Nested Lists- Python

Question

I need to write the body of a Python method that does the following:

1)takes a list, where list[0] is a string and list[1] is either a list which looks the same or None

2)print every string of the list

I have to use the while loop and not use list comprehension or flatten.

def pick_cherries_iter(field):
    """"e.g.
    >>> cherry_field = ['cherry1', ['cherry2', ['cherry3', ['cherry4', ['Yay!!!', None]]]]]
    >>> pick_cherries_iter(cherry_field)
    cherry1
    cherry2
    cherry3
    cherry4
    Yay!!!"""

    _______________________
    _______________________
    _______________________
    _______________________
    while _________________:
        _______________________
        _______________________
        _______________________

I know that for the example above I can print cheery1 if I print cherry_field[0] or cherry1 for cherry_field[1][0] or cherry2 for cherry_filed[1][1][0] etc, however I am not sure how to go through these elements using the while loop.

Possible duplicate of Flatten a list of strings and lists of strings and lists in Python — KenHBS
– KenHBS, Commented Nov 19, 2018 at 13:59
is there a guarantee in the assignment that each list is at most 2 elements long? — Ilia Gilmijarow
– Ilia Gilmijarow, Commented Nov 19, 2018 at 14:10

Dinesh Suthar · Accepted Answer · 2018-11-21 06:24:38Z

0

I think this should work for you. Please check it.

Using While Loop:

def pick_cherry(field):
    """"e.g.
    >>> cherry_field = ['cherry1', ['cherry2', ['cherry3', ['cherry4', ['Yay!!!',None]]]]]
    >>> pick_cherry(cherry_field)
    cherry1
    cherry2
    cherry3
    cherry4
    Yay!!!"""

    while field[1] != None:
        temp = field[0]
        print temp
        field = field[1]
    print field[0]

Using Flatten(and Recursion):

flatten_field = []

def pick_cherry(field):
    if field[1] != None:
        flatten_field.append(field[0])
        pick_cherry(field[1])
    else:
        flatten_field.append(field[0])

def flatten_func(field):
    """"e.g.
    >>> cherry_field = ['cherry1', ['cherry2', ['cherry3', ['cherry4', ['Yay!!!',None]]]]]
    >>> flatten_func(cherry_field)
    cherry1
    cherry2
    cherry3
    cherry4
    Yay!!!"""

    pick_cherry(field)

    for item in flatten_field:
        print item

edited Nov 21, 2018 at 6:24

answered Nov 19, 2018 at 14:31

Dinesh Suthar

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1 Comment

M.Papapetros Over a year ago

If I had to use flatten (and recursion), how could this be done?

Lulz · Accepted Answer · 2018-11-19 14:32:13Z

0

I would do this recursively because you have no way of knowing if an element is a list or not.

#!/usr/bin/python -E

cherry_field = ['cherry1', ['cherry2', ['cherry3', ['cherry4', ['Yay!!!', None]]]]]
def print_list(field):
    i = 0
    list_length = len(field)
    while i < list_length:
     if field[i] is not None and type(field[i]) is not list:
        print(field[i])
     else:
        if field[i] is not None:
           print_list(field[i])
     i += 1
     if i < list_length and type(field[i]) is list:
        print_list(field[i])
        i += 1


def pick_cherries(field):
    if type(field) is list:
       print_list(field)

pick_cherries(cherry_field)

edited Nov 19, 2018 at 14:32

answered Nov 19, 2018 at 14:18

Lulz

11 bronze badge

5 Comments

M.Papapetros Over a year ago

Does your answer include list comprehension?

Lulz Over a year ago

I just tested on a variety of nested lists, made a couple minor changes, and it seems to work on everything so far. That's the advantage of recursion.

Lulz Over a year ago

For example, it can also do this list: cherry_field = ['cherry1', ['cherry2', ['cherry3', ['cherry4', ['Yay!!!', None]]]],'cherry5']

M.Papapetros Over a year ago

Your answer using recursion is just fine, my question is whether you are using list coprehension or not.

Lulz Over a year ago

Umm, no. Referring back to the OP's question, I do not see that requirement.

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