我有一个带有偶数和奇数的未排序数字列表。我需要按排序顺序隔离奇数和偶数。
例如：
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预期产出：
List = [5,6,4,7,11,14,12,1,3]
我的程序是将奇数和偶数分开。
[4,6,12,14,1,3,5,7,11]
我正在尝试使用插入排序对列表进行排序，无法获得正确的输出。任何方式轻松排序
答案 0 :(得分：4)
使用＆＃34; smart＆＃34; list.sort / sorted的关键功能：
>>> list(sorted(lst, key=lambda x: [x % 2, x]))
[4, 6, 12, 14, 1, 3, 5, 7, 11]
将偶数数字映射到值[0, n]，将奇数映射到值[1, n]，以便偶数数字按照自然顺序排在第一位。
答案 1 :(得分：2)
Short_list =[5,7,3,2,8,1,0,10,9,4,6]
def sort_list(my_list):
even_list = []
odd_list = []
for i in my_list:
if i % 2 == 0:
even_list.append(i)
else:
odd_list.append(i)
even_list.sort(),odd_list.sort()
even_list.extend(odd_list)
return even_list
print(sort_list(Short_list))
[0，2，4，6，8，10，1，3，5，7，9]
答案 2 :(得分：1)
我们可以先对n%2（n模2）进行排序，对于奇数，它将为0，对于偶数，将为1，然后是数字本身：
L = [5,6,4,7,11,14,12,1,3]
out = sorted(L, key = lambda n:(n%2, n))
print(out)
# [4, 6, 12, 14, 1, 3, 5, 7, 11]
我们用作键的元组首先根据它们的第一个项目进行排序，然后再按照它们的第二项进行排序。
它也适用于负数......
答案 3 :(得分：0)
制作一个evens列表和赔率列表，然后合并：
lst = [5,6,4,7,11,14,12,1,3]
even = sorted([i for i in lst if i%2 == 0])
odd = sorted([i for i in lst if i%2])
print(even + odd)
或使用filter，lambda：
lst = [5,6,4,7,11,14,12,1,3]
lst.sort()
even = list(filter(lambda x: not x%2, lst))
odd = list(filter(lambda x: x%2, lst))
print(even + odd)
答案 4 :(得分：0)
只需使用列表理解基础
>>> arr = [5,6,4,7,11,14,12,1,3]
>>> evens = sorted([e for e in arr if e % 2 ==0])
>>> odds = sorted([e for e in arr if e % 2 !=0])
>>> print(evens + odds)
[4, 6, 12, 14, 1, 3, 5, 7, 11]
答案 5 :(得分：0)
一个简单的解决方案：
public void getParticularBook(String nameOfBook){
String bookDetails = "";
Iterator<Book> iterator = allBooks.iterator();
while(iterator.hasNext()) {
Book b = iterator.next();
if(b.getTitle().equalsIgnoreCase(nameOfBook)){
bookDetails = b.toString();
}
}
System.out.println(bookDetails);
}
public void getBooksDataOnRange(int from, int to){
String bookDetails = "";
Iterator<Book> iterator = allBooks.iterator();
while(iterator.hasNext()) {
Book b = iterator.next();
if(b.getIssueYear() >= from && b.getIssueYear() <= to){
bookDetails = b.toString();
}
}
if(bookDetails.isEmpty()){
System.out.println("No books in range of: " + from + "-" + to);
}
System.out.println(bookDetails);
}
public void getBooksDataOnType(String type){
String bookDetails = "";
Iterator<Book> iterator = allBooks.iterator();
while(iterator.hasNext()) {
Book b = iterator.next();
if(b.getType().equalsIgnoreCase(type)){
bookDetails = b.toString();
}
}
if(bookDetails.isEmpty()){
System.out.println("No books of type: " + type);
}
System.out.println(bookDetails);
}
答案 6 :(得分：0)
如果您乐意使用第三方库，则可以使用 <Switch
android:id="@+id/switch1"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_marginEnd="8dp"
android:layout_marginStart="8dp"
android:layout_marginTop="152dp"
android:text="Switch1"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toTopOf="parent" />
<Switch
android:id="@+id/switch2"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_marginEnd="8dp"
android:layout_marginStart="8dp"
android:layout_marginTop="8dp"
android:text="Switch2"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toBottomOf="@+id/switch1" />
//////////Java Code///////////
public class MainActivity extends AppCompatActivity {
Switch gSwitch,switch1, switch2;
boolean state;
int id;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
switch1 = findViewById(R.id.switch1);
switch1 = findViewById(R.id.switch2);
switch1.setOnCheckedChangeListener(checkedChangeListener);
switch2.setOnCheckedChangeListener(checkedChangeListener);
}
CompoundButton.OnCheckedChangeListener checkedChangeListener = new CompoundButton.OnCheckedChangeListener() {
@Override
public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
id = buttonView.getId();
gSwitch = findViewById(id);// generic switch;
state = isChecked;
// call your method here for example:
