'How to modify a slice within for loop? [duplicate]
I've got a slice of articles in my reading list. Each Article has the attribute "FeedURL" that has the URL of the feed the article came from. When I unsubscribe from a feed, I want to be able to remove every Article that contains that Feed's URL.
type Article struct {
FeedURL string
URL string // should be unique
// ... more data
}
func unsubscribe(articleList []Article, url string) []Article {
// how do I remove every Article from articleList that contains url?
}
func main() {
myArticleList := []Article{
Article{"http://blog.golang.org/feed.atom", "http://blog.golang.org/race-detector"},
Article{"http://planet.python.org/rss20.xml", "http://archlinux.me/dusty/2013/06/29/creating-an-application-in-kivy-part-3/"},
Article{"http://planet.python.org/rss20.xml", "http://feedproxy.google.com/~r/cubicweborg/~3/BncbP-ap0n0/2957378"},
// ... much more examples
}
myArticleList = unsubscribe(myArticleList, "http://planet.python.org/rss20.xml")
fmt.Printf("%+v", myArticleList)
}
What is the efficient way of solving this problem?
At first my code looked like this for unsubscribe:
func unsubscribe(articleList []Article, url string) []Article {
for _, article := range articleList {
if article.FeedURL == url {
articleList = append(articleList[:i], articleList[i+1:]...)
}
}
return articleList
}
But then I realized that this would change the slice and make the for loop unpredictable.
What is an efficient and pretty way to accomplish this?
Solution 1:[1]
PeterSO's answer is gets the job done, and with efficiency.
But, I might go with something simple like this
func unsubscribe(articleList []Article, url string) (filtered []Article) {
filtered = articleList[:0] // optional. reuses already-allocated memory.
for _, article := range articleList {
if article.FeedURL != url {
filtered = append(filtered, article)
}
}
return
}
which only takes about a two seconds to read, and comprehend.
The idea works fine with pointers to articles too and, like PeterSO said, if your Article struct is big, that may be a good thing to do.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
