Related
Below is Bokeh 1.4.0 code that tries to draw a HexTile map of the input dataframe, with axes, and tries to place labels on each hex.
I've been stuck on this for two days solid, reading bokeh doc, examples and github known issues, SO, Bokeh Discourse and Red Blob Games's superb tutorial on Hexagonal Grids, and trying code. (I'm less interested in raising Bokeh issues for the future, and far more interested in pragmatic workarounds to known limitations to just get my map code working today.) Plot is below, and code at bottom.
Here are the issues, in rough decreasing order of importance (it's impossible to separate the root-cause and tell which causes which, due to the way Bokeh handles glyphs. If I apply one scale factor or coord transform it fixes one set of issues, but breaks another, 'whack-a-mole' effect):
The label placement is obviously wrong, but I can't seem to hack up any variant of either (x,y) coords or (q,r) coords to work. (I tried combinations of figure(..., match_aspect=True)), I tried 1/sqrt(2) scaling the (x,y)-coords, I tried Hextile(... size, scale) params as per redblobgames, e.g. size = 1/sqrt(3) ~ 0.57735).
Bokeh forces the origin to be top left, and y-coords to increase as you go down, however the default axis labels show y or r as being negative. I found I still had to use p.text(q, -r, .... I suppose I have to manually patch the auto-supplied yaxis labels or TickFormatter to be positive.
I use np.mgrid to generate the coord grid, but I still seem to have to assign q-coords right-to-left: np.mgrid[0:8, (4+1):0:-1]. Still no matter what I do, the hexes are flipped L-to-R
(Note: empty '' counties are placeholders to get the desired shape, hence the boolean mask [counties!=''] on grid coords. This works fine and I want to leave it as-is)
The source (q,r) coords for the hexes are integers, and I use 'odd-r' offset coords (not axial or hexagonal coords). No matter what HexTile(..., size, scale) args I use, one or both dimensions in the plot is wrong or squashed. Or whether I include the 1/sqrt(2) factor in coord transform.
My +q-axis is east and my +r-axis should be 120° SSE
Ideally I'd like to have my origin at bottom-left (math plot style, not computer graphics). But Bokeh apparently doesn't support that, I can live without that. However defaulting the y-axis labels to negative, while requiring a mix of positive and negative coords, is confusing. Anyway, how to hack an automatic fix to that with minimum grief? (manual p.yrange = Range1d(?, ?)?)
Bokeh's approach to attaching (hex) glyphs to plots is a hard idiom to use. Ideally I simply want to reference (q,r)-coords everywhere for hexes, labels, axes. I never want to see (x,y)-coords appearing on axes, label coords, tick-marks, etc. but seems Bokeh won't allow you. I guess you have to manually hack the axes and ticks later. Also, the plot<->glyph interface doesn't allow you to expose a (q,r) <-> (x,y) coord transform function, certainly not a bidirectional one.
The default axes don't seem to have any accessors to automatically find their current extent/limits; p.yaxis.start/end are empty unless you specified them. The result from p.yaxis.major_tick_in,p.yaxis.major_tick_out is also wrong, for this plot it gives (2,6) for both x and y, seems to be clipping those to the interior multiples of 2(?). How to automatically get the axes' extent?
My current plot:
My code:
import pandas as pd
import numpy as np
from math import sqrt
from bokeh.plotting import figure
from bokeh.models import ColumnDataSource
from bokeh.models.glyphs import HexTile
from bokeh.io import show
# Data source is a list of county abbreviations, in (q,r) coords...
counties = np.array([
['TE','DY','AM','DN', ''],
['DL','FM','MN','AH', ''],
['SO','LM','CN','LH', ''],
['MO','RN','LD','WH','MH'],
['GA','OY','KE','D', ''],
['', 'CE','LS','WW', ''],
['LC','TA','KK','CW', ''],
['KY','CR','WF','WX', ''],
])
#counties = counties[::-1] # UNUSED: flip so origin is at bottom-left
# (q,r) Coordinate system is “odd/even-r” horizontal Offset coords
r, q = np.mgrid[0:8, (4+1):0:-1]
q = q[counties!='']
r = r[counties!='']
sqrt3 = sqrt(3)
# Try to transform odd-r (q,r) offset coords -> (x,y). Per Red Blob Games' tutorial.
x = q - (r//2) # this may be slightly dubious
y = r
counties_df = pd.DataFrame({'q': q, 'r': r, 'abbrev': counties[counties!=''], 'x': x, 'y': y })
counties_ds = ColumnDataSource(ColumnDataSource.from_df(counties_df)) # ({'q': q, 'r': r, 'abbrev': counties[counties != '']})
p = figure(tools='save,crosshair') # match_aspect=True?
