What does yield do in python

Shortcut to understanding yield

When you see a function with yield statements, apply this easy trick to understand what will happen:

  1. Insert a line result = [] at the start of the function.
  2. Replace each yield expr with result.append(expr).
  3. Insert a line return result at the bottom of the function.
  4. Yay – no more yield statements! Read and figure out code.
  5. Compare function to the original definition.

This trick may give you an idea of the logic behind the function, but what actually happens with yield is significantly different than what happens in the list-based approach. In many cases, the yield approach will be a lot more memory efficient and faster too. In other cases, this trick will get you stuck in an infinite loop, even though the original function works just fine. Read on to learn more…

Don’t confuse your Iterables, Iterators, and Generators

First, the iterator protocol – when you write

for x in mylist:
...loop body...

Python performs the following two steps:


  1. Gets an iterator for mylist:
  2. Call iter(mylist) -> this returns an object with a next() method (or __next__() in Python 3).
  3. [This is the step most people forget to tell you about]


  1. Uses the iterator to loop over items:
  2. Keep calling the next() method on the iterator returned from step 1. The return value from next() is assigned to x and the loop body is executed. If an exception StopIteration is raised from within next(), it means there are no more values in the iterator and the loop is exited.

The truth is Python performs the above two steps anytime it wants to loop over the contents of an object – so it could be a for loop, but it could also be code like otherlist.extend(mylist) (where otherlist is a Python list).

Here mylist is iterable because it implements the iterator protocol. In a user-defined class, you can implement the __iter__() method to make instances of your class iterable. This method should return an iterator. An iterator is an object with a next() method. It is possible to implement both __iter__() and next() on the same class, and have __iter__() return self. This will work for simple cases, but not when you want two iterators looping over the same object at the same time.

So that’s the iterator protocol, many objects implement this protocol:

  1. Built-in lists, dictionaries, tuples, sets, files.
  2. User-defined classes that implement __iter__().
  3. Generators.

Note that a for loop doesn’t know what kind of object it’s dealing with – it just follows the iterator protocol, and is happy to get item after item as it calls next(). Built-in lists return their items one by one, dictionaries return the keys one by one, files return the lines one by one, etc. And generators return… well that’s where yield comes in:

def f123():
yield 1
yield 2
yield 3

for item in f123():
print item

Instead of yield statements, if you had three return statements in f123() only the first would get executed, and the function would exit. But f123() is no ordinary function. When f123() is called, it does not return any of the values in the yield statements! It returns a generator object. Also, the function does not really exit – it goes into a suspended state. When the for loop tries to loop over the generator object, the function resumes from its suspended state at the very next line after the yield it previously returned from, executes the next line of code, in this case, a yield statement, and returns that as the next item. This happens until the function exits, at which point the generator raises StopIteration and the loop exits.

So the generator object is sort of like an adapter – at one end it exhibits the iterator protocol, by exposing __iter__() and next() methods to keep the for loop happy. At the other end, however, it runs the function just enough to get the next value out of it, and puts it back in suspended mode.

Why Use Generators?

Usually, you can write code that doesn’t use generators but implements the same logic. One option is to use the temporary list ‘trick’ I mentioned before. That will not work in all cases, for e.g. if you have infinite loops, or it may make inefficient use of memory when you have a really long list. The other approach is to implement a new iterable class SomethingIter that keeps the state in instance members and performs the next logical step in its next() (or __next__() in Python 3) method. Depending on the logic, the code inside the next() method may end up looking very complex and be prone to bugs. Here generators provide a clean and easy solution.

Could not GET ‘play-services-location/maven-metadata.xml’. Received status code 502 from server: Bad Gateway

It looks like a temporary issue, the server with these libraries is down. I have the same problem now with Room:

Could not GET 'https://google.bintray.com/exoplayer/androidx/room/room-common/maven-metadata.xml'. Received status code 502 from server: Bad Gateway

You can try using offline mode if you’re using Android Studio, then it will use the cached version of this library if you have it until it is fixed.

UPD. I switched to an alpha version of a flutter lib which caused this (workmanager) and it works well now. As far as I understand it was depending on an old version of Android Room library which is not available anymore since Bintray is not available. The new version of Room is available as it’s being downloaded thorough another link. So for you the solution could be updating to a newer version of Flutter location package or forking it and changing the version of play-services-location to the most recent one.

Failed to start DevTools: Dart DevTools exited with code 255

I am getting this error code in my vscode. How to fix it? I am using flutter v2.5.3.

You can try fixing it by running this in the terminal:

Just copy and paste the below code into the terminal and run it.

dart pub global activate devtools -v 2.8.0

which downgrades the version to 2.8.0 (that works fine). Found answer on GitHub

Android hide/close soft keyboard programmatically?

Solution 1

To help clarify this madness, I’d like to begin by apologizing on behalf of all Android users for Google’s downright ridiculous treatment of the soft keyboard. The reason there are so many answers, each different, for the same simple question is that this API, like many others in Android, is horribly designed. I can think of no polite way to state it.

I want to hide the keyboard. I expect to provide Android with the following statement: Keyboard.hide(). The end. Thank you very much. But Android has a problem. You must use the InputMethodManager to hide the keyboard. OK, fine, this is Android’s API to the keyboard. BUT! You are required to have a Context in order to get access to the IMM. Now we have a problem. I may want to hide the keyboard from a static or utility class that has no use or need for any Context. or And FAR worse, the IMM requires that you specify what View (or even worse, what Window) you want to hide the keyboard FROM.

