根据条件将列表分解为列表列表

Question

I have the following lists x1, x2, x3, which i want to chunk into the respective outputs mentioned below:

x1 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'f']

expected_out1 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']]

x2 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'f', 'req']

expected_out2 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f'], ['req']]

x3 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'req']

expected_out3 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req'], ['req']]

I wrote the following code to address these scenarios:

import numpy as np
def split_basedon_condition(b):
    num_arr = np.array(b)
    arrays = np.split(num_arr, np.where(num_arr[:-1] == "req")[0])
    return [i for i in [i.tolist() for i in arrays] if i != []]

But I'm getting the following results:

split_basedon_condition(x1)
actual_out1 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']] # expected

split_basedon_condition(x2)
actual_out2 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f', 'req']] # not expected

split_basedon_condition(x3)
actual_out3 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'req']] # not expected

Answer 1

Reason is in this line:

arrays = np.split(num_arr, np.where(num_arr[:-1] == "req")[0])

By doing num_arr[:-1] you are considering num_arr with jettisoned last element, thus this behavior when "req" is last element. Replace above line with:

arrays = np.split(num_arr, np.where(num_arr == "req")[0])

and it will work as excepted for all test cases you provided.

As side note if you are allowed to use external python libraries other than numpy , you might harness more_itertools.split_before for that task.

Answer 2

Here's much faster solution pure python.

def split(arr, pred):
    j = len(arr)
    for i,_ in filter(lambda x: x[1] == pred, zip(range(j-1, -1, -1), reversed(arr))):
        yield arr[i:j]
        j = i

list(split(['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'req'], 'req'))
# [['req'], ['req'], ['req', 'd', 'e'], ['req', 'a', 'b', 'c']]

Answer 3

If the first one always is "req" .This is one-line:

def func(l):
    return list(map(lambda x: x.insert(0, "req") or x, map(list, "".join(l[1:]).split("req"))))

Result:

[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']]
[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f'], ['req']]
[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req'], ['req']]