Python – Setting the Windows taskbar icon in PyQt

iconspyqtpythontaskbarwindows

I'm working on an applcation in Python's PyQt4 and cannot find how to change the taskbar icon. I made my .ui files in Qt's Designer, where I can change the windowIcon properties. But that is not what I am looking for. I want to change the look of the application's icon in windows taskbar. For now it is Python logo in a window icon.

I found some information on SO: link but it's not helping me much.

I tried:

app = QtGui.QApplication([])
app.setWindowIcon(QtGui.QIcon('chip_icon_normal.png'))
app.exec_()

But the icon remains unchanged.

What i want to change, showing the picture:
(This is done calling the setWindowIcon on main window/ dialog, or the application, as shown above.)

Taskbar icon change

Best Answer

This problem is caused by some peculiarities in how taskbar icons are handled on the Windows platform.

See this answer for details, along with a workaround using ctypes.

New answer, powershell

TCP

Get-Process -Id (Get-NetTCPConnection -LocalPort YourPortNumberHere).OwningProcess

UDP

Get-Process -Id (Get-NetUDPEndpoint -LocalPort YourPortNumberHere).OwningProcess

Old answer, cmd

 C:\> netstat -a -b

(Add -n to stop it trying to resolve hostnames, which will make it a lot faster.)

Note Dane's recommendation for TCPView. It looks very useful!

-a Displays all connections and listening ports.

-b Displays the executable involved in creating each connection or listening port. In some cases well-known executables host multiple independent components, and in these cases the sequence of components involved in creating the connection or listening port is displayed. In this case the executable name is in [] at the bottom, on top is the component it called, and so forth until TCP/IP was reached. Note that this option can be time-consuming and will fail unless you have sufficient permissions.

-n Displays addresses and port numbers in numerical form.

-o Displays the owning process ID associated with each connection.

Python – Finding the index of an item in a list

>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

Check for it first with item in my_list (clean, readable approach), or
Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)