/compute transparent pixel, np.nan instead of zero

[chapmanjacobd]

Looking over the expression parser I'm not sure if requesting nan squares is possible.

v1 * (v1 >= {filter_slider1[0]}) * (v1 <= {filter_slider1[1]})

This works great in general but I'm wondering if it would be possible to show a transparent pixel if the expression evaluates to 0.

in pseudo-code:

f"v1 if (v1 >= {filter_slider1[0]}) * (v1 <= {filter_slider1[1]}) else np.nan"

[j08lue]

Nice case, @chapmanjacobd.

Maybe we could make this possible by adding a parameter that lets you specify a NaN / transparent value and mask with that after the expression got evaluated. By default, this could be the raster NaN value.

[dionhaefner]

Now that #222 is merged, both invalid values (nan / inf) and masked values will be transparent. Most of the masked_* functions from numpy.ma are now allowed in expressions.

So your expression should reduce to a one-liner:

masked_outside(v1, lower, upper)

Could you try this and report back?

[chapmanjacobd]

works great!!

    expr1 = f"masked_outside(v1, {filter_slider1[0]}, {filter_slider1[1]})"
    expr2 = f"masked_outside(v2, {filter_slider2[0]}, {filter_slider2[1]})"
    # expr = expr1
    expr = f"({expr1}) + ({expr2})"