java 比较两个 String[] 数组并打印出不同的字符串

Compare two String[] arrays and print out the strings which differ

提问by Martin Erlic

I've got a list of all the file names in a folder and a list of files which have been manually "checked" by a developer. How would I go about comparing the two arrays such that we print out only those which are not contained in the master list.

我有一个文件夹中所有文件名的列表和一个由开发人员手动“检查”的文件列表。我将如何比较两个数组，以便我们只打印出那些未包含在主列表中的数组。

public static void main(String[] args) throws java.lang.Exception {
        String[] list = {"my_purchases", "my_reservation_history", "my_reservations", "my_sales", "my_wallet", "notifications", "order_confirmation", "payment", "payment_methods", "pricing", "privacy", "privacy_policy", "profile_menu", "ratings", "register", "reviews", "search_listings", "search_listings_forms", "submit_listing", "submit_listing_forms", "terms_of_service", "transaction_history", "trust_verification", "unsubscribe", "user", "verify_email", "verify_shipping", "404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu", "main_searchbar", "primary_navbar"};
        String[] checked = {"404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu"};

        ArrayList<String> ar = new ArrayList<String>();

            for(int i = 0; i < checked.length; i++)
                {
                    if(!Arrays.asList(list).contains(checked[i]))
                    ar.add(checked[i]);
                }
    }

回答by SatyaTNV

Change your loop to :

将您的循环更改为：

ArrayList<String> ar = new ArrayList<String>();

for(int i = 0; i < checked.length; i++) {
      if(!Arrays.asList(list).contains(checked[i]))
      ar.add(checked[i]);
}

ArrayListarshould be outside of the forloop. Otherwise arwill be created every time when element of checkedarray exists in list.

ArrayListar应该在for循环之外。否则ar每次当checked数组元素存在于list.

Edit:

编辑：

if(!Arrays.asList(list).contains(checked))

With this statement you are checking whether the checkedreference is not the element of list. It should be checked[i]to check whether the element of checkedexists in listor not.

使用此语句，您将检查checked引用是否不是 的元素list。应该是checked[i]检查 的元素是否checked存在list。

If you want to print elements in listthat are not in checked. Then use :

如果要打印list不在checked. 然后使用：

for(int i = 0; i < list.length; i++) {
      if(!Arrays.asList(checked).contains(list[i]))
      ar.add(list[i]);
}
System.out.println(ar);

回答by Luke Wahlmeier

Your updated solution seems kind of odd to me, not sure why you would add list[i] to the result list. Generally this sounds like something hashsets are made for:

您更新后的解决方案对我来说似乎有点奇怪，不知道为什么要将 list[i] 添加到结果列表中。通常，这听起来像是哈希集的用途：

String[] list = { "my_purchases", "my_reservation_history","my_reservations","my_sales", "my_wallet", "notifications", "order_confirmation", "payment", "payment_methods", "pricing", "privacy", "privacy_policy", "profile_menu", "ratings", "register", "reviews", "search_listings", "search_listings_forms", "submit_listing", "submit_listing_forms", "terms_of_service", "transaction_history", "trust_verification", "unsubscribe", "user", "verify_email", "verify_shipping", "404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu", "main_searchbar", "primary_navbar"};
String[] checked = { "404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu"};

HashSet<String> s1 = new HashSet<String>(Arrays.asList(checked));
s1.removeAll(Arrays.asList(list));
System.out.println(s1);

回答by Pat Myron

for (String s: checked) {      // go through all in second list
    if (! list.contains(s)) {  // if string not in master list
        System.out.println(s); // print that string
    }
}

回答by Philipp

First of all, I think your code has some errors:

首先，我认为您的代码有一些错误：

• s1 is not defined
• ar is not defined
• you mean to use Arrays.toString instead of Array.toString

• s1 未定义
• ar 未定义
• 你的意思是使用 Arrays.toString 而不是 Array.toString

So I fixed your code (using Java 8) and it should work like that:

所以我修复了你的代码（使用 Java 8），它应该像这样工作：

public static void main(String[] args) throws java.lang.Exception {
    String[] list = {"my_purchases", "my_reservation_history", "my_reservations", "my_sales", "my_wallet", "notifications", "order_confirmation", "payment", "payment_methods", "pricing", "privacy", "privacy_policy", "profile_menu", "ratings", "register", "reviews", "search_listings", "search_listings_forms", "submit_listing", "submit_listing_forms", "terms_of_service", "transaction_history", "trust_verification", "unsubscribe", "user", "verify_email", "verify_shipping", "404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu", "main_searchbar", "primary_navbar"};
    String[] checked = {"404", "account_menu", "auth", "base", "dashboard_base", "dashboard_menu", "fiveohthree", "footer", "header", "header_menu", "listings_menu"};

    final List<String> result = Stream.of(list)
            .filter(listEntry -> Stream.of(checked)
                    .filter(checkedEntry -> checkedEntry.equals(listEntry)).findFirst().orElse(null) == null)
            .collect(Collectors.toList());

    System.out.println(result);
}

If you don't want to use Java 8, you have to replace the usage of Streams and filters and collect with the appropriate functions in Java 7 (see e.g., Satya's post).

如果您不想使用 Java 8，则必须替换 Streams 和过滤器的使用，并使用 Java 7 中的适当函数进行收集（参见例如 Satya 的帖子）。

Anyways, I should mention that there are better (regarding performance) implementations to solve your problem, e.g.,

无论如何，我应该提到有更好的（关于性能）实现来解决您的问题，例如，

• you could sort your lists prior to searching for duplicates,
• you could use, e.g., hash-based implementations to increase the speed when searching for duplicates,
• you could move the code outside of the inner loop,
• and many more

• 您可以在搜索重复项之前对列表进行排序，
• 您可以使用例如基于哈希的实现来提高搜索重复项时的速度，
• 你可以将代码移到内部循环之外，
• 还有很多