Java map string contains

How to Check If a Key Exists in a Map

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1. Overview

In this brief tutorial, we’ll look at ways to check if a key exists in a Map.

Specifically, we’ll focus on containsKey and get.

2. containsKey

Returns true if this map contains a mapping for the specified key

We can see that this method is a pretty good candidate for doing what we want.

Let’s create a very simple map and verify its contents with containsKey:

@Test public void whenKeyIsPresent_thenContainsKeyReturnsTrue() < Mapmap = Collections.singletonMap("key", "value"); assertTrue(map.containsKey("key")); assertFalse(map.containsKey("missing")); >

Simply put, containsKey tells us whether the map contains that key.

3. get

Now, get can sometimes work, too, but it comes with some baggage, depending on whether or not the Map implementation supports null values.

Again, taking a look at Map‘s JavaDoc, this time for Map#put, we see that it will only throw a NullPointerException:

if the specified key or value is null and this map does not permit null keys or values

Since some implementations of Map can have null values (like HashMap), it’s possible for get to return null even though the key is present.

So, if our goal is to see whether or not a key has a value, then get will work:

@Test public void whenKeyHasNullValue_thenGetStillWorks() < Mapmap = Collections.singletonMap("nothing", null); assertTrue(map.containsKey("nothing")); assertNull(map.get("nothing")); >

But, if we are just trying to check that the key exists, then we should stick with containsKey.

4. Conclusion

In this article, we looked at containsKey. We also took a closer look at why it’s risky to use get for verifying a key’s existence.

As always, check out the code examples over on Github.

Slow MySQL query performance is all too common. Of course it is. A good way to go is, naturally, a dedicated profiler that actually understands the ins and outs of MySQL.

The Jet Profiler was built for MySQL only, so it can do things like real-time query performance, focus on most used tables or most frequent queries, quickly identify performance issues and basically help you optimize your queries.

Critically, it has very minimal impact on your server’s performance, with most of the profiling work done separately — so it needs no server changes, agents or separate services.

Basically, you install the desktop application, connect to your MySQL server, hit the record button, and you’ll have results within minutes:

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How can I contains String value with map.key?

Question: Given an array of strings and char, return strings containing char (no built in functions)? Given a list of Strings, return an array containing the same Strings in the same order Solution 1: You shouldn’t remove elements from the inside your loop.

How can I contains String value with map.key?

String[] mass = new String[]//each element contains substring p1,p2,p3,p4 

Map map = new HashMap<>(); map.put("p1","blablabla"); map.put("p2","blablabla"); map.put("p3","blablabla"); map.put("p4","blablabla"); 

And I want contains each element of mass with key of map

How can I realize this? Maby not Map?

public void method(String[] mass, Map map) < for (String s : mass ) < if (s.contains(map.key)) form.setField(map.key,map.value); >> 

I want create map like config and pass like parameter to method. I have some massives of String and I need parse it. cange only 1 parameter p1 or p999

As I can see, you’re asking that for each String in your mass variable, is there any substring, which represents a key in Map

If that’s the scenario, then you’d want to get multiple loops, one to iterate over your String and another to iterate over the your Map .

for (String s : mass ) < for (Map.Entryentry : map.entrySet()) < if (s.contains(entry.getKey())) < form.setField(entry.getKey(), entry.getValue()); break; >> > 

Here is an solution using streams:

for (String s : mass ) < Optionalkey = map.keySet().parallelStream().filter(s::contains).findFirst(); if(key.isPresent()) < form.setField(key.get(), map.get(key.get())); >> 

The idea is to filter the map for every string adn check if there is a key which is contained in the current string.

Map array return key in a string format in JavaScript, I have a function that map and create a new array from a given array. After I map the array to have a key: «value», but the map function return me the «key»: «value». How can I get or map the key not in a string format ?

Given a list of Strings, return an Array containing the same Strings in the same order

I can’t figure out why the last value of stringList (ie, Banana) is not being printed out or stored in arrayForTheList for that matter.

Given a list of Strings, return an array containing the same Strings in the same order

list2Array( ["Apple", "Orange", "Banana"] ) -> list2Array( ["Red", "Orange", "Yellow"] ) -> list2Array( ["Left", "Right", "Forward", "Back"] ) -> public String[] list2Array(List stringList) < String[] arrayForTheList = new String[(stringList.size())]; for (int i = 0; i < stringList.size() ; i++) < arrayForTheList[i] = stringList.remove(0); System.out.println(arrayForTheList[i]); >return arrayForTheList; > 

You shouldn’t remove elements from the List inside your loop. You can clear the List after the loop if you have to.

for (int i = 0; i < stringList.size() ; i++) < arrayForTheList[i] = stringList.get(i); System.out.println(arrayForTheList[i]); >stringList.clear(); 

Keep in mind that your code clear List which it receive as parameter.

You can simply implement this by using the toArray() method of lists. For example:

String[] arrayForTheList = (String[])stringList.toArray(); 

Javascript map method on array of string elements, 8 Answers. The map method will literally ‘map’ a function call onto each element in the array, take this as a simple example of increasing the value of each integer in an array by 1: var items = [1,2,3]; items.map (function (item) < return item + 1; >); // returns [2,3,4] In your case, you are trying to use map to accept or reject a string …

Given an array of strings and char, return strings containing char? I’ve gotten O(N^2), but is there a way to optimize?

Given an array of strings and char, return strings containing char (no built in functions)? I’ve gotten O(N^2), but is there a way to optimize? Perhaps a way utilizing a dict? Thanks! Edit: Strings are unsorted/unique to list.

def find_char(array, char_to_find): strings = [] for i in xrange(****(array)): notFound = True for j in xrange(****(array[i])): if array[i][j] == char_to_find and notFound: strings.append(array[i]) notFound = False return strings array = ['bob', 'yo', 'hello', 'yes'] print find_char(array, 'o') 

return [hit for hit in array if char_to_find in hit] 

Also, how does your solution come up as O(N^2)? I see it as O(N*M), where M is a fair statistic of the string length, and N is the length of array .

I got more info from the complexity searches:

1. The in operator is linear, O(m) (length of the string being searched). In «good» cases, it’s O(m/t) , where t is length of the string we want to find. That’s 1 in this application.
2. List comprehension is faster than append , but still O(n) .
3. We seem stuck with taking each word in the list individually: O(n) .

That leaves us still with O(n^2 * m) .

Now . is it worth implementing a linked list of strings to get O(1) insertion (front of the list)? It has a higher overhead, but drops the complexity to O(n*m) .

@Bi Rico, thanks. I misread my own link (why did you post it a second time?). append is, indeed, O(n) , so the original problem does have complexity O(m*n) .

A few tips. Convert the string into a set and then check for membership in the set. That becomes an O(1) operation. I’m not sure the runtime for in when it comes to strings (I’d guess O(n)). Still, the time to convert the each string to a set is linear in the size of the string (as pointed out in the comments) so while this may run faster than your solution by eliminating the inner for loop, it does not guarantee and asymptotic runtime better than O(N^2)

def find_char(array, char_to_find): strings = [] for my_string in array: str_chars = set(my_string) if char_to_find in str_chars: strings += [my_string] return strings array = ['bob', 'yo', 'hello', 'yes'] print find_char(array, 'o') 

Given an array of strings, return another array containing, According to your code, the element at the 0 index of ‘longestWord’ will be the longest word. Since you need all the elements that are longest, take the length of that first element in ‘longestWord’ variable, use a for loop to check all the elements which have length equal to the first element of ‘longestWord’, if the …

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