当内部列表的数量未知时，如何将列表列表送入* args？

Question

I've got a function which takes positional arguments in this format. Editing it isn't feasible for my use case (long story).

import pandas as pd
import numpy as np
import random

def dummy_func(*args):

    outlist = []

    for arg in args:
        arg.append(42)

        outlist.append(arg)

    return outlist

Normally, function accepts any number of lists of integers in the following format and processes them like this:

s1 = [random.randrange(1, 101, 1) for _ in range(10)]
s2 = [random.randrange(1, 101, 1) for _ in range(10)]
s3 = [random.randrange(1, 101, 1) for _ in range(10)]

dummy_func(s1,s2,s3)

I'm trying to build a function that assembles a list of lists of integers to feed to the dummy_func listed above. The lists are all the same length, but we're feeding them to the function in batches and we don't know how many lists will be in each batch.

def analyse_some_series():

    # Generate a random number of lists
    count = np.random.choice(10)

    for i in range(count):

        mylist = [random.randrange(1, 101, 1) for _ in range(10)]

        outer_list.append(mylist)

    # This does not work for some reason
    finalfunc(outer_list)

For some reason, when you feed the list of lists to the function, it does not unwrap and process the list of lists. Instead it just sees one list object in the wrong format.

Now normally I would just do something like:

x,y,z = series_list

But that won't work here since we don't know how many lists are going to be generated.

Any help you could offer would be greatly appreciated.

Answer 1

Call it as finalfunc(*outer_list) .

The asterisk does the unpacking for you.

Note that the asterisk in the function signature means that multiple items are packed into a single list, but in calling the function with a list it means that items of a list are unpacked into individual arguments. Hence, the asterisk means opposite things in the two places, but it does the only sensible packing/unpacking related job in both situations so I think it's reasonable to use the same symbol.