Yes, you can get it from the [`File`](http://docs.oracle.com/javase/7/docs/api/java/io/File.html) object by using [`File.toPath()`](http://docs.oracle.com/javase/7/docs/api/java/io/File.html#toPath()). Keep in mind that this is only for Java 7+. Java versions 6 and below do not have it.

From [the documentation](http://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html):

&gt; Paths associated with the default `provider` are generally interoperable
&gt; with the `java.io.File` class. Paths created by other providers are
&gt; unlikely to be interoperable with the abstract path names represented
&gt; by `java.io.File`. **The `toPath` method may be used to obtain a Path from
&gt; the abstract path name represented by a java.io.File object.** The
&gt; resulting Path can be used to operate on the same file as the
&gt; `java.io.File` object. **In addition, the `toFile` method is useful to
&gt; construct a `File` from the `String` representation of a `Path`.**

(emphasis mine)

So, for [`toFile`](http://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html#toFile()):

&gt; Returns a `File` object representing this path.

And [`toPath`](http://docs.oracle.com/javase/7/docs/api/java/io/File.html#toPath()):
&gt; Returns a `java.nio.file.Path` object constructed from the this abstract path.

You likely want [`File.toPath()`][1].


  [1]: http://docs.oracle.com/javase/7/docs/api/java/io/File.html#toPath%28%29

As many have suggested, JRE v1.7 and above has File.toPath();

    File yourFile = ...;
    Path yourPath = yourFile.toPath();

On Oracle&#39;s [jdk 1.7 documentation][1] which is also mentioned in other posts above, the following equivalent code is described in the description for toPath() method, which may work for JRE v1.6;

    File yourFile = ...;
    Path yourPath = FileSystems.getDefault().getPath(yourFile.getPath());

  [1]: https://docs.oracle.com/javase/7/docs/api/java/io/File.html#toPath%28%29

I need to find unique rows in a `numpy.array`.

For example:

    &gt;&gt;&gt; a # I have
    array([[1, 1, 1, 0, 0, 0],
           [0, 1, 1, 1, 0, 0],
           [0, 1, 1, 1, 0, 0],
           [1, 1, 1, 0, 0, 0],
           [1, 1, 1, 1, 1, 0]])
    &gt;&gt;&gt; new_a # I want to get to
    array([[1, 1, 1, 0, 0, 0],
           [0, 1, 1, 1, 0, 0],
           [1, 1, 1, 1, 1, 0]])

I know that i can create a set and loop over the array, but I am looking for an efficient pure `numpy` solution. I believe that there is a way to set data type to void and then I could just use `numpy.unique`, but I couldn&#39;t figure out how to make it work.



Find unique rows in numpy.array

I&#39;m trying to do some things on the ActionBar in Android.

I&#39;ve already added new items in the right side of the action bar.

How can I change the left side of the action bar? I want to change the icon and the text, and I want to add a &quot;Back Button&quot; in the action bar for the other screens

![Android Action Bar][1]


  [1]: http://i.stack.imgur.com/9w4cI.png

How do I change the android actionbar title and icon

Is it possible to get a Path object from a `java.io.File`?

I know you can convert a path to a file using `toFile()` method, but I couldn&#39;t find the opposite conversion. Is there a way to do this in Java 6 or lower?


Get java.nio.file.Path object from java.io.File

<p>Is it possible to get a Path object from a <code>java.io.File</code>?</p>
<p>I know you can convert a path to a file using <code>toFile()</code> method, but I couldn't find the opposite conversion. Is there a way to do this in Java 6 or lower?</p>


I want to implement custom `ActionBar` which must look like this:

![enter image description here][1]


  [1]: http://i.stack.imgur.com/vrjd7.png

So questions:

 1. How can I implement a button like custom view: just some image?
 2. How can I draw a line on the top of the `ActionBar`?
 3. And how can I implement buttons without separator lines: add tabs on the `ActionBar` or what?

How can I implement custom Action Bar with custom buttons in Android?

