Question

使用merge函数连接2个data.frame我发现它的参数sort没有任何改变。例如：

id_df <- structure(list(id = c("click", "event", "funnel", "impression", 
                               "tracker", "visibility"), 
                        id_Havas = c("a1", "a2", "a3", "a4", "a5", "a6")), 
                   .Names = c("id", "my_id"), class = "data.frame", 
                   row.names = c(NA, -6L))

my_df <- data.frame("id" = c("click", "click", "impression", "visibility", "click"),
                     stringsAsFactors = F)

结果：

my_df

# id
# 1      click
# 2      click
# 3 impression
# 4 visibility
# 5      click

merge(my_df, id_df, by = "id", all.x = TRUE, sort = FALSE)
# id my_id
# 1      click    a1
# 2      click    a1
# 3      click    a1
# 4 impression    a4
# 5 visibility    a6

merge(my_df, id_df, by = "id", all.x = TRUE, sort = TRUE)
# id my_id
# 1      click    a1
# 2      click    a1
# 3      click    a1
# 4 impression    a4
# 5 visibility    a6

我错过了什么吗？

Answer 1

要保留原始订单，您可以使用match：

my_df$my_id <- id_df$my_id[match(my_df$id, id_df$id)]
my_df
#          id my_id
#1      click    a1
#2      click    a1
#3 impression    a4
#4 visibility    a6
#5      click    a1

根据您的具体情况，merge和match之间的基准比较，考虑到my_df的60000种不同的ID和100000行：

f_merge <- function(){merge(my_df, id_df, by = "id", all.x = TRUE, sort = FALSE)}
f_match <- function(){my_df$my_id <- id_df$my_id[match(my_df$id, id_df$id)]}
microbenchmark(f_match(), f_merge(), unit="relative")
 #     expr      min       lq     mean   median       uq      max neval cld
 #f_match()  1.00000  1.00000  1.00000  1.00000  1.00000 1.000000   100  a 
 #f_merge() 41.16602 46.42379 26.62328 47.59711 17.28836 7.176999   100   b

sort中的参数merge如何运作：

来自?merge，＆＃34;价值＆＃34;部分，你可以阅读：

默认情况下，行按字典顺序在公共列上排序，但对于sort = FALSE，行的顺序是未指定的。

因此，所有常见名称都是＆＃34;放在一起＆＃34;但如果sort=FALSE，则不会对唯一名称进行排序。

<强> 实施例

my_df <- data.frame("id" = c("impression", "click", "click", "impression", "visibility", "click"), stringsAsFactors = F)

merge(my_df, id_df, by = "id", all.x = TRUE, sort = FALSE)
#          id my_id
#1 impression    a4
#2 impression    a4
#3      click    a1
#4      click    a1
#5      click    a1
#6 visibility    a6

merge(my_df, id_df, by = "id", all.x = TRUE, sort = TRUE)
#          id my_id
#1      click    a1
#2      click    a1
#3      click    a1
#4 impression    a4
#5 impression    a4
#6 visibility    a6

R合并奇怪的行为

1 个答案: