合并 2 个向量并删除所有重复项答案

【问题标题】：Merging 2 vectors and removing all repetitions合并 2 个向量并删除所有重复项
【发布时间】：2017-02-03 05:41:24
【问题描述】：

假设有一个向量：

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，得到下面的？

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..

【问题讨论】：

你能展示其他模式吗？
两个向量的长度总是相同的。您始终使用破折号保留元素，并且只使用一次重复的元素。任何没有破折号的元素总是两个向量中的重复元素
c("lol", "it","is", "now", "jab-time").和 c("lol", "it-is", "now", "jab", "time")。合并到 c("lol", "it-is", "now", "jab-time")。一个向量 js 中的最后一个元素始终是虚线元素。

标签： r

【解决方案1】：

如果值的顺序不是问题，我们可以试试这个：

v <- union(v1, v2)
Filter(function(x) length(grep(x, v))==1, v)
# [1] "ab"      "bc"      "cd"      "ef"      "sm"      "ez-de"   "yolo-da"

【讨论】：

【解决方案2】：

逐步解决方案；理解后可以减少步骤

# 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)

【讨论】：

【解决方案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

【讨论】：