Comments (2)
Yes, assuming you know the breaks.
df = pd.DataFrame({'x': [1, 10, 100, 1000, 10000, 100000]})
(ggplot(df)
+ geom_point(aes('x', 'x'))
+ scale_x_log10(labels=['$10^1$', '$10^3$', '$10^5$'])
)
Otherwise, you can write a function (takes breaks and returns labels enclosed in $$
) to do the labelling.
def label_func(breaks):
return ['$10^{}$'.format(int(np.log10(i))) for i in breaks]
(ggplot(df)
+ geom_point(aes('x', 'x'))
+ scale_x_log10(labels=label_func)
)
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When label have Negative index
, @has2k1 solution should be changed into:
def label_to_scentific_mode_func(breaks):
return ['$10^{{{}}}$'.format(int(np.log10(i))) for i in breaks]
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