Projection operations (C#)
Projection refers to the operation of transforming an object into a new form that often consists only of those properties that will be subsequently used. By using projection, you can construct a new type that is built from each object. You can project a property and perform a mathematical function on it. You can also project the original object without changing it.
The standard query operator methods that perform projection are listed in the following section.
Methods
| Method names | Description | C# query expression syntax | More information |
|---|---|---|---|
| Select | Projects values that are based on a transform function. | select |
Enumerable.Select Queryable.Select |
| SelectMany | Projects sequences of values that are based on a transform function and then flattens them into one sequence. | Use multiple from clauses |
Enumerable.SelectMany Queryable.SelectMany |
| Zip | Produces a sequence of tuples with elements from 2-3 specified sequences. | Not applicable. | Enumerable.Zip Queryable.Zip |
Select
The following example uses the select clause to project the first letter from each string in a list of strings.
List<string> words = new() { "an", "apple", "a", "day" };
var query = from word in words
select word.Substring(0, 1);
foreach (string s in query)
Console.WriteLine(s);
/* This code produces the following output:
a
a
a
d
*/
SelectMany
The following example uses multiple from clauses to project each word from each string in a list of strings.
List<string> phrases = new() { "an apple a day", "the quick brown fox" };
var query = from phrase in phrases
from word in phrase.Split(' ')
select word;
foreach (string s in query)
Console.WriteLine(s);
/* This code produces the following output:
an
apple
a
day
the
quick
brown
fox
*/
Zip
There are several overloads for the Zip projection operator. All of the Zip methods work on sequences of two or more possibly heterogenous types. The first two overloads return tuples, with the corresponding positional type from the given sequences.
Consider the following collections:
// An int array with 7 elements.
IEnumerable<int> numbers = new[]
{
1, 2, 3, 4, 5, 6, 7
};
// A char array with 6 elements.
IEnumerable<char> letters = new[]
{
'A', 'B', 'C', 'D', 'E', 'F'
};
To project these sequences together, use the Enumerable.Zip<TFirst,TSecond>(IEnumerable<TFirst>, IEnumerable<TSecond>) operator:
foreach ((int number, char letter) in numbers.Zip(letters))
{
Console.WriteLine($"Number: {number} zipped with letter: '{letter}'");
}
// This code produces the following output:
// Number: 1 zipped with letter: 'A'
// Number: 2 zipped with letter: 'B'
// Number: 3 zipped with letter: 'C'
// Number: 4 zipped with letter: 'D'
// Number: 5 zipped with letter: 'E'
// Number: 6 zipped with letter: 'F'
Important
The resulting sequence from a zip operation is never longer in length than the shortest sequence. The numbers and letters collections differ in length, and the resulting sequence omits the last element from the numbers collection, as it has nothing to zip with.
The second overload accepts a third sequence. Let's create another collection, namely emoji:
// A string array with 8 elements.
IEnumerable<string> emoji = new[]
{
"🤓", "🔥", "🎉", "👀", "⭐", "💜", "✔", "💯"
};
To project these sequences together, use the Enumerable.Zip<TFirst,TSecond,TThird>(IEnumerable<TFirst>, IEnumerable<TSecond>, IEnumerable<TThird>) operator:
foreach ((int number, char letter, string em) in numbers.Zip(letters, emoji))
{
Console.WriteLine(
$"Number: {number} is zipped with letter: '{letter}' and emoji: {em}");
}
// This code produces the following output:
// Number: 1 is zipped with letter: 'A' and emoji: 🤓
// Number: 2 is zipped with letter: 'B' and emoji: 🔥
// Number: 3 is zipped with letter: 'C' and emoji: 🎉
// Number: 4 is zipped with letter: 'D' and emoji: 👀
// Number: 5 is zipped with letter: 'E' and emoji: ⭐
// Number: 6 is zipped with letter: 'F' and emoji: 💜
Much like the previous overload, the Zip method projects a tuple, but this time with three elements.
The third overload accepts a Func<TFirst, TSecond, TResult> argument that acts as a results selector. Given the two types from the sequences being zipped, you can project a new resulting sequence.
foreach (string result in
numbers.Zip(letters, (number, letter) => $"{number} = {letter} ({(int)letter})"))
{
Console.WriteLine(result);
}
// This code produces the following output:
// 1 = A (65)
// 2 = B (66)
// 3 = C (67)
// 4 = D (68)
// 5 = E (69)
// 6 = F (70)
With the preceding Zip overload, the specified function is applied to the corresponding elements numbers and letter, producing a sequence of the string results.
Select versus SelectMany
The work of both Select and SelectMany is to produce a result value (or values) from source values. Select produces one result value for every source value. The overall result is therefore a collection that has the same number of elements as the source collection. In contrast, SelectMany produces a single overall result that contains concatenated sub-collections from each source value. The transform function that is passed as an argument to SelectMany must return an enumerable sequence of values for each source value. These enumerable sequences are then concatenated by SelectMany to create one large sequence.
The following two illustrations show the conceptual difference between the actions of these two methods. In each case, assume that the selector (transform) function selects the array of flowers from each source value.
This illustration depicts how Select returns a collection that has the same number of elements as the source collection.
This illustration depicts how SelectMany concatenates the intermediate sequence of arrays into one final result value that contains each value from each intermediate array.
Code example
The following example compares the behavior of Select and SelectMany. The code creates a "bouquet" of flowers by taking the first two items from each list of flower names in the source collection. In this example, the "single value" that the transform function Select<TSource,TResult>(IEnumerable<TSource>, Func<TSource,TResult>) uses is itself a collection of values. This requires the extra foreach loop in order to enumerate each string in each sub-sequence.
class Bouquet
{
public List<string> Flowers { get; set; }
}
static void SelectVsSelectMany()
{
List<Bouquet> bouquets = new()
{
new Bouquet { Flowers = new List<string> { "sunflower", "daisy", "daffodil", "larkspur" }},
new Bouquet { Flowers = new List<string> { "tulip", "rose", "orchid" }},
new Bouquet { Flowers = new List<string> { "gladiolis", "lily", "snapdragon", "aster", "protea" }},
new Bouquet { Flowers = new List<string> { "larkspur", "lilac", "iris", "dahlia" }}
};
IEnumerable<List<string>> query1 = bouquets.Select(bq => bq.Flowers);
IEnumerable<string> query2 = bouquets.SelectMany(bq => bq.Flowers);
Console.WriteLine("Results by using Select():");
// Note the extra foreach loop here.
foreach (IEnumerable<String> collection in query1)
foreach (string item in collection)
Console.WriteLine(item);
Console.WriteLine("\nResults by using SelectMany():");
foreach (string item in query2)
Console.WriteLine(item);
/* This code produces the following output:
Results by using Select():
sunflower
daisy
daffodil
larkspur
tulip
rose
orchid
gladiolis
lily
snapdragon
aster
protea
larkspur
lilac
iris
dahlia
Results by using SelectMany():
sunflower
daisy
daffodil
larkspur
tulip
rose
orchid
gladiolis
lily
snapdragon
aster
protea
larkspur
lilac
iris
dahlia
*/
}
See also
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