I understand `Enum` is Serializable. Hence, it is safe to do so. (selectedCountry is `enum Country`)

Original enum without customer member variables
-----------------------------------------------

    public enum Country {
        Australia,
        Austria,
        UnitedState;
    }

Fragment
--------

    @Override
    public void onActivityCreated (Bundle savedInstanceState) {
        super.onActivityCreated(savedInstanceState);
        if (savedInstanceState != null) {
            selectedCountry = (Country)savedInstanceState.getSerializable(SELECTED_COUNTRY_KEY);
        }
    }
    
    @Override
    public void onSaveInstanceState(Bundle savedInstanceState) {
        savedInstanceState.putSerializable(SELECTED_COUNTRY_KEY, selectedCountry);
    }

However, what if I have non-serializable members in custom enum class? For instance,

Original enum customer member variables
---------------------------------------

    package org.yccheok;
    
    import org.yccheok.R;
    
    /**
     *
     * @author yccheok
     */
    public enum Country {
        Australia(R.drawable.flag_au),
        Austria(R.drawable.flag_at),
        UnitedState(R.drawable.flag_us);
     
        Country(int icon) {
            this.icon = icon;
            nonSerializableClass = new NonSerializableClass(this.toString());
        }
        
        public int getIcon() {
            return icon;
        }
        
        public static class NonSerializableClass {
            public NonSerializableClass(String dummy) { this.dummy = dummy; }
            public String dummy;
        }
        
        private final int icon;
        
        public NonSerializableClass nonSerializableClass;
    }

I tested. It works. (I tested by printing out all the value of member variables before and after serialization. They are same before and after) 

**However, I do not understand why it works? As I do not provide proper `readObject` and `writeObject`, as required by `Serializable` interface.**

As pointed in Effective Java *Item 75: Consider using a custom serialized form*, do I need to provide my own `readObject` and `writeObject`, if I have custom member variables in my enum?

Is custom enum Serializable too?

I&#39;m putting values into the hashmap which is of the form,

	Map&lt;Long, Double&gt; highLowValueMap=new HashMap&lt;Long, Double&gt;();
	highLowValueMap.put(1l, 10.0);
	highLowValueMap.put(2l, 20.0);

I want to create a list by using `values()` method of map.

    List&lt;Double&gt; valuesToMatch=new ArrayList&lt;&gt;();
    valuesToMatch=(List&lt;Double&gt;) highLowValueMap.values();
or 

    List&lt;Double&gt; valuesToMatch=(List&lt;Double&gt;) highLowValueMap.values();

However, it throws an exception:

&gt; Exception in thread &quot;main&quot; java.lang.ClassCastException:  
&gt; java.util.HashMap$Values cannot be cast to java.util.List

But it allows me to pass it in to the creation of a list:

    List&lt;Double&gt; valuesToMatch  = new ArrayList&lt;Double&gt;( highLowValueMap.values());



why HashMap Values are not cast in List?

I am new to Maven, I have a Java based web project with maven configured in my MyEclipse.   
Now if I modified any java files then do I need to do `Run as -&gt; Mvn install` or `Mvn package`?

Mvn install or Mvn package

Let&#39;s say I have a Java `Person` class:

    class Person {
        String name;
        String email;
    }

With REST Assured, you can deserialize this JSON object

    {&quot;name&quot;:&quot;Bob&quot;, &quot;email&quot;:&quot;bob@email.com&quot;} 

to a Java `Person` instance using

    Person bob = given().when().get(&quot;person/Bob/&quot;).as(Person.class);

How does one use REST Assured to deserialize this JSON array

    [{&quot;name&quot;:&quot;Bob&quot;, &quot;email&quot;:&quot;bob@email.com&quot;}, 
     {&quot;name&quot;:&quot;Alice&quot;, &quot;email&quot;:&quot;alice@email.com&quot;}, 
     {&quot;name&quot;:&quot;Jay&quot;, &quot;email&quot;:&quot;jay@email.com&quot;}]

into a `List&lt;Person&gt;`? For example, this would be handy:

    List&lt;Person&gt; persons = given().when().get(&quot;person/&quot;).as(...);

REST Assured - Generic List deserialization

To read some text file, in C or Pascal, I always use the following snippets to read the data until EOF:

    while not eof do begin
      readline(a);
      do_something;
    end;

Thus, I wonder how can I do this simple and fast in Python? 

