If you don't set explicit bounds on your axis, its range will be a DataRange1d, with bounds automatically computed from whatever you plot. In this case, setting the range's flipped attribute will flip it without requiring you to set explicit bounds:
from bokeh.plotting import figure, show fig = figure() # Do some plotting calls with fig... fig.y_range.flipped = True show(fig)
The following will flip the y-axis for a scatter plot.
p = figure() xmin = data[xval].min() xmax = data[xval].max() ymin = data[yval].min() ymax = data[yval].max() # Note that ymin and ymax are in reverse order in y_range. p.scatter(xval, yval, x_range = (xmin, xmax), y_range = (ymax, ymin)) show(p)
I am trying to reverse the y axis and set anycodings_plot range for both x and y in a Bokeh scatter anycodings_plot plot. I am using:,If you want to set explicit bounds, see anycodings_bokeh this answer on another question. As Don anycodings_bokeh Smythe's answer mentions, you can set anycodings_bokeh the bounds in reverse order to invert anycodings_bokeh any axis type.,Any idea how axes can be reversed in a Bokeh anycodings_plot scatterplot with a pandas dataframe input,The following will flip the y-axis for a anycodings_bokeh scatter plot.
I am trying to reverse the y axis and set anycodings_plot range for both x and y in a Bokeh scatter anycodings_plot plot. I am using:
BokehPlot.bokeh_scatter(data = df, x_range = (min_utc, max_utc), y_range = (min_val, max_val))
I get an error:
TypeError: bokeh_scatter() got an unexpected keyword argument 'x_range'If you don't set explicit bounds on your anycodings_bokeh axis, its range will be a DataRange1d, anycodings_bokeh with bounds automatically computed from anycodings_bokeh whatever you plot. In this case, setting anycodings_bokeh the range's flipped attribute will flip anycodings_bokeh it without requiring you to set explicit anycodings_bokeh bounds:
from bokeh.plotting import figure, show fig = figure() # Do some plotting calls with fig... fig.y_range.flipped = True show(fig)
The following will flip the y-axis for a anycodings_bokeh scatter plot.
p = figure() xmin = data[xval].min() xmax = data[xval].max() ymin = data[yval].min() ymax = data[yval].max() # Note that ymin and ymax are in reverse order in y_range. p.scatter(xval, yval, x_range = (xmin, xmax), y_range = (ymax, ymin)) show(p)
I've tried reversing the range to [10, 0] but this results in displacing the image, not fliping the y axis. The only way I've found to plotting it right is by flipping the numpy array lenna[::-1] before sending it to plot (but I guess it should be a better way).,I've used matplotlib and I've never wondered about this before because in matplotlib the default origin is the top left pixel (my case use). Searching a bit I found that as you suggest they also use a keyword origin= in the call to imshow to change between the two options:,Results with the image plotted upside down, as seen in the attached images (image is also converted to grayscale but this is not relevant):,Well I guess that's points to making the auto-fiipping configurable as well. But unless I am misunderstanding, not flipping is not consistent with the mental model above? If every pixel of an image has its own position, then flipping the range implies that the image would flip.
import bokeh.plotting as plt
from skimage.io
import imread
lenna = imread('/tmp/lenna.png', flatten = True)
p = plt.figure(x_range = [0, 10], y_range = [0, 10])
p.image(image = [lenna], x = 0, y = 0, dw = 10, dh = 10, palette = 'Greys9')
plt.show(p)import numpy as np
import PIL.Image
lena_img = PIL.Image.open('lena.png')
width_pixel, height_pixel = 512, 512
size_new = (width_pixel, height_pixel)
lena_img = lena_img.resize(size_new).convert('RGBA')
a = np.array(lena_img)
# stack each tuple(R, G, B, A) to a UINT32, then reshape to(width, height)
img = a.view(dtype = np.uint32).reshape(a.shape[: -1])
# reverse y axis manualy
img = img[::-1]
...
# reverse y_range
plot = figure(x_range = (1, width_pixel), y_range = (height_pixel, 1), title = 'LENA', tools = 'reset,save', plot_height = height_pixel, plot_width = width_pixel)
# enjoy
plot.image_rgba(image = [img], x = 0, y = height_pixel, dw = width_pixel, dh = height_pixel)invert_axis = None #(
default)
invert_axis = 0 # or 'y'
invert_axis = 1 # or 'x'
invert_axis = (0, 1) # or('y', 'x') or 'both'for the option invert_yaxis=True it seems that there is a bug, the option works with matplotlib backend but don’t with bokeh backend even if the option exist in bokeh.,In order to invert the y axis you can also use : plot.opts(invert_yaxis=True) For your 2nd request I’m not sure of what you want to do. Maybe a hv.Area between a line at y=3000 and the edge of your plot will do the job. If you want more help can you provide a small reproducible example please ?,Thank you very much @AurelienSciarra for your quick reply. In my case plots.opts(invert_yaxis=True) didnt work:,My aim would be to have a plot similar to the one at the bottom (generated with matplotlib):
I generated the figure above with the command:
plot = myXarrayData.hvplot.contourf(z = 'V', x = 'refdist', y = 'depth',
levels = datatoplot['cf_levels'],
clabel = datatoplot['cbtitle'], label = datatoplot['pltitle'])After digging the github issue log, I found the solution of using the recent implemented transform argument:
reversedepth = -hv.dim('depth')
plot = myXarrayData.hvplot.contourf(z = 'V', x = 'refdist', y = 'depth',
transforms = dict(depth = reversedepth)
)The two commands below generated the same plots…
import xarray as xr
import hvplot.xarray
import holoviews as hv
data = np.random.rand(3, 4)
refdist = np.array(range(100, 400, 100))
depth = np.array(range(5, 25, 5))
ds = xr.Dataset({
"V": (["refdist", "depth"], data),
},
coords = {
"refdist": (refdist),
"depth": (depth),
},
)
plot = ds['V'].hvplot.contourf(z = 'V', x = 'refdist', y = 'depth')
plotplot.opts(invert_yaxis = True)
From this code:
Y = 20 + np.random.randint(10, size = (len(refdist)))
Y2 = 40 * np.ones(Y.shape)
AreaBelowPlot = hv.Area((refdist, -Y, -Y2), vdims = ['y', 'y2'])
(pl * AreaBelowPlot).opts(
hv.opts.Area(fill_color = 'saddlebrown')
)