data.table存在与函数变量同名列时如何消除变量名歧义

报错核心原因 #

data.table 在[]内执行表达式时，默认优先查找表内列，其次才会查找外部/函数环境的变量。之前写的corpus[get(source)=="027021335"]触发first argument has length > 1错误，和base包内置的source()函数无关：执行时get()会先拿到表内长度为10的source列，把整列作为变量名传入get()，不符合get()要求首个参数为单长度变量名的规则，因此报错。

可直接使用的正确写法 #

以下三种写法都能正确引用外部/函数环境中定义的source变量（该变量存储需要过滤的列名字符串，和dplyr中!!rlang::sym(source)的逻辑完全对应），不会被表内同名列干扰：

• 显式指定查找环境，通用性最强

# source为函数/外部环境中定义的变量，值为要过滤的列名，例：source <- "idref"
corpus[get(source, envir = parent.frame()) == "027021335"]

• 使用data.table专属..前缀，明确告知解析器去上层环境查找变量

corpus[get(..source) == "027021335"]

• 绕开data.table内部作用域规则，零冲突写法

corpus[corpus[[source]] == "027021335"]

验证示例 #

使用提供的dput示例数据测试，以上三种写法均能正确返回idref == "027021335"的对应行：

library(data.table)
# 读入示例数据
corpus <- structure(list(idref = c("027021335", "182132870", "221468579", 
"034574654", "069546592", "159340950", "169800458", "028529413", 
"076605442", "026762889"), iddoc = c(97466L, 101100L, 103772L, 
110039L, 134077L, 55693L, 38787L, 39304L, 73483L, 74350L), nom = c("Méhaut", 
"Favre", "Guerdjikova", "Diebolt", "Giraud-Héraud", "Charlier", 
"Moumni", "Henni", "Bonnel", "Callens"), prenom = c("Philippe", 
"Karine", "Ani", "Claude", "Eric", "Christophe", "Nicolas", "Ahmed", 
"Patrick", "Stéphane"), order = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 
0), role = c("supervisor", "supervisor", "supervisor", "supervisor", 
"supervisor", "supervisor", "supervisor", "supervisor", "supervisor", 
"supervisor"), Annee_soutenance = c("2011", "2014", "2018", "2009", 
"2006", "2015", "2012", "2008", "2009", "2010"), source = c("as.character(idref)", 
"as.character(idref)", "as.character(idref)", "as.character(idref)", 
"as.character(idref)", "as.character(idref)", "as.character(idref)", 
"as.character(idref)", "as.character(idref)", "as.character(idref)"
), time_variable = c("as.character(idref)", "as.character(idref)", 
"as.character(idref)", "as.character(idref)", "as.character(idref)", 
"as.character(idref)", "as.character(idref)", "as.character(idref)", 
"as.character(idref)", "as.character(idref)")), row.names = c(NA, 
-10L), class = c("data.table", "data.frame"))

# 定义外部同名变量
source <- "idref"

# 执行任意一种写法，均返回目标行
corpus[get(..source) == "027021335"]

内容的提问来源于stack exchange，提问作者Homard