Java Reference
In-Depth Information
Employees in descending order by last name then first:
Jason Red 5000.00 IT
Matthew Indigo 3587.50 Sales
Luke Indigo 6200.00 IT
James Indigo 4700.77 Marketing
Ashley Green 7600.00 IT
Wendy Brown 4236.40 Marketing
Jason Blue 3200.00 Sales
Fig. 17.12
|
Sorting
Employee
s by last name then first name. (Part 2 of 2.)
You previously used
map
operations to perform calculations on
int
values and to convert
String
s to uppercase letters. In both cases, the resulting streams contained values of the same
types as the original streams. Figure 17.13 shows how to map objects of one type (
Employee
)
to objects of a different type (
String
). Lines 77-81 perform the following tasks:
• Line 77 creates a
Stream<Employee>
.
• Line 78 maps the
Employees
to their last names using the instance method refer-
ence
Employee::getName
as method
map
's
Function
argument. The result is a
Stream<String>
.
• Line 79 calls
Stream
method
distinct
on the
Stream<String>
to eliminate any
duplicate
String
objects in a
Stream<String>
.
• Line 80 sorts the unique last names.
• Finally, line 81 performs a terminal
forEach
operation that processes the stream
pipeline and outputs the unique last names in sorted order.
Lines 86-89 sort the
Employees
by last name then first name, then
map
the
Employee
s to
String
s with
Employee
instance method
getName
(line 88) and display the sorted names
in a terminal
forEach
operation.
75
// display unique employee last names sorted
76
System.out.printf(
"%nUnique employee last names:%n"
);
77
list.stream()
78
.map(Employee::getLastName)
.distinct()
79
80
.sorted()
81
.forEach(System.out::println);
82
83
// display only first and last names
84
System.out.printf(
85
"%nEmployee names in order by last name then first name:%n"
);
86
list.stream()
87
.sorted(lastThenFirst)
88
.map(Employee::getName)
89
.forEach(System.out::println);
90
Fig. 17.13
|
Mapping
Employee
objects to last names and whole names. (Part 1 of 2.)