This is what makes hiding the keyboard so challenging. Dear Google: When I’m looking up the recipe for a cake, there is no RecipeProvider on Earth that would refuse to provide me with the recipe unless I first answer WHO the cake will be eaten by AND where it will be eaten!!

This sad story ends with the ugly truth: to hide the Android keyboard, you will be required to provide 2 forms of identification: a Context and either a View or a Window.

I have created a static utility method that can do the job VERY solidly, provided you call it from an Activity.

public static void hideKeyboard(Activity activity) {
    InputMethodManager imm = (InputMethodManager) activity.getSystemService(Activity.INPUT_METHOD_SERVICE);
    //Find the currently focused view, so we can grab the correct window token from it.
    View view = activity.getCurrentFocus();
    //If no view currently has focus, create a new one, just so we can grab a window token from it
    if (view == null) {
        view = new View(activity);
    imm.hideSoftInputFromWindow(view.getWindowToken(), 0);

Be aware that this utility method ONLY works when called from an Activity! The above method calls getCurrentFocus of the target Activity to fetch the proper window token.

But suppose you want to hide the keyboard from an EditText hosted in a DialogFragment? You can’t use the method above for that:

hideKeyboard(getActivity()); //won't work

This won’t work because you’ll be passing a reference to the Fragment‘s host Activity, which will have no focused control while the Fragment is shown! Wow! So, for hiding the keyboard from fragments, I resort to the lower-level, more common, and uglier:

public static void hideKeyboardFrom(Context context, View view) {
    InputMethodManager imm = (InputMethodManager) context.getSystemService(Activity.INPUT_METHOD_SERVICE);
    imm.hideSoftInputFromWindow(view.getWindowToken(), 0);

Below is some additional information gleaned from more time wasted chasing this solution:

About windowSoftInputMode

There’s yet another point of contention to be aware of. By default, Android will automatically assign initial focus to the first EditText or focusable control in your Activity. It naturally follows that the InputMethod (typically the soft keyboard) will respond to the focus event by showing itself. The windowSoftInputMode attribute in AndroidManifest.xml, when set to stateAlwaysHidden, instructs the keyboard to ignore this automatically-assigned initial focus.


Almost unbelievably, it appears to do nothing to prevent the keyboard from opening when you touch the control (unless focusable=”false” and/or focusableInTouchMode=”false” are assigned to the control). Apparently, the windowSoftInputMode setting applies only to automatic focus events, not to focus events triggered by touch events.

Therefore, stateAlwaysHidden is VERY poorly named indeed. It should perhaps be called ignoreInitialFocus instead.


UPDATE: More ways to get a window token

If there is no focused view (e.g. can happen if you just changed fragments), there are other views that will supply a useful window token.

These are alternatives for the above code if (view == null) view = new View(activity); These don’t refer explicitly to your activity.

Inside a fragment class:

view = getView().getRootView().getWindowToken();

Given a fragment fragment as a parameter:

view = fragment.getView().getRootView().getWindowToken();

Starting from your content body:

view = findViewById(android.R.id.content).getRootView().getWindowToken();

UPDATE 2: Clear focus to avoid showing keyboard again if you open the app from the background

Add this line to the end of the method:


React Router Programmatically Change route

React-Router v6+ Answer

You can use the new useNavigate hook. useNavigate hook returns a function that can be used for programmatic navigation. Example from the react-router documentation


import { useNavigate } from "react-router-dom";

function SignupForm() {
  let navigate = useNavigate();

  async function handleSubmit(event) {
    await submitForm(event.target);
    navigate("../success", { replace: true });

  return &lt;form onSubmit={handleSubmit}&gt;{/* ... */}&lt;/form&gt;;

React-Router 5.1.0+ Answer (using hooks and React >16.8)

You can use the useHistory hook on Functional Components and Programmatically navigate:

import { useHistory } from "react-router-dom";

function HomeButton() {
  let history = useHistory();
  // use history.push('/some/path') here

React-Router 4.0.0+ Answer

In 4.0 and above, use the history as a prop of your component.

class Example extends React.Component {
// use `this.props.history.push('/some/path')` here

NOTE: this.props.history does not exist in the case your component was not rendered by . You should use to have this.props.history in YourComponent


React-Router 3.0.0+ Answer

In 3.0 and above, use the router as a prop of your component.

class Example extends React.Component {
// use `this.props.router.push('/some/path')` here

React-Router 2.4.0+ Answer


In 2.4 and above, use a higher order component to get the router as a prop of your component.

import { withRouter } from 'react-router';

class Example extends React.Component {
// use `this.props.router.push('/some/path')` here

// Export the decorated class
var DecoratedExample = withRouter(Example);

// PropTypes
Example.propTypes = {
router: React.PropTypes.shape({
push: React.PropTypes.func.isRequired

How to use requirements.txt to install all dependencies in a python project

Solution 1

If you are using Linux OS:

  1. Remove matplotlib==1.3.1 from requirements.txt
  2. Try to install with sudo apt-get install python-matplotlib
  3. Run pip install -r requirements.txt (Python 2), or pip3 install -r requirements.txt (Python 3)
  4. pip freeze > requirements.txt

If you are using Windows OS:

  1. python -m pip install -U pip setuptools
  2. python -m pip install matplotlib


Solution 2


pip install -r requirements.txt for python 2.x

pip3 install -r requirements.txt for python 3.x (in case multiple versions are installed)