Question

我想结合使用mutate()和ifelse()和&。但是，R没有意识到这一变化，但我没有出错。

因此，必须有一个错字。这是我使用的代码：

library(dplyr)
dat %>% 
  mutate(City=ifelse(grepl("\\(030)|30|^\\+4930|(30)|^\\+49 30|^0049030|^\\+49030|0049030|^4930|^4930|^030", 
                             `Business Phone`) & Country == "Germany", "Berlin", City))

目标是如果"Berlin"具有`Business Phone`中的模式并且grepl()为Country，则推算"Germany"。

这里有一个小的dput：

structure(list(Country = c("Germany", "Germany", "Germany", "Germany", 
"Germany", "Germany", "Germany", "Germany", "Germany", "Germany"
), City = c(NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), `Business Phone` = c("+49 3020618791360", "+49 (30) 24729320", 
"+49 (30) 29034056", "+49 (30) 31422940", "+49 (30) 78893131", 
"+49 30 2060708870", "+49 (30) 84452575", "+49 (30) 38629224", 
"+49 (30) 93923158", "+49 (30) 36288666")), row.names = c(NA, 
-10L), class = c("tbl_df", "tbl", "data.frame"))

Answer 1

您可以使用正则表达式和条件赋值。

dat$City[grepl("^\\+?(0?0)?(49)0?\\s?\\(?(30).+|^\\(?(0?(30)).+", 
      dat$`Business Phone`) & dat$Country == "Germany"] <- "Berlin"

> dat
# A tibble: 10 x 3
   Country City   `Business Phone` 
   <chr>   <chr>  <chr>            
 1 Germany Berlin +49 3020618791360
 2 Germany Berlin +49 (30) 24729320
 3 Germany Berlin +49 (30) 29034056
 4 Germany Berlin +49 (30) 31422940
 5 Germany Berlin +49 (30) 78893131
 6 Germany Berlin +49 30 2060708870
 7 Germany Berlin +49 (30) 84452575
 8 Germany Berlin +49 (30) 38629224
 9 Germany Berlin +49 (30) 93923158
10 Germany Berlin +49 (30) 36288666

Test包含以下数据：

nums <- c("+49 (30) 78893131", "+49 30 2060708870", "+42 (30) 36288666 ", 
"+19 (30) 36288666 ", "+49 (20) 36288666", "30456745674", "+493045674567", 
"(30)45674567", "+49 3045674567", "004903045674567", "+4903045674567", 
"004903045674567", "493045674567", "493045674567", "03045674567", 
"+49 (20) 36288666", "004920000", "204054756", "3145675678", 
"49403235678", "49030345435", "20456745674", "+193045674567", 
"(20)45674567", "+41 3045674567", "004103045674567", "+4103045674567", 
"004104945674567", "413045674567", "413045674567", "01045674567"
)

> nums[grepl("^\\+?(0?0)?(49)0?\\s?\\(?(30).+|^\\(?(0?(30)).+", nums)]
 [1] "+49 (30) 78893131" "+49 30 2060708870" "30456745674"       "+493045674567"    
 [5] "(30)45674567"      "+49 3045674567"    "004903045674567"   "+4903045674567"   
 [9] "004903045674567"   "493045674567"      "493045674567"      "03045674567"      
[13] "49030345435"

模式匹配并使用`＆`和ifelse进行突变

1 个答案: