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My first post on Python language as I am learning it.

I have a shape file that has around 10000 polygons.

I am trying to generate code like below that creates Polygon1, Polygon2 all the way to Polygon10000 using syntax like this:

polygon1 = shape(shapes[0])
polygon2 = shape(shapes[1])
polygon3 = shape(shapes[2])
polygon4 = shape(shapes[3])
.
.
polygon10000 = shape(shapes[9999])

So all I am trying to do is to write code that is much more smaller than having to write 10000 lines of code like above.

I came up with some syntax but none of this is really working:

Method 1- Just prints the required syntax in the log but does not execute it so I have to copy the output after the code runs (from console) and then paste it in the code and then run that code 

        for x in range(1,10):
            print('polygon' '%d =' ' shape(shapes[' '%d' '])' % (x, x-1 ))

Method 2 - Does the job but still need to write 10000 lines of code to create all 10000 polygons

        def automate(n):
            return shape(shapes[n])


        polygon1 = automate(0)
        polygon2 = automate(1)
        .
        .
        polygon10000 = automate(9999)

Any suggestions on doing this in a quicker and shorter way would be highly appreciated..

Thank you, Tina

martineau
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Tina
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  • Why would you ever want to do this? Dont create dynamic variables, use *containers*, like `list` or `dict` objects. You probably just want a `list`. Indeed, you already *have* a list. Just use something like `polygon = list(map(shape, shapes))` and now `polygon` is a list of shapes. You get the first one using `polygon[0]`, etc – juanpa.arrivillaga Nov 18 '19 at 22:31

1 Answers1

1

How about something like:

polygons = []
for i in range(10000):
    polygons.append(shape(shapes[i])

Then you can reference the polygons as polygons[j] where j is an index from 0 to 9999.

John Anderson
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