glyph = HexTile(orientation='pointytop', q='x', r='y', size=0.76, fill_color='#f6f699', line_color='black') # q,r,size,scale=??!?!!? size=0.76 is an empirical hack.
p.add_glyph(counties_ds, glyph)
p.xaxis.minor_tick_line_color = None
p.yaxis.minor_tick_line_color = None
print(f'Axes: x={p.xaxis.major_tick_in}:{p.xaxis.major_tick_out} y={p.yaxis.major_tick_in}:{p.yaxis.major_tick_out}')
# Now can't manage to get the right coords for text labels
p.text(q, -r, text=["(%d, %d)" % (q,r) for (q, r) in zip(q, r)], text_baseline="middle", text_align="center")
# Ideally I ultimately want to fix this and plot `abbrev` column as the text label
show(p)
There is an axial_to_cartesian function that will just compute the hex centers for you. You can then attach the labels in a variety of orientations and anchoring from these.
Bokeh does not force the origin to be anywhere. There is one axial to cartesian mapping Bokeh uses, exactly what is given by axial_to_cartesian. The position of the Hex tiles (and hence the cartesian coordinates that the axes display) follows from this. If you want different ticks, Bokeh affords lots of control points over both tick location and tick labelling.
There is more than one convention for Axial coords. Bokeh picked the one that has the r-axis tile "up an to the left", i.e. the one explicitly shown here:
https://docs.bokeh.org/en/latest/docs/user_guide/plotting.html#hex-tiles
Bokeh expects up-and-to-the-left axial coords. You will need to convert whatever coordinate system you have to that. For "squishing" you will need to set match_aspect=True to ensure the "data space" aspect ratio matches the "pixel space" aspect ratio 1-1.
Alternatively, if you don't or can't use auto-ranging you will need to set the plot size carefully and also control the border sizes with min_border_left etc to make sure the borders are always big enough to accommodate any tick labels you have (so that the inner region will not be resized)
I don't really understand this question, but you have absolute control over what ticks visually appear, regardless of the underlying tick data. Besides the built-in formatters, there is FuncTickFormatter that lets you format ticks any way you want with a snippet of JS code. [1] (And you also have control of where ticks are located, if you want that.)
[1] Please note the CoffeeScript and from_py_func options are both deprecated and being removed in then next 2.0 release.
Again, you'll want to use axial_to_cartesian to position anything other then Hex tiles. No other glyphs in Bokeh understand axial coordinates (which is why we provide the conversion function).
You misunderstood what major_tick_in and major_tick_out are for. They are literally how far the ticks visually extend inside and outside the plot frame, in pixels.
Auto-ranging (with DataRange1d) is only computed in the browser, in JavaScript, which is why the start/end are not available on the "Python" side. If you need to know the start/end, you will need to explicitly set the start/end, yourself. Note, however that match_aspect=True only function with DataRange1d. If you explicitly set start/end manually, Bokeh will assume you know what you want, and will honor what you ask for, regardless of what it does to aspect.
Below are my solution and plot. Mainly per #bigreddot's advice, but there's still some coordinate hacking needed:
Expecting users to pass input coords as axial instead of offset coords is a major limitation. I work around this. There's no point in creating a offset_to_cartesian() because we need to negate r in two out of three places:
My input is even-r offset coords. I still need to manually apply the offset: q = q + (r+1)//2
I need to manually negate r in both the axial_to_cartesian() call and the datasource creation for the glyph. (But not in the text() call).
