[FIXED] How to flatten a list containing lists with different numbers of dictionaries?

Issue

I have a python code like below:

import numpy as np
import multiprocessing as mp

def func(first, sec):
    results = []
    for x in range(first):
        result_dict = {"a": x, "b": x**sec, "c": x/sec}
        results.append(result_dict.copy())
    return results

with mp.Pool(mp.cpu_count()) as pool:
    res = pool.starmap(func, [(a, 8) for a in range(1, 4)])

Then I flatten the res with this piece of code:

res = np.asarray(res)
res = res.reshape(-1)
res = np.array(res).tolist()

After that when I print the res the output is like below:

[[{'a': 0, 'b': 0, 'c': 0.0}],
 [{'a': 0, 'b': 0, 'c': 0.0}, {'a': 1, 'b': 1, 'c': 0.125}],
 [{'a': 0, 'b': 0, 'c': 0.0},
  {'a': 1, 'b': 1, 'c': 0.125},
  {'a': 2, 'b': 256, 'c': 0.25}]]

But I want the output to be like this:

[{'a': 0, 'b': 0, 'c': 0.0},
 {'a': 0, 'b': 0, 'c': 0.0},
 {'a': 1, 'b': 1, 'c': 0.125},
 {'a': 0, 'b': 0, 'c': 0.0},
 {'a': 1, 'b': 1, 'c': 0.125},
 {'a': 2, 'b': 256, 'c': 0.25}]

Do you have any idea how to change the code to get the desired result for res ?

Solution

The chain.from_iterable of itertools library lets you chain sub-lists to one list.

import itertools

res = [[{'a': 0, 'b': 0, 'c': 0.0}],
 [{'a': 0, 'b': 0, 'c': 0.0}, {'a': 1, 'b': 1, 'c': 0.125}],
 [{'a': 0, 'b': 0, 'c': 0.0},
  {'a': 1, 'b': 1, 'c': 0.125},
  {'a': 2, 'b': 256, 'c': 0.25}]]


print(list(itertools.chain.from_iterable(res)))

Answered By - Aviv Yaniv