Question

说有一个向量：

v1 <- c("ab", "bc", "cd", "ef", "yolo", "da", "sm", "ez-de") 
v2 <- c("ab", "bc", "cd", "ef", "yolo-da", "sm", "ez", "de")

如何合并上面的两个向量，以便我们得到以下内容？

c("ab", "bc", "cd", "ef", "yolo-da", "sm", "ez-de")

请注意，上面的两个向量具有相同的长度..

Answer 1

如果不关注值的顺序，我们可以尝试这样做：

driver.manage().window().maximize();    
driver.findElement(By.id("T1")).sendKeys("test@gmail.com");
driver.findElement(By.id("T2")).sendKeys("test@1234");

WebDriverWait wait = new WebDriverWait(driver, 20);
WebDriverWait wait1 = new WebDriverWait(driver, 20);

wait.until(ExpectedConditions.visibilityOf(driver.findElement(By.id("loginbtn"))));
System.out.println("Is Visible");
wait1.until(ExpectedConditions.elementToBeClickable(driver.findElement(By.id("loginbtn"))));
System.out.println("Is clickable");

driver.findElement(By.id("loginbtn")).click();

Answer 2

解决方案的逐步方法;一旦理解，步骤可以减少

# case 1. 
a=c("ab", "bc", "cd", "ef", "yolo", "da", "sm", "ez-de") 
b=c("ab", "bc", "cd", "ef", "yolo-da", "sm", "ez", "de")
# [1] "ab"      "bc"      "cd"      "ef"      "sm"      "yolo-da" "ez-de" 

# case 2.
a = c("lol", "it","is", "now", " jab-time")
b = c("lol", "it-is", " now", "jab", " time")
# [1] "lol"      "now"      "it-is"    "jab-time"

a = trimws(a)  # since observed that case 2 . "now" had whitespaces
b = trimws(b)  # these 2 steps are unnecessary, just check if that was a typo

c = intersect(a, b)  # extract the common values from both vectors
a = a[!(a %in% c)]   # keep only those which are not there in c
b = b[!(b %in% c)]   # keep only those which are not there in c

d = grep("-", c(a, b), value = TRUE)  # this returns only those having "-" in it

ans <- c(c , d)

Answer 3

我的方法：

library(dplyr)

a=c("ab", "bc", "cd", "ef", "yolo", "da", "sm", "ez-de") 
b=c("ab", "bc", "cd", "ef", "yolo-da", "sm", "ez", "de")

ab <- c(a,b)
ab_unique <- unique(ab)

ab
ab_unique

合并2个向量并删除所有重复

3 个答案: