我需要计算一行的列之间的相似性,并尝试使用columnsimilarities()方法来获得结果.
public static void main(String[] args) {
SparkConf sparkConf = new SparkConf().setAppName("CollarberativeFilter").setMaster("local");
JavaSparkContext sc = new JavaSparkContext(sparkConf);
SparkSession spark = SparkSession.builder().appName("CollarberativeFilter").getOrCreate();
double[][] array = {{5,0,5}, {0,10,0}, {5,0,5}};
LinkedList<Vector> rowsList = new LinkedList<Vector>();
for (int i = 0; i < array.length; i++) {
Vector currentRow = Vectors.dense(array[i]);
rowsList.add(currentRow);
}
JavaRDD<Vector> rows = sc.parallelize(rowsList);
// Create a RowMatrix from JavaRDD<Vector>.
RowMatrix mat = new RowMatrix(rows.rdd());
CoordinateMatrix simsPerfect = mat.columnSimilarities();
RowMatrix mat2 = simsPerfect.toRowMatrix();
List<Vector> vs2 = mat2.rows().toJavaRDD().collect();
List<Vector> vs = mat.rows().toJavaRDD().collect();
System.out.println("mat"); …Run Code Online (Sandbox Code Playgroud) 我正在使用git kraken客户端gui,我需要对存储库执行强制推送.如果有人可以指导我对git kraken进行强制推动的步骤,那将是非常感激的.
我已经使用firebase-ui成功设置了Firebase电子邮件/密码身份验证。
var uiConfig = {
signInSuccessUrl: '<?php echo $url; ?>',
signInOptions: [
// Leave the lines as is for the providers you want to offer your users.
firebase.auth.GoogleAuthProvider.PROVIDER_ID,
firebase.auth.FacebookAuthProvider.PROVIDER_ID,
firebase.auth.EmailAuthProvider.PROVIDER_ID
],
// Terms of service url.
tosUrl: '<your-tos-url>'
};
// Initialize the FirebaseUI Widget using Firebase.
var ui = new firebaseui.auth.AuthUI(firebase.auth());
// The start method will wait until the DOM is loaded.
ui.start('#firebaseui-auth-container', uiConfig);
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但是出于安全原因,我希望用户确认她/他的电子邮件。但是从上面的代码中,它不会向用户发送验证邮件。因此,如果用户未验证他/她的帐户邮件,我已经使用以下方法将验证邮件发送给用户。
firebase.auth().onAuthStateChanged(function(user) {
if (user && user.uid != currentUid) {
if (firebase.auth().currentUser.emailVerified) {
currentUid = user.uid; …Run Code Online (Sandbox Code Playgroud) 我使用CodeIgniter作为我的Web框架,并从控制器中使用,这是我$this->load->view('Apartments_Hotels', $data第一次加载数据。但是现在我需要在视图中使用ajax函数将新数据重新加载到页面中。
视图中的Ajax函数
$(document).ready(function() {
$(".filter_checkbox").click(function(){
var filter = [];
$.each($("input[name='filter']:checked"), function(){
filter.push($(this).val());
});
$.ajax({
async: false,
type: 'POST',
data: {
filter: filter
},
url: "<?php echo base_url(); ?>index.php/ListingsController/showFilterListings",
success: function (response) {
}
});
});
});
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控制器功能
public function showFilterListings(){echo $this->load->view('Apartments_Hotels', $data, TRUE);
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}
但这是行不通的。任何人都可以使用ajax函数一次又一次地帮助我加载视图
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