Python Try Catch Block inside lambda

问题

Is it possible to use try catch 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.

回答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 that takes a value and converts it to an integer, and returns 0 if this can't be done.

回答2:

In this specific instance, you can avoid using a try block like this:

lambda s: int(s) if s.isdigit() else 0

The isdigit() string method returns true if all the characters of s are digits. (If you need to accept negative numbers, you will have to do some extra checking.)

回答3:

Depending on your need, another approach could be to keep the try:catch outside lambda fn

toint  = lambda x : int(x)
strval = ['3', '']

for s in strval:
    try:
        print 2 + toint(s)
    except ValueError:
        print 2

Output:

5
2

回答4:

While there is no general way for handling exceptions in a lambda expression, you can achieve it in a restricted way for at least one kind of exception; throwing a StopIteration from a part of an expression and catching it in another part is achievable; see:

from random import randrange

list((lambda:(yield from (randrange(0,2) or next(iter(())) for _ in (None,))))())

where next(iter(())) raises a StopIteration while yield from will catch it; the expression above randomly returns [] or [1] according to the inner random value (a 0 will raise the exception and a 1 will be normally evaluated).

You can read more about it ta http://baruchel.github.io/python/2018/06/20/python-exceptions-in-lambda/ .

来源：https://stackoverflow.com/questions/12451531/python-try-catch-block-inside-lambda

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