Create Lon and Lat points from UTM zone and lower-left corner

[dpath2o]

I'm hoping someone here might be monitoring this and help me to create a list of geographic points (lon/lat) from the UTM zone and a lower-left corner. That is I have gridded matrix (shape is (2527, 2695)) of bathymetry depths that I want to plot in geographic context. The metadata that I have is the UTM zone, cell size, units (meters), a lower-left corner in meters (340700,2468000) spheroid, datum, shifting value and some other information that I don't think is of consequence to my goal plotting this information. If I can get the lon/lat associated with each grid point then I can plot it using various methods that I know, I just want to accurately place the information on the globe, which does not appear trivial from my vantage point.

Thanks in advance for any insights/help.

[snowman2]

What is the cell size?

Step 1: Build the transform

Based on what you have, you can construct the affine/transform: https://github.com/sgillies/affine

This is helpful for determining the order: https://gdal.org/user/raster_data_model.html#affine-geotransform

from affine import Affine

resolution_x = 100  # right
resolution_y = -100  # down
left = 340700
bottom = 2468000
top = bottom - resolution_y * height
transform  = Affine.translation(left, top) * Affine.scale(resolution_x, resolution_y)

Step 2: Get the CRS from UTM Zone

https://pyproj4.github.io/pyproj/stable/build_crs.html#projected-crs

from pyproj.crs import ProjectedCRS
from pyproj.crs.coordinate_operation import UTMConversion

crs = ProjectedCRS(conversion=UTMConversion(14))

Step 3: Assign the transform/affine & CRS to raster

rioxarray can make this simple:

import rioxarray
import xarray

rds = xarray.DataArray(data,  dims=("y", "x"))
rds.rio.write_transform(transform, inplace=True)
rds.rio.write_crs(crs, inplace=True)

Step 4: Re-project to geographic

rds = rds.rio.reproject("EPSG:4326")

[dpath2o]

Hey Snowman2, thanks for asking. The cell size is 100 metres

[snowman2]

Just updated the original comment.

[dpath2o]

Thank you for that very succinct answer Snowman2. Tonight, I look forward to giving this a crack!