Dear Matplotlib developers,

running the following program

import pylab

x=pylab.arange(0,1e-8,1e-9)+1.0

pylab.plot(x)

pylab.show()

with matplotlib 0.70.1 results in a graph that shows:

(i) correctly the linear increase of x

(ii) but the labels on the y-axis of the plot all show "1e", apart from the lowest one, which shows (correctly) just "1".

All works fine when I subtract the mean of x but there seems to be a problem with labelling axes for plotted data which is not close to zero but shows only small variations.

By the way: thank you for providing matplotlib.

Cheers,

Hans

Hi Hans,

Let me ask, what do you think would be the correct labels? As I remember,

Matlab will go out to 4 decimal points, so your ticks would be labeled

1,1.0000,1.0000.

I have been meaning to do some work with the tick formatters. It has been

suggested here that all significant digits are preserved, so for example,

your ticks would be 1,1,1 (If you want to go out to 9 sig digits, I think you

would have to write your own custom formatter, which is discussed on this

list and in the user's guide). If you ticks are 1,1.01,1.02, the labels would

be 1.00,1.01,1.02. I think this results in the best looking result, but I'd

like to know others opinions before I start. Also, for ticks like

1e5,2e5,3e5, I am intending to make the ticks 1,2,3, and have a x10^5 just

outside the axis or in the last label. Comments, suggestions?

Darren

## ยทยทยท

On Thursday 24 February 2005 05:53 am, Hans Fangohr wrote:

Dear Matplotlib developers,

running the following program

import pylab

x=pylab.arange(0,1e-8,1e-9)+1.0

pylab.plot(x)

pylab.show()

with matplotlib 0.70.1 results in a graph that shows:

(i) correctly the linear increase of x

(ii) but the labels on the y-axis of the plot all show "1e", apart from

the lowest one, which shows (correctly) just "1".

All works fine when I subtract the mean of x but there seems to be a

problem with labelling axes for plotted data which is not close to zero

but shows only small variations.

By the way: thank you for providing matplotlib.

Cheers,

Hans

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