Dynamically generate function arguments python

Question

I am working in Python 2.7 and need to generate a list of functions each of which has a different default argument but where the functions are explained. This is not the best explanation, but I think the example below clarifies my question. Is there a way of doing this?

function_list = []
for i in range(10):
    function_list.append(lambda x: myfunc(i,x))

def myfunc(a,b):
    print a + b

This doesn't work, as, in this example, all functions in the function list return 9 + x

In the example above "x" is a separate variable to be passed to the function in a different way.

I realise that this is a pretty ugly construction but have inherited a large amount of code around this that would need to be changed to avoid this.

AndreyF · Accepted Answer · 2018-01-24 12:22:45Z

3

You can use i as a default value for a lambda with two arguments. This way it will fit better to the question description:

need to generate a list of functions each of which has a different default argument

for the following code:

function_list = []
for i in range(10):
    function_list.append(lambda x, y=i: myfunc(x,y))

for f in function_list:
    f(5)

The output will be:

also you can use f with two arguments:

for j,f in enumerate(function_list):
    f(j+2,5)

and you'll get:

edited Jan 24, 2018 at 12:22

answered Jan 24, 2018 at 11:57

AndreyF

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