'Is it possible to store a KeyValuePair of a dictionary without the use of a foreach loop?
The syntax for iterating over a dictionary with a foreach loop is:
foreach (KeyValuePair<key, value> item in dictionary)
Inside the foreach loop the key is accessed with item.key and the value with item.value.
This got me thinking, can this be used without the use of a foreach loop as a convenient (although niche) way to represent a specific dictionary pair?
I am not looking for some weird work arounds, like running a foreach loop and saving the KeyValuePair into a variable once the target key is reached, because at this point it would be more convenient to just use 2 variables.
Solution 1:[1]
Like this
var dic = new Dictionary<string, int>();
dic["a"] = 42;
KeyValuePair<string, int> keyVal;
foreach(var kv in dic) {
keyVal = kv; << gets the last entry from the dictioanry
}
Note that the dictionary does not store KeyValuePairs, it creates one for the enumeration, so the simple thing to do is this (because we are not expensively recreating something)
var dic = new Dictionary<string, int>();
dic["a"] = 42;
KeyValuePair<string, int> keyVal = new KeyValuePair<string, int>("a", dic["a"]);
this is more efficient than the (neat) LINQ Sinlge method
Solution 2:[2]
The IDictionary<TKey, TValue> interface implements IEnumerable<KeyValuePair<TKey,TValue>>. This means you can simply use Single() to get the entry you want.
IDictionary<string, int> dict = ...;
KeyValuePair<string, int> entry = dict.Single(it => it.Key == "yourKey");
Solution 3:[3]
try this
var dict = new Dictionary<string, string>() {
{"hi","Hello World!"},
{"adieu","Goodby"}
};
string hi = dict["hi"]; //Hello World!
or if you want a list
List<KeyValuePair<string,string>> list = dict.ToList();
result
[{"Key":"hi","Value":"Hello World!"},{"Key":"adieu","Value":"Goodby"}]
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 | |
| Solution 2 | Progman |
| Solution 3 |