What is the perfect counterpart in Python for &quot;while not EOF&quot;

Recently I am using Python module os, when I tried to change the permission of a file, I did not get the expected result. For example, I intended to change the permission to rw-rw-r--,

    os.chmod(&quot;/tmp/test_file&quot;, 664)

The ownership permission is actually -w--wx--- (230)

    --w--wx--- 1 ag ag 0 Mar 25 05:45 test_file

However, if I change 664 to 0664 in the code, the result is just what I need, e.g.

    os.chmod(&quot;/tmp/test_file&quot;, 0664)

The result is:

    -rw-rw-r-- 1 ag ag 0 Mar 25 05:55 test_file

Could anybody help explaining why does that leading 0 is so important to get the correct result?

Python module os.chmod(file, 664) does not change the permission to rw-rw-r-- but -w--wx----

I&#39;m trying to use `file_get_contents` together with `stream_context_create` to make POST requests. My code so far:

		$options = array(&#39;http&#39; =&gt; array(
			&#39;method&#39;  =&gt; &#39;POST&#39;,
			&#39;content&#39; =&gt; $data,
			&#39;header&#39;  =&gt; 
				&quot;Content-Type: text/plain\r\n&quot; .
		    	&quot;Content-Length: &quot; . strlen($data) . &quot;\r\n&quot;
		));
		$context  = stream_context_create($options);
		$response = file_get_contents($url, false, $context);

It works fine, however, when an HTTP error occurs, it spits out a warning:

    file_get_contents(...): failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request

and returns false. Is there a way to:

- suppress a warning (I&#39;m planning to throw my own exception in case of failure)
- obtain the error information (at least, the response code) from the stream



HTTP requests with file_get_contents, getting the response code

This line:

    final FileOutputStream outputStream = new FileOutputStream(name);

results in a `FileNotFoundException` with the message being `/2ozjfFQzwv: open failed: EROFS (Read-only file system)` where &quot;2ozjfFQzwv&quot; is what I passed as the name of the file.

I have tried this with and without the `WRITE_INTERNAL_STORAGE` permission. How do I create this file for writing?

Alternatively, I just want to be able to give an image to a new activity, and it is too large to serialize it in an extra. Is there an easier way than writing it to a file then reading from it again? All the questions on here seem to be about writing to an SD card, which I don&#39;t want to do because many people don&#39;t have SD card slots.

Trying to create a file in Android: open failed: EROFS (Read-only file system)

I have a JAR file and I need to get the name of all classes inside this JAR file. How can I do that?

I googled it and saw something about JarFile or Java `ClassLoader` but I have no idea how to do it.


How to get names of classes inside a jar file?

How can I extract path and filename from a variable?

    Setlocal EnableDelayedExpansion
    set file=C:\Users\l72rugschiri\Desktop\fs.cfg

I want to do that without using any function or any GOTO.
is it possible?



Batch Extract path and filename from a variable

I can&#39;t seem to get any search results that explain how to do this.

All I want to do is be able to know if a given path is a file or a directory (folder).

Node.js check if path is file or directory

I want to create a program that converts files. I would like the user to be able to place the executable file in any directory, and when executing that program (double-clicking on the .exe) I want the program to process **all the files within the current folder where the exe file exists**. How can the program determine the path in which it is currently executing?

I tried `System.Windows.Forms.Application.StartupPath` but that seems to be the wrong way.

Any ideas?

Get current folder path

I tend to use only forward slashes for paths (&#39;/&#39;) and python is happy with it also on windows.
In the description of os.path.join it says that is the correct way if you want to go cross-platform. But when I use it I get mixed slashes:

    import os

    a = &#39;c:/&#39;
    b = &#39;myFirstDirectory/&#39;
    c = &#39;mySecondDirectory&#39;
    d = &#39;myThirdDirectory&#39;
    e = &#39;myExecutable.exe&#39;


    print os.path.join(a, b, c, d, e)

    # Result:
    c:/myFirstDirectory/mySecondDirectory\myThirdDirectory\myExecutable.exe

Is this correct? Should I check and correct it afterward or there is a better way?

Thanks

EDIT:
I also get mixed slashes when asking for paths

    import sys
    for item in sys.path:
        print item
        
    # Result:
    C:\Program Files\Autodesk\Maya2013.5\bin
    C:\Program Files\Autodesk\Maya2013.5\mentalray\scripts\AETemplates
    C:\Program Files\Autodesk\Maya2013.5\Python
    C:\Program Files\Autodesk\Maya2013.5\Python\lib\site-packages
    C:\Program Files\Autodesk\Maya2013.5\bin\python26.zip\lib-tk
    C:/Users/nookie/Documents/maya/2013.5-x64/prefs/scripts
    C:/Users/nookie/Documents/maya/2013.5-x64/scripts
    C:/Users/nookie/Documents/maya/scripts
    C:\Program Files\Nuke7.0v4\lib\site-packages
    C:\Program Files\Nuke7.0v4/plugins/modules

mixed slashes with os.path.join on windows

How can I create a [`java.nio.file.Path`][1] object from a `String` object in Java 7?

