How about:

if lastPen is None or np.any(pen != lastPen):

How about:

if lastPen is None or np.any(pen != lastPen):

This is obviously the more elegant solution 😅
Edited it.
@pijyoi Let me know if there is something else I need to do 😉

This LGTM, thanks for the PR @adrianlubitz and thanks for the review/feedback @pijyoi

@j9ac9k When will this be uploaded to PyPI as a new Version? I want to use it as a dependency in CoBaIR and PyPI only allows dependencies from PyPI itself. Looking forward to Version 0.13.4 📦 🥳

Looking at the code, I don't think that passing in red,green,blue,alpha,width as a numpy record array to specify the pens seems a good idea. (It's also possible to pass in a regular numpy array to specify r,g,b,a, but that's not documented in the docstring)
The lastPen variable is a sort of optimization to avoid recreating QPens, but it only works if the same-colored pens are adjacent.

I would think a more useful input parameter would be a Python list of (either QPen or something that mkPen accepts). This way, the user can use the same pre-created QPen object within the list. (Similar to what BarGraphItem code is doing)

Prior to this PR, use of multiple pens wouldn't have worked anyway. Would it be a good opportunity to fix the api now?

Would it be a good opportunity to fix the api now?

Now is always better than later.

Prior to this PR, use of multiple pens wouldn't have worked anyway.

Now I am puzzled. https://github.com/pyqtgraph/pyqtgraph/blob/master/pyqtgraph/examples/GraphItem.py specifies multiple pens and does work on 0.13.3. So what was this PR fixing?

In [40]: lines = np.array([
    ...:     (255,0,0,255,1),
    ...:     (255,0,255,255,2),
    ...:     (255,0,255,255,3),
    ...:     (255,255,0,255,2),
    ...:     (255,0,0,255,1),
    ...:     (255,255,255,255,4),
    ...:     ], dtype=[('red',np.ubyte),('green',np.ubyte),('blue',np.ubyte),('alpha',np.ubyte),('width',float)])

In [41]: lines[4] == None
Out[41]: False

In [42]: lines[4] != None
Out[42]: True

In [43]: np.any(lines[4] != None)
Out[43]: True

Prior to this PR, use of multiple pens wouldn't have worked anyway.

Now I am puzzled. https://github.com/pyqtgraph/pyqtgraph/blob/master/pyqtgraph/examples/GraphItem.py specifies multiple pens and does work on 0.13.3. So what was this PR fixing?

In [40]: lines = np.array([
    ...:     (255,0,0,255,1),
    ...:     (255,0,255,255,2),
    ...:     (255,0,255,255,3),
    ...:     (255,255,0,255,2),
    ...:     (255,0,0,255,1),
    ...:     (255,255,255,255,4),
    ...:     ], dtype=[('red',np.ubyte),('green',np.ubyte),('blue',np.ubyte),('alpha',np.ubyte),('width',float)])

In [41]: lines[4] == None
Out[41]: False

In [42]: lines[4] != None
Out[42]: True

In [43]: np.any(lines[4] != None)
Out[43]: True

Hi @pijyoi,
I haven't had much time lately but now I came to the point where I had some time to investigate.
My mistake was that I had my array nested one level to deep. It looked something like that:

lines = np.array([
    np.array([(255, 0, 0, 255, 1)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
    np.array([(255, 0, 255, 255, 2)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
    np.array([(255, 0, 255, 255, 3)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
    np.array([(255, 255, 0, 255, 2)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
    np.array([(255, 0, 0, 255, 1)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
    np.array([(255, 255, 255, 255, 4)], dtype=[
             ('red', np.ubyte), ('green', np.ubyte), ('blue', np.ubyte), ('alpha', np.ubyte), ('width', float)]),
])

With the solution from the the PR nested arrays work as well and additionally I could fix my issue and can use pyqtgraph 0.13.3 as dependency.

Thanks 😄