The call needs to be: axial_to_cartesian(q, -r, size=2/3, orientation='pointytop')
Need p = figure(match_aspect=True ...) to prevent squishing
I need to manually create my x,y axes to get the range right
Solution:
import pandas as pd
import numpy as np
from math import sqrt
from bokeh.plotting import figure
from bokeh.models import ColumnDataSource, Range1d
from bokeh.models.glyphs import HexTile
from bokeh.io import curdoc, show
from bokeh.util.hex import cartesian_to_axial, axial_to_cartesian
counties = np.array([
['DL','DY','AM','', ''],
['FM','TE','AH','DN', ''],
['SO','LM','CN','MN', ''],
['MO','RN','LD','MH','LH'],
['GA','OY','WH','D' ,'' ],
['' ,'CE','LS','KE','WW'],
['LC','TA','KK','CW','' ],
['KY','CR','WF','WX','' ]
])
counties = np.flip(counties, (0)) # Flip UD for bokeh
# (q,r) Coordinate system is “odd/even-r” horizontal Offset coords
r, q = np.mgrid[0:8, 0:(4+1)]
q = q[counties!='']
r = r[counties!='']
# Transform for odd-r offset coords; +r-axis goes up
q = q + (r+1)//2
#r = -r # cannot globally negate 'r', see comments
# Transform odd-r offset coords (q,r) -> (x,y)
x, y = axial_to_cartesian(q, -r, size=2/3, orientation='pointytop')
counties_df = pd.DataFrame({'q': q, 'r': -r, 'abbrev': counties[counties!=''], 'x': x, 'y': y })
counties_ds = ColumnDataSource(ColumnDataSource.from_df(counties_df)) # ({'q': q, 'r': r, 'abbrev': counties[counties != '']})
p = figure(match_aspect=True, tools='save,crosshair')
glyph = HexTile(orientation='pointytop', q='q', r='r', size=2/3, fill_color='#f6f699', line_color='black') # q,r,size,scale=??!?!!?
p.add_glyph(counties_ds, glyph)
p.x_range = Range1d(-2,6)
p.y_range = Range1d(-1,8)
p.xaxis.minor_tick_line_color = None
p.yaxis.minor_tick_line_color = None
p.text(x, y, text=["(%d, %d)" % (q,r) for (q, r) in zip(q, r)],
text_baseline="middle", text_align="center")
show(p)
This question is very similar to those answered here,
matplotlib values under cursor
In a matplotlib figure window (with imshow), how can I remove, hide, or redefine the displayed position of the mouse?
Interactive pixel information of an image in Python?
except that instead of pixel data (x,y,z) I have various measurements associated with (x,y) coordinates that I'd like to portray on a line plot. Specifically, the (x,y) are spatial positions (lat, lon) and at each (lat,lon) point there is a collection of data (speed, RPM, temp, etc.). I just sketched up something quickly to illustrate, a scatter plot with connecting lines, and then when you hover over a data point it displays all of the "z" values associated with that data point.
Is there an easy way to do something like this?
You could probably build on something like this example. It doesn't display the information inside the figure (for now only using a print() statement), but it demonstrates a simple method of capturing clicks on scatter points and showing information for those points:
import numpy as np
import matplotlib.pylab as pl
pl.close('all')
n = 10
lat = np.random.random(n)
lon = np.random.random(n)
speed = np.random.random(n)
rpm = np.random.random(n)
temp = np.random.random(n)
def on_click(event):
i = event.ind[0]
print('lon={0:.2f}, lat={1:.2f}: V={2:.2f}, RPM={3:.2f}, T={4:.2f}'\
.format(lon[i], lat[i], speed[i], rpm[i], temp[i]))
fig=pl.figure()
pl.plot(lon, lat, '-')
pl.scatter(lon, lat, picker=True)
fig.canvas.mpl_connect('pick_event', on_click)
Clicking around a bit gives me:
lon=0.63, lat=0.58: V=0.51, RPM=0.00, T=0.43
lon=0.41, lat=0.07: V=0.95, RPM=0.59, T=0.98
lon=0.86, lat=0.13: V=0.33, RPM=0.27, T=0.85
I am trying to plot contour lines of pressure level. I am using a netCDF file which contain the higher resolution data (ranges from 3 km to 27 km). Due to higher resolution data set, I get lot of pressure values which are not required to be plotted (rather I don't mind omitting certain contour line of insignificant values). I have written some plotting script based on the examples given in this link http://matplotlib.org/basemap/users/examples.html.
After plotting the image looks like this
From the image I have encircled the contours which are small and not required to be plotted. Also, I would like to plot all the contour lines smoother as mentioned in the above image. Overall I would like to get the contour image like this:-
Possible solution I think of are
Find out the number of points required for plotting contour and mask/omit those lines if they are small in number.
or
Find the area of the contour (as I want to omit only circled contour) and omit/mask those are smaller.
or
Reduce the resolution (only contour) by increasing the distance to 50 km - 100 km.
