当存在与变量同名的列时，在函数中消除变量名称的歧义（data.table）

Question

我有一个带有名为source的变量的函数。 该函数可以正常工作，但如果应用该函数的数据框有一个也名为source的列，则它不起作用。

一个使用 dplyr 和过滤的简单示例：以下两行有效，但它们根据列进行过滤（我想过滤函数中定义的变量名）：

corpus %>% dplyr::filter(!!source=="027021335")
corpus %>% dplyr::filter(source=="027021335")

以下行工作并正确使用函数中定义的变量：

corpus %>% dplyr::filter(!!rlang::sym(source)=="027021335")

如何使用 data.table() 实现相同的目标？ 我尝试了许多c() 、 get()和..的组合，但没有成功。 我认为corpus[get(source)=="027021335"]应该可以工作，但事实并非如此，因为它返回的first argument has length > 1错误。

编辑：我认为我收到此错误的一个可能原因是，除了作为变量和列名的源之外，还有一个 source() 函数是 base r。

使用 dput（语料库）的语料库：

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"), .internal.selfref = <pointer: 0x000001ee3a911ef0>)

Answer 1

使用data.table开发版本 (1.14.3)，这可以通过新的env参数来完成，请参阅data.table 上的编程：

data.table::update.dev.pkg()
source = "idref"
corpus[source=="027021335",env=list(source=source)]

       idref iddoc    nom   prenom order       role Annee_soutenance              source       time_variable
1: 027021335 97466 Méhaut Philippe     0 supervisor             2011 as.character(idref) as.character(idref)