在R中的两个条件下合并两列

我正在尝试将两列合并为一列

挑战在于，我只想在两个条件下将它们结合在一起

• 如果工资栏有什么内容，我们不加入
• 如果comp列没有货币符号

  （

  ），这意味着没有数值，则它们不会合并

我第一次尝试了paste（），但我认为这不是正确的情况，它也没有连接两个数据帧，因为它们只是同一数据帧中的两列。有没有办法实现这个条件连接

数据集：

                        salary                                    comp
19              £9.18 per hour                                    <NA>
20 £35,000 to £50,000 per year          Self employed, Commission Only
21 £29,000 to £30,000 per year                                    <NA>
22 £20,000 to £35,000 per year                                    <NA>
23 £29,000 to £31,000 per year                                    <NA>
24          £65 to £80 per day                                    <NA>
25          £65 to £75 per day Opportunity to become a permanent role.
26                        <NA>                                    <NA>
27                        <NA>              £18,000 + Bonus & Benefits
28                        <NA>                NMW + Overtime and Bonus
29                        <NA>                 Depending on experience
30                        <NA>                    £21,892 - £24,157 pa
31                        <NA>              £38,890 -£44,503 per annum

预期产出：

                        salary                                    comp
19              £9.18 per hour                                    <NA>
20 £35,000 to £50,000 per year          Self employed, Commission Only
21 £29,000 to £30,000 per year                                    <NA>
22 £20,000 to £35,000 per year                                    <NA>
23 £29,000 to £31,000 per year                                    <NA>
24          £65 to £80 per day                                    <NA>
25          £65 to £75 per day Opportunity to become a permanent role.
26                        <NA>                                    <NA>
27  £18,000 + Bonus & Benefits              £18,000 + Bonus & Benefits
28                        <NA>                NMW + Overtime and Bonus
29                        <NA>                 Depending on experience
30        £21,892 - £24,157 pa                    £21,892 - £24,157 pa
31  £38,890 -£44,503 per annum              £38,890 -£44,503 per annum

salary comp
每小时19英镑9.18英镑
20.自营职业者每年35000至50000英镑，仅佣金
21每年29000至30000英镑
22每年20000至35000英镑
23每年29000至31000英镑
24每天65至80英镑
25每天65至75英镑成为永久性角色的机会。
26
27 18000英镑+奖金和福利18000英镑+奖金和福利
28 NMW+加班费和奖金
29取决于经验
30 21892英镑-24157英镑/年21892英镑-24157英镑/年
31每年38890-44503英镑每年38890-44503英镑

联接涉及两个不同的表，它们根据某个键合并在一起。您的预期输出只是根据某些条件替换列salary中的值，这可以通过简单的

mutate

：

dataset %>% 
  mutate(salary = if_else(is.na(salary) & grepl(pattern = "£", x=comp), comp, salary))

---

这应该可以做到

str_detect（）

内部

ifelse（）

来自

{stringr}

。如果这是您需要的，请告诉我。：-）

库（tidyverse）
df%
变异（salary=ifelse（is.na（salary）和str_detect（comp，”），comp，salary））

dataset %>% 
  mutate(salary = if_else(is.na(salary) & grepl(pattern = "£", x=comp), comp, salary))

library(tidyverse)

df <- structure(list(salary = c("£9.18 per hour", "£35,000 to £50,000 per year", 
                                "£29,000 to £30,000 per year", "£20,000 to £35,000 per year", 
                                "£29,000 to £31,000 per year", "£65 to £80 per day", "£65 to £75 per day", 
                                NA, NA, NA, NA, NA, NA), comp = c(NA, "Self employed, Commission Only", 
                                                                  NA, NA, NA, NA, "Opportunity to become a permanent role.", NA, 
                                                                  "£18,000 + Bonus & Benefits", "NMW + Overtime and Bonus", "Depending on experience", 
                                                                  "£21,892 - £24,157 pa", "£38,890 -£44,503 per annum")), row.names = 19:31, class = "data.frame")

as_tibble(df) %>%
  mutate(salary = ifelse(is.na(salary) & str_detect(comp, "£"), comp, salary))