I am able to successfully get the points using SO thread Python: find contour lines from matplotlib.pyplot.contour()
But I am not able to implement any of the suggested solution above using those points.
Any solution to implement the above suggested solution is really appreciated.
Edit:-
# Andras Deak
I used print 'diameter is ', diameter line just above del(level.get_paths()[kp]) line to check if the code filters out the required diameter. Here is the filterd messages when I set if diameter < 15000::
diameter is 9099.66295612
diameter is 13264.7838257
diameter is 445.574234531
diameter is 1618.74618114
diameter is 1512.58974168
However the resulting image does not have any effect. All look same as posed image above. I am pretty sure that I have saved the figure (after plotting the wind barbs).
Regarding the solution for reducing the resolution, plt.contour(x[::2,::2],y[::2,::2],mslp[::2,::2]) it works. I have to apply some filter to make the curve smooth.
Full working example code for removing lines:-
Here is the example code for your review
#!/usr/bin/env python
from netCDF4 import Dataset
import matplotlib
matplotlib.use('agg')
import matplotlib.pyplot as plt
import numpy as np
import scipy.ndimage
from mpl_toolkits.basemap import interp
from mpl_toolkits.basemap import Basemap
# Set default map
west_lon = 68
east_lon = 93
south_lat = 7
north_lat = 23
nc = Dataset('ncfile.nc')
# Get this variable for later calucation
temps = nc.variables['T2']
time = 0 # We will take only first interval for this example
# Draw basemap
m = Basemap(projection='merc', llcrnrlat=south_lat, urcrnrlat=north_lat,
llcrnrlon=west_lon, urcrnrlon=east_lon, resolution='l')
m.drawcoastlines()
m.drawcountries(linewidth=1.0)
# This sets the standard grid point structure at full resolution
x, y = m(nc.variables['XLONG'][0], nc.variables['XLAT'][0])
# Set figure margins
width = 10
height = 8
plt.figure(figsize=(width, height))
plt.rc("figure.subplot", left=.001)
plt.rc("figure.subplot", right=.999)
plt.rc("figure.subplot", bottom=.001)
plt.rc("figure.subplot", top=.999)
plt.figure(figsize=(width, height), frameon=False)
# Convert Surface Pressure to Mean Sea Level Pressure
stemps = temps[time] + 6.5 * nc.variables['HGT'][time] / 1000.
mslp = nc.variables['PSFC'][time] * np.exp(9.81 / (287.0 * stemps) * nc.variables['HGT'][time]) * 0.01 + (
6.7 * nc.variables['HGT'][time] / 1000)
# Contour only at 2 hpa interval
level = []
for i in range(mslp.min(), mslp.max(), 1):
if i % 2 == 0:
if i >= 1006 and i <= 1018:
level.append(i)
# Save mslp values to upload to SO thread
# np.savetxt('mslp.txt', mslp, fmt='%.14f', delimiter=',')
P = plt.contour(x, y, mslp, V=2, colors='b', linewidths=2, levels=level)
# Solution suggested by Andras Deak
for level in P.collections:
for kp,path in enumerate(level.get_paths()):
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter < 15000: # threshold to be refined for your actual dimensions!
#print 'diameter is ', diameter
del(level.get_paths()[kp]) # no remove() for Path objects:(
#level.remove() # This does not work. produces ValueError: list.remove(x): x not in list
plt.gcf().canvas.draw()
plt.savefig('dummy', bbox_inches='tight')
plt.close()
After the plot is saved I get the same image
You can see that the lines are not removed yet. Here is the link to mslp array which we are trying to play with http://www.mediafire.com/download/7vi0mxqoe0y6pm9/mslp.txt
If you want x and y data which are being used in the above code, I can upload for your review.
Smooth line
You code to remove the smaller circles working perfectly. However the other question I have asked in the original post (smooth line) does not seems to work. I have used your code to slice the array to get minimal values and contoured it. I have used the following code to reduce the array size:-
slice = 15
CS = plt.contour(x[::slice,::slice],y[::slice,::slice],mslp[::slice,::slice], colors='b', linewidths=1, levels=levels)
The result is below.