// BackgroundWorker backgroundWorker = new BackgroundWorker(this);
// backgroundWorker.execute("switch", state, id);
}
};
布尔索引。
numpy.lexsort以反向方式排序，即在numpy之前考虑A % 2：
A答案 7 :(得分：0)
def even_odd_sort(list):
evens=[]
odds=[]
for i in list:
if(i%2==0):
evens.append(i)
else:
odds.append(i)
evens.sort()
odds.sort()
return evens+odds
答案 8 :(得分：0)
此过程涉及在单独的临时列表中将值分为偶数和奇数，然后按升序对它们进行数字排序，最后将所有值汇总到一个列表中。下图显示了两个使用C＃的项目，一个包含所有过程，而第二个更为实用，但是该过程适用于您的项目。
步骤：
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace WindowsFormsApplication1
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
combodata1();
}
public void combodata1() {
combo1.DataSource = new String[] {"5","10","15","20","25","30","35","40"};
}
private void button2_Click(object sender, EventArgs e)
{
generate();
}
List<String> li1 = new List<String>();
List<String> li2 = new List<String>();
List<String> li3 = new List<String>();
List<String> li4 = new List<String>();
List<String> li5 = new List<String>();
public void generate() {
li1.Clear();
int r = Convert.ToInt32(combo1.SelectedValue)-1;
Random ran = new Random();
for(int i=0;i<=r;i++){
int num1 = ran.Next(100);
li1.Add(Convert.ToString(num1));
}
lista1.DataSource = null;
lista1.DataSource = li1;
}
private void button1_Click(object sender, EventArgs e)
{
String te1 = data1.Text;
if (te1.Equals("") == false)
{
add();
}
else {
MessageBox.Show("- You need to add integer numbers only, empty space or letters are not allowed.","Message");
}
}
public void add() {
String dato = data1.Text;
li1.Add(dato);
lista1.DataSource = null;
lista1.DataSource = li1;
data1.Text = "";
}
private void button3_Click(object sender, EventArgs e)
{
int n = lista1.Items.Count;
if(n>0){
split();
}
else {
MessageBox.Show("- Original List must present values.","Message");
}
}
public void split() {
int n = lista1.Items.Count-1;
for (int i = 0; i <= n;i++){
double a = Convert.ToDouble(li1.ElementAt(i));
double b = (a / 2);
double c = b - Math.Floor(b);
if (c == 0){
li2.Add(Convert.ToString(a));
}
else {
li3.Add(Convert.ToString(a));
}
}
sort_evenly();
}
public void sort_evenly() {
int w = li2.Count;
if (w > 0)
{
int n = li2.Count - 1;
int a = Convert.ToInt32(li2.ElementAt(0));
for (int i = 1; i <= n; i++)
{
int b = Convert.ToInt32(li2.ElementAt(i));
if (a > b) { a = b * 1; }
if (a < b) { a = a * 1; }
}
li4.Add(Convert.ToString(a));
li2.Remove(Convert.ToString(a));
sort_evenly();
}
else {
sort_oddly();
}
}
public void sort_oddly() {
int w = li3.Count;
if (w > 0)
{
int n = li3.Count - 1;
int a = Convert.ToInt32(li3.ElementAt(0));
for (int i = 1; i <= n; i++)
{
int b = Convert.ToInt32(li3.ElementAt(i));
if (a < b) { a = a * 1; }
if (a > b) { a = b * 1; }
}
li5.Add(Convert.ToString(a));
li3.Remove(Convert.ToString(a));
sort_oddly();
}
else {
sortedlist();
}
}
String even = "";
String odd = "";
String result = "";
public void sortedlist() {
even = "";
odd = "";
result = "";
int n = li4.Count-1;
int k = li5.Count-1;
for(int i=0;i<=n;i++){
String hyphen = "-";
if(i==n){hyphen="";}
even = even + Convert.ToString(li4.ElementAt(i)) + hyphen;
}
for(int i=0;i<=k;i++){
String hyphen = "-";
if(i==k){hyphen="";}
odd = odd + Convert.ToString(li5.ElementAt(i)) + hyphen;
}
result = even + "-" + odd;
res1.Text = result;
li2.Clear();
li3.Clear();
li4.Clear();
li5.Clear();
}
private void button4_Click(object sender, EventArgs e)
{
clear();
}
public void clear() {
li2.Clear();
li3.Clear();
li4.Clear();
li5.Clear();
res1.Text = "";
}
private void button5_Click(object sender, EventArgs e)
{
clearall();
}
public void clearall() {
data1.Text = "";
li1.Clear();
li2.Clear();
li3.Clear();
li4.Clear();
li5.Clear();
lista1.DataSource = null;
res1.Text = "";
}
}
}
答案 9 :(得分：0)
在蟒蛇中：
data = [100, 1, 2, 3, 4, 5, 6, 65, 89, 7, 8, 9, 10]
print(sorted([e for e in data if e % 2 == 0]) + sorted([e for e in data if e % 2 != 0]))