I.e.

    String textPath = &quot;c:/dir1/dir2/dir3&quot;;
    Path path = ?;

where `?` is the missing code that uses `textPath`.

  [1]: http://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html

Create a Path from String in Java7

I am a new comer to the android world and with a .NET background. I am trying to install the Android Studio but I am having a mismatch on Android.com about the JDK version required for developing Android applications.

From the [SDK web page][1] it states that **JDK 6** is required (under system requirements section). [Jim Wilson][2] also insist in [his course on Pluralsight][3] (which is quite new, from last May) that using version 7 with the Android SDK will create a lot of headache and makes it clear to only use JDK 6. 

Although in the [Android studio page][4] (Under installing step 2) they explain how to set an environment variable indicating the correct JDK location. In here they use a path of **JDK 7**

&gt; Select `Start menu &gt; Computer &gt; System Properties &gt; Advanced System Properties`. Then open `Advanced tab &gt; Environment Variables` and add a new system variable `JAVA_HOME` that points to your JDK folder, for example `C:\Program Files\Java\jdk1.7.0_21`.

I am pretty sure both the ADT &amp; Eclipse bundle and the Android Studio bundle are using the same SDK but now I am confused about which JDK version I need to install. My sole purpose is to avoid problems as much as possible while getting my first steps into the Android development.

Disclaimer: I am not related by any mean to Pluralsight. I am just a normal user following the courses.

  [1]: http://developer.android.com/sdk/index.html
  [2]: http://pluralsight.com/training/Authors/Details/jim-wilson
  [3]: http://www.pluralsight.com/training/Courses/TableOfContents/android-dotnet-developers-getting-started
  [4]: http://developer.android.com/sdk/installing/studio.html#Installing

Which JDK version (Language Level) is required for Android Studio?

On my machine I have two Java versions installed: (1.6 and 1.7 installed manually by me).  I need both of them for different projects. But for Maven I need 1.7, but my Maven uses the 1.6 Java version.

How can I set Maven to use 1.7?


How to set specific Java version to Maven?

Preparing for the Oracle Certified Associate Java SE 8 Programmer 1 exam, I came across the following paragraph about the ternary expression in the official Study Guide:

&gt;**Ternary Expression Evaluation**
&gt;&lt;br/&gt;
&gt;As of Java 7, only one of the right-hand expressions of the ternary operator will be evaluated at runtime. In a manner similar to the short-circuit operators, if one of the two right-hand expressions in a ternary operator performs a side effect, then it may not be applied at runtime. Let&#39;s illustrate this principle with the following example: [...]

It says that only one of the two expressions is evaluated, demonstrating with the following example:

    int y = 1;
    int z = 1;
    int a = y &lt; 10 ? y++ : z++;

Here, only `y` increments, but `z` does not, as you would expect.

What I am stumbling across is the beginning of the paragraph (marked in yellow) where it says &quot;As of Java 7, ...&quot;. I tested the same code with Java 1.6 and I can&#39;t find a difference in the behavior. I expected Java 1.6 to evaluate both expressions just from the information given in the paragraph. Does anyone have an idea what they wanted to say with &quot;As of Java 7,  ...&quot;?

**Edit:** To avoid confusion: It boils down to the question, Since they write &#39;As of Java 7&#39;, was there anything that changed concerning the ternary operator, when switching from Java 6 to Java 7?


  [1]: http://i.stack.imgur.com/PaazW.jpg

Content Type	Original Author	Original Content on Stackoverflow
Question	Karlovsky120	View Question on Stackoverflow
Solution 1 - Java	Vivin Paliath	View Answer on Stackoverflow
Solution 2 - Java	wchargin	View Answer on Stackoverflow
Solution 3 - Java	hertzsprung	View Answer on Stackoverflow
Solution 4 - Java	dArtagnon	View Answer on Stackoverflow

Get java.nio.file.Path object from java.io.File

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How do I change the android actionbar title and icon

Find unique rows in numpy.array

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