After searching for few hours I found this SO thread having simmilar issue:-
Regridding regular netcdf data
But none of the solution provided over there works.The questions similar to mine above does not have proper solutions. If this issue is solved then the code is perfect and complete.
General idea
Your question seems to have 2 very different halves: one about omitting small contours, and another one about smoothing the contour lines. The latter is simpler, since I can't really think of anything else other than decreasing the resolution of your contour() call, just like you said.
As for removing a few contour lines, here's a solution which is based on directly removing contour lines individually. You have to loop over the collections of the object returned by contour(), and for each element check each Path, and delete the ones you don't need. Redrawing the figure's canvas will get rid of the unnecessary lines:
# dummy example based on matplotlib.pyplot.clabel example:
import matplotlib
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
delta = 0.025
x = np.arange(-3.0, 3.0, delta)
y = np.arange(-2.0, 2.0, delta)
X, Y = np.meshgrid(x, y)
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
# difference of Gaussians
Z = 10.0 * (Z2 - Z1)
plt.figure()
CS = plt.contour(X, Y, Z)
for level in CS.collections:
for kp,path in reversed(list(enumerate(level.get_paths()))):
# go in reversed order due to deletions!
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter<1: # threshold to be refined for your actual dimensions!
del(level.get_paths()[kp]) # no remove() for Path objects:(
# this might be necessary on interactive sessions: redraw figure
plt.gcf().canvas.draw()
Here's the original(left) and the removed version(right) for a diameter threshold of 1 (note the little piece of the 0 level at the top):
Note that the top little line is removed while the huge cyan one in the middle doesn't, even though both correspond to the same collections element i.e. the same contour level. If we didn't want to allow this, we could've called CS.collections[k].remove(), which would probably be a much safer way of doing the same thing (but it wouldn't allow us to differentiate between multiple lines corresponding to the same contour level).
To show that fiddling around with the cut-off diameter works as expected, here's the result for a threshold of 2:
All in all it seems quite reasonable.
Your actual case
Since you've added your actual data, here's the application to your case. Note that you can directly generate the levels in a single line using np, which will almost give you the same result. The exact same can be achieved in 2 lines (generating an arange, then selecting those that fall between p1 and p2). Also, since you're setting levels in the call to contour, I believe the V=2 part of the function call has no effect.
import numpy as np
import matplotlib.pyplot as plt
# insert actual data here...
Z = np.loadtxt('mslp.txt',delimiter=',')
X,Y = np.meshgrid(np.linspace(0,300000,Z.shape[1]),np.linspace(0,200000,Z.shape[0]))
p1,p2 = 1006,1018
# this is almost the same as the original, although it will produce
# [p1, p1+2, ...] instead of `[Z.min()+n, Z.min()+n+2, ...]`
levels = np.arange(np.maximum(Z.min(),p1),np.minimum(Z.max(),p2),2)
#control
plt.figure()
CS = plt.contour(X, Y, Z, colors='b', linewidths=2, levels=levels)
#modified
plt.figure()
CS = plt.contour(X, Y, Z, colors='b', linewidths=2, levels=levels)
for level in CS.collections:
for kp,path in reversed(list(enumerate(level.get_paths()))):
# go in reversed order due to deletions!
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter<15000: # threshold to be refined for your actual dimensions!
del(level.get_paths()[kp]) # no remove() for Path objects:(
# this might be necessary on interactive sessions: redraw figure
plt.gcf().canvas.draw()
plt.show()
Results, original(left) vs new(right):
Smoothing by resampling
I've decided to tackle the smoothing problem as well. All I could come up with is downsampling your original data, then upsampling again using griddata (interpolation). The downsampling part could also be done with interpolation, although the small-scale variation in your input data might make this problem ill-posed. So here's the crude version:
import scipy.interpolate as interp #the new one
# assume you have X,Y,Z,levels defined as before
# start resampling stuff
dN = 10 # use every dN'th element of the gridded input data
my_slice = [slice(None,None,dN),slice(None,None,dN)]
# downsampled data
X2,Y2,Z2 = X[my_slice],Y[my_slice],Z[my_slice]
# same as X2 = X[::dN,::dN] etc.
# upsampling with griddata over original mesh
Zsmooth = interp.griddata(np.array([X2.ravel(),Y2.ravel()]).T,Z2.ravel(),(X,Y),method='cubic')
# plot
plt.figure()
CS = plt.contour(X, Y, Zsmooth, colors='b', linewidths=2, levels=levels)
You can freely play around with the grids used for interpolation, in this case I just used the original mesh, as it was at hand. You can also play around with different kinds of interpolation: the default 'linear' one will be faster, but less smooth.
