..2 = c(NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_), NA..3 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_), 
NA..4 = c(NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_), NA..5 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..6 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..7 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..8 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..9 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..10 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..11 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..12 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..13 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..14 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..15 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..16 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..17 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..18 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..19 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..20 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..21 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..22 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..23 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..24 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..25 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..26 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..27 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..28 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..29 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..30 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..31 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), NA..32 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), NA..33 = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), NA..34 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
)), class = "data.frame", row.names = c(NA, -6L))

我试过这个代码，但它不工作:

df2[,49] <- NA #
names(df2)[49] <- "JobDescrip" 
for (i in 1 : nrow(df2)) {
if (df2$externalCode[i] == '^1' && df2$longDesciptions.sectionId[i]==
'^P') {
df2[i,49] <- "A"
} 
if (df2$externalCode[i] == '^1') {
df2[i,49] <- "B"
} 
else {
df2[i,49] <- ""
}
}
Error in if (df2$externalCode[i] == "^1" && df2$longDesciptions.sectionId[i] ==  : 
missing value where TRUE/FALSE needed

我知道这类问题已经被问过很多次了，但我找不到一个适合我数据的解决方案。任何帮助将不胜感激!

Answer 1

这是您可以考虑的tidyverse方法。我会考虑其他的矢量化方法，而不是循环。

在本例中，您可以使用dplyr中的mutate来添加新列，并使用case_when而不是多个if语句来添加逻辑。如果第一次求值为假，则测试第二次求值，依此类推。

如果您使用grepl，您可以检查字符串是否包含"Plage"(您可以考虑其他正则表达式模式的替代方案)。使用substr可以查看字符串中的特定字符。

library(dplyr)
df %>%
mutate(job_descrip = case_when(
grepl("Plage", longDesciptions.desc.en_US) & substr(externalCode, 1, 1) == "1" ~ "A",
substr(externalCode, 1, 1) == "1" ~ "B",
TRUE ~ NA_character_
))