Calling iterator from another function in python

Question

Let's say i have a list of lists. i want to iterate through every index in every list, and use it in another func. So:

def func_a(self):
    for i in range(len(big_list)):
        current_list = big_list[i]
            for j in range(len(current_list):
                char_iter = iter(current_list)

def func_b(self):
    <<here i want to use the next char from iter>>
    <<how do i do it>>

Things I have tried:

next(func) -> Error
next(char_iter) - > Last Char

note: i write this funcs in a class.

So what you want func_a to do is just iterate through everything and the func_b will use each value coming out of func_a? — Joel
– Joel, Commented Jan 16, 2015 at 13:40
Note that for i in range(len(big_list)): current_list = big_list[i] is a poor way to do iteration. Much better to do for current_list in big_list — Joel
– Joel, Commented Jan 16, 2015 at 13:48

tobias_k · Accepted Answer · 2015-01-16 13:46:47Z

3

Not entirely sure what you are asking. As I understand the question, you want func_a to iterate though all the elements in all the nested lists, and make it so that other functions (particularly func_b) can reuse those.

In this case, you should loop through the lists and yield the chars, making func_a a generator function.

def func_a(self):
    for current_list in big_list:
        for char in current_list:
            yield char

You can then use it like this in func_b

def func_b(self):
    for char in func_a():
        print char

or using next:

def func_b(self):
    gen = func_a()
    print next(gen)
    print next(gen)

answered Jan 16, 2015 at 13:46

tobias_k

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Comments

Joel · Accepted Answer · 2015-01-16 13:46:52Z

1

This should give you the basic idea. Give more clarification if it's not enough.

big_list = [[1,2,3],[3,4]]

def func_a():
    for current_list in big_list:
        for value in current_list:
            yield value

y = func_a()


for x in y:
    print x
    if x>2:
        break

print 'broke'
for x in y:
    print x

answered Jan 16, 2015 at 13:46

Joel

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Comments

deets · Accepted Answer · 2015-01-16 13:47:43Z

0

You should not use index-based access, Python works best if you always just iterate, as it allows for lazy evaluation, as well as processing streams instead of random access containers.

I suggest using a generator:

def g(big_list):
    # no need for complicated indexing
    for current_list in big_list:
        # again, no need for index-based access
        for char in current_list:
            yield char


for char in g(the_big_list):
    func_b(char)

answered Jan 16, 2015 at 13:47

deets

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Comments

fnl · Accepted Answer · 2015-01-16 14:03:44Z

0

def func_a(self):
    return [j for i in big_list for j in i]

def func_b(self):
    flattened_big_list = func_a()

edited Jan 16, 2015 at 14:03

answered Jan 16, 2015 at 13:46

fnl

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5 Comments

Joel Over a year ago

here func_a actually creates the entire list. I don't think that's what OP wants.

tobias_k Over a year ago

You can change those outer [...] to (...) to make it a generator.

fnl Over a year ago

@Joel Maybe... but its just such a nice one-liner you can do with Python list comprehensions I just had to show it off... :)

fnl Over a year ago

@tobias_k ah, right! But to be honest, if you really want to iterate, a generator with for ... for ... yield as suggested by others is easier to read.

fnl Over a year ago

@Joel made my case of list-comprehension-based list flattening semantically more sensible

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