Result after downsampling(left) and upsampling(right):
Of course you should still apply the small-line-removal algorithm after this resampling business, and keep in mind that this heavily distorts your input data (since if it wasn't distorted, then it wouldn't be smooth). Also, note that due to the crude method used in the downsampling step, we introduce some missing values near the top/right edges of the region under consideraton. If this is a problem, you should consider doing the downsampling based on griddata as I've noted earlier.
This is a pretty bad solution, but it's the only one that I've come up with. Use the get_contour_verts function in this solution you linked to, possibly with the matplotlib._cntr module so that nothing gets plotted initially. That gives you a list of contour lines, sections, vertices, etc. Then you have to go through that list and pop the contours you don't want. You could do this by calculating a minimum diameter, for example; if the max distance between points is less than some cutoff, throw it out.
That leaves you with a list of LineCollection objects. Now if you make a Figure and Axes instance, you can use Axes.add_collection to add all of the LineCollections in the list.
I checked this out really quick, but it seemed to work. I'll come back with a minimum working example if I get a chance. Hope it helps!
Edit: Here's an MWE of the basic idea. I wasn't familiar with plt._cntr.Cntr, so I ended up using plt.contour to get the initial contour object. As a result, you end up making two figures; you just have to close the first one. You can replace checkDiameter with whatever function works. I think you could turn the line segments into a Polygon and calculate areas, but you'd have to figure that out on your own. Let me know if you run into problems with this code, but it at least works for me.
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
def checkDiameter(seg, tol=.3):
# Function for screening line segments. NB: Not actually a proper diameter.
diam = (seg[:,0].max() - seg[:,0].min(),
seg[:,1].max() - seg[:,1].min())
return not (diam[0] < tol or diam[1] < tol)
# Create testing data
x = np.linspace(-1,1, 21)
xx, yy = np.meshgrid(x,x)
z = np.exp(-(xx**2 + .5*yy**2))
# Original plot with plt.contour
fig0, ax0 = plt.subplots()
# Make sure this contour object actually has a tiny contour to remove
cntrObj = ax0.contour(xx,yy,z, levels=[.2,.4,.6,.8,.9,.95,.99,.999])
# Primary loop: Copy contours into a new LineCollection
lineNew = list()
for lineOriginal in cntrObj.collections:
# Get properties of the original LineCollection
segments = lineOriginal.get_segments()
propDict = lineOriginal.properties()
propDict = {key: value for (key,value) in propDict.items()
if key in ['linewidth','color','linestyle']} # Whatever parameters you want to carry over
# Filter out the lines with small diameters
segments = [seg for seg in segments if checkDiameter(seg)]
# Create new LineCollection out of the OK segments
if len(segments) > 0:
lineNew.append(mpl.collections.LineCollection(segments, **propDict))
# Make new plot with only these line collections; display results
fig1, ax1 = plt.subplots()
ax1.set_xlim(ax0.get_xlim())
ax1.set_ylim(ax0.get_ylim())
for line in lineNew:
ax1.add_collection(line)
plt.show()
FYI: The bit with propDict is just to automate bringing over some of the line properties from the original plot. You can't use the whole dictionary at once, though. First, it contains the old plot's line segments, but you can just swap those for the new ones. But second, it appears to contain a number of parameters that are in conflict with each other: multiple linewidths, facecolors, etc. The {key for key in propDict if I want key} workaround is my way to bypass that, but I'm sure someone else can do it more cleanly.
I want to generate a surface which should look like a hemisphere.. What I have done so far is to read an already existing BEM mesh and try to show the scalar values on it. But now I have to show the scalar values on a hemisphere instead of the Bem mesh. And I don't know how to generate using a triangular mesh that looks like an hemisphere.
