Python try except block inside lambda

Question

Is it possible to use a try except block inside of a lambda function? I need the lambda function to convert a certain variable into an integer, but not all of the values will be able to be converted into integers.

Answer 1

Nope. A Python lambda can only be a single expression. Use a named function.

It is convenient to write a generic function for converting types:

def tryconvert(value, default, *types):
    for t in types:
        try:
            return t(value)
        except (ValueError, TypeError):
            continue
    return default

Then you can write your lambda:

lambda v: tryconvert(v, 0, int)

You could also write tryconvert() so it returns a function that takes the value to be converted; then you don't need the lambda:

def tryconvert(default, *types):
    def convert(value):
        for t in types:
            try:
                return t(value)
            except (ValueError, TypeError):
                continue
        return default
    # set name of conversion function to something more useful
    namext = ("_%s_" % default) + "_".join(t.__name__ for t in types)
    if hasattr(convert, "__qualname__"): convert.__qualname__ += namext
    convert.__name__ += namext
    return convert

Now tryconvert(0, int) returns a function convert_0_int that takes a value and converts it to an integer, and returns 0 if this can't be done. You can use this function right away (not saving a copy):

mynumber = tryconert(0, int)(value)

Or save it to call it later:

intconvert = tryconvert(0, int)
# later...
mynumber = intconvert(value)