This hemisphere needs to contain a set of N number of points(x,y,z)[using the mlab.triangular_mesh] and at each vertex I need to represent N data(float) either as a value or using variations in colormap(eg: blue(lowest value of the data) to red(highest value of the data)). data=its an array of size 2562, a set of float values, could be randomly generated as its part of another codes. Points were part of another set of code too.its of shape(2562,3). but the shape is not a hemisphere
This was the program I used for viewing using the BEM surface
fname = data_path + '/subjects/sample/bem/sample-5120-5120-5120-bem-sol.fif'
surfaces = mne.read_bem_surfaces(fname, add_geom=True)
print "Number of surfaces : %d" % len(surfaces)
head_col = (0.95, 0.83, 0.83) # light pink
colors = [head_col]
try:
from enthought.mayavi import mlab
except:
from mayavi import mlab
mlab.figure(size=(600, 600), bgcolor=(0, 0, 0))
for c, surf in zip(colors, surfaces):
points = surf['rr']
faces = surf['tris']
s=data
mlab.triangular_mesh(points[:, 0], points[:, 1], points[:, 2],faces,color=c, opacity=1,scalars=s[:,0])
#mesh= mlab.triangular_mesh(x,y,z,triangles,representation='wireframe',opacity=0) #point_data=mesh.mlab_source.dataset.point_data
#point_data.scalars=t
#point_data.scalars.name='Point data'
#mesh2= mlab.pipeline.set_active_attribute(mesh,point_scalars='Point data')
As others have pointed out your question is not very clear, and does not include an easily reproducible example -- your example would take considerable work for us to reproduce and you have not described the steps you have taken very clearly.
What you are trying to do is easy. Scalars can be defined for each vertex (i.e., each VTK point):
surf = mlab.triangular_mesh(x,y,z,triangles)
surf.mlab_source.scalars = t
And you need to set a flag to get them to appear, which I think might be your problem:
surf.actor.mapper.scalar_visibility=True
Here is some code to generate a half-sphere. It produces a VTK polydata. I'm not 100% sure if the mayavi source is the same source type as triangular_mesh but I think it is.
res = 250. #desired resolution (number of samples on sphere)
phi,theta = np.mgrid[0:np.pi:np.pi/res, 0:np.pi:np.pi/res]
x=np.cos(theta) * np.sin(phi)
y=np.sin(theta) * np.sin(phi)
z=np.cos(phi)
mlab.mesh(x,y,z,color=(1,1,1))
I am working in image processing right now in python using numpy and scipy all the time. I have one piece of code that can enlarge an image, but not sure how this works.
So please some expert in scipy/numpy in python can explain to me line by line. I am always eager to learn.
import numpy as N
import os.path
import scipy.signal
import scipy.interpolate
import matplotlib.pyplot as plt
import matplotlib.cm as cm
def enlarge(img, rowscale, colscale, method='linear'):
x, y = N.meshgrid(N.arange(img.shape[1]), N.arange(img.shape[0]))
pts = N.column_stack((x.ravel(), y.ravel()))
xx, yy = N.mgrid[0.:float(img.shape[1]):1/float(colscale),
0.:float(img.shape[0]):1/float(rowscale)]
large = scipy.interpolate.griddata(pts, img.flatten(), (xx, yy), method).T
large[-1,:] = large[-2,:]
large[:,-1] = large[:,-2]
return large
Thanks a lot.
First, a grid of empty points is created with point per pixel.
x, y = N.meshgrid(N.arange(img.shape[1]), N.arange(img.shape[0]))
The actual image pixels are placed into the variable pts which will be needed later.
pts = N.column_stack((x.ravel(), y.ravel()))
After that, it creates a mesh grid with one point per pixel for the enlarged image; if the original image was 200x400, the colscale set to 4 and rowscale set to 2, the mesh grid would have (200*4)x(400*2) or 800x800 points.
xx, yy = N.mgrid[0.:float(img.shape[1]):1/float(colscale),
0.:float(img.shape[0]):1/float(rowscale)]
Using scipy, the points in pts variable are interpolated into the larger grid. Interpolation is the manner in which missing points are filled or estimated usually when going from a smaller set of points to a larger set of points.
large = scipy.interpolate.griddata(pts, img.flatten(), (xx, yy), method).T
I am not 100% certain what the last two lines do without going back and looking at what the griddata method returns. It appears to be throwing out some additional data that isn't needed for the image or performing a translation.
large[-1,:] = large[-2,:]
large[:,-1] = large[